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UG MGT

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Substitute values into an algebraic expression and calculate the result.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Evaluate the following expression,

\\[\\simplify{p^{n}+{a}*r*t+{c}},\\]

when $p = \\var{pval}$, $r = \\var{rval}$, and $t = \\var{tval}$.

", "advice": "

In order to evaluate $\\simplify{p^{n}+{a}*r*t+{c}},$ with the given values, $p = \\var{pval}$, $r = \\var{rval}$, and $t = \\var{tval}$, we replace each instance of that letter with its corresponding value and then apply the rules of BIDMAS:

\\[\\var{pval}^\\var{n}+\\var{a}\\times \\var{rval} \\times \\var{tval} + \\var{c}\\]

Which gives the answer $\\var{ans}$.

Follow this link for more help on tackling these kind of questions.

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Rearrange a specific formula. No randomisation.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Rearrange the following equation, to make $y$ the subject:

\\[{cy -b = 3x}\\]

", "advice": "

In order to rearrange the equation so that it is in terms of $y$, we must first add $b$ to both sides, and then divide both sides of the equation by $c$:

\\begin{split} cy-b &= 3x \\\\ cy &= 3x + b \\\\ y &=\\frac{3x+b}{c} \\end{split}

Use this link to find some resources which will help you revise this topic.

$y=$ [[0]]

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Solve linear equations with unkowns on both sides. Including brackets and fractions.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "", "advice": "

Given $\\simplify{{m}w-{n} = {p}w+{q}}$, we can get all the $w$'s on the left hand side and all the numbers on the right hand side, and then divide both sides by the coefficient of $w$ to get $w$ by itself.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n



$\\simplify{{m}w+{n}}$	$=$	$\\simplify{{p}w+{q}}$

$\\simplify[!cancelTerms,unitFactor]{{m}w-{n}-{p}w}$	$=$	$\\simplify[!cancelTerms,unitFactor]{{p}w+{q}-{p}w}$

$\\simplify{{m-p}w-{n}}$	$=$	$\\var{q}$

$\\var{m-p}w-\\var{n}+\\var{n}$	$=$	$\\var{q}+\\var{n}$

$\\var{m-p}w$	$=$	$\\var{q+n}$

$\\displaystyle{\\frac{\\var{m-p}w}{\\var{m-p}}}$	$=$	$\\displaystyle{\\frac{\\var{q+n}}{\\var{m-p}}}$

$w$	$=$	$\\displaystyle{\\simplify{{q+n}/{m-p}}} = \\var{precround(ansA,1)} \\text{ to 1 dp}$

Use this link to find resources to help you revise how to solve linear equations

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Solve $\\simplify{({m}w-{n}) = {p}w+{q}}$

$w=$ [[0]]

Questions testing understanding of the precedence of operators using BIDMAS, applied to integers. These questions only test DMAS. That is, only Division/Multiplcation and Addition/Subtraction.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Evaluate the following expression:

", "advice": "

BIDMAS stands for:

Brackets

Indices

Division

Multiplication

Addition

Subtraction

And is a way for us to remember guidance about the order in which calculations are carried out to ensure that everyone doing teh same sum gets the same answer. In this case the first thing that is in the question is Multiplication.

First work through the expression from left to right, evaluating any multiplication as you come to them. You should be left with an expression involving only pluses and minuses. Evaluate this expression, again working from left to right. Thus:

\\[\\var{a}-\\var{b} \\times \\var{c}\\]

\\[=\\var{a}-\\var{b*c}\\]

\\[=\\var{a-b*c}\\]

Use this link to find some resources which will help you revise this topic.

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Calculate

$\\var{a}-\\var{b} \\times\\var{c}$

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Questions testing understanding of the precedence of operators using BIDMAS, applied to integers. These questions only test DMAS. That is, only Division/Multiplcation and Addition/Subtraction.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Evaluate the following expression:

", "advice": "

BIDMAS stands for:

Brackets

Indices

Division

Multiplication

Addition

Subtraction

First work through the expression from left to right, evaluating any division as you come to it. You should be left with an expression involving only pluses and minuses. Evaluate this expression, again working from left to right. Thus:

\\[\\var{h}-\\var{a2*b2} \\div \\var{b2}\\]

\\[=\\var{h}-\\var{a2}\\]

\\[=\\var{h-a2}\\]

Use this link to find some resources which will help you revise this topic.

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$\\var{h}-\\var{a2*b2} \\div \\var{b2}$

", "minValue": "{h-a2}", "maxValue": "{h-a2}", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "NA17 Worded proportion", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Proportional calculation in context.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

If {n1} adult cinema tickets cost £{n1price}, how much would it cost for {n2} adults to buy tickets to the cinema?

", "advice": "

If {n1} tickets cost £{n1price} then we can calculate that one ticket costs

\\[£\\var{n1price}\\div\\var{n1}=£\\var{price}.\\]

This means that {n2} tickets cost $\\var{n2}\\times£\\var{price}=£\\var{n2price}$

Use this link to find some resources which will help you revise this topic.

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£[[0]]

Given a student discount, calculate a discounted price.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

{pname} is buying a new {item}. The price of the model he picked is £{price}. On a website with discounts for students, he found a voucher for a discount of {percentage}%.

", "advice": "

There are multiple methods to approach this problem. The first method involves working out the discounted price as a percentage of the original, while the second method calculates the value of the discount and subtracts that from the listed price.

Method 1

There is a {percentage}% decrease in price. This means that the new price will be {100-percentage}% of the old price.

\\[\\begin{align} \\frac{\\var{100-percentage}}{100} \\times \\var{price} &= \\var{dpformat((100-percentage)/100*price,4)} \\\\&= \\var{dpformat((100-percentage)/100*price, 2)}\\text{.} \\end{align}\\]

Or, using the multiplier method,

\\[\\begin{align} \\var{(100-percentage)/100} \\times \\var{price} &= \\var{dpformat((100-percentage)/100*price,4)}\\\\&= \\var{dpformat((100-percentage)/100*price, 2)}\\text{.} \\end{align}\\]

When we are talking about money, it is usually assumed that we will round the answer to 2 decimal places.

Method 2

We find the discount first. This is

\\[\\frac{\\var{percentage}}{100} \\times \\var{price} = \\var{dpformat((percentage)/100*price,4)}\\text{.}\\]

Or using a decimal multiplier,

\\[\\var{(percentage)/100} \\times \\var{price} = \\var{dpformat((percentage)/100*price,4)}\\text{.}\\]

Then we subtract the discount from the original price to get the new price:

\\[ \\var{price} - \\var{dpformat(discount,2)} = \\var{dpformat(price - discount, 2)}\\text{.} \\]

Use this link to find some resources which will help you revise this topic.

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Names.

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Price of an item.

", "templateType": "anything", "can_override": false}, "item": {"name": "item", "group": "Ungrouped variables", "definition": "random(\"TV\", \"laptop\", \"smartphone\", \"PC\", \"gaming console\", \"fridge\")", "description": "

The bought item.

", "templateType": "anything", "can_override": false}, "percentage": {"name": "percentage", "group": "Ungrouped variables", "definition": "random(5..40 #5)", "description": "

Discount percentage.

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What will the discounted price of the {item} be?

Round your answer to the nearest penny.

£ [[0]]

Your answer does not make sense in real life, we cannot divide a penny any further. Shops always round their prices for items. That is why you should have rounded your answer to $\\var{precround((100-percentage)/100*price, 2)}$.

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Calculate the percentage increase (as a percentage) given a number and the size of the increase.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

What is the percentage increase in a class of {total} if {additional} more are added to it?

Give your answer to 2 decimal places.

", "advice": "

To calculate a percentage increase you need to find how much the increase is as a percentage of the original number. In this question the increase is {additional} and the original number is {total} so the percentage is

\\[ \\frac{\\var{additional}}{\\var{total}}\\times100\\%=\\var{dpformat(additional/total,4)}\\times 100\\%=\\var{dpformat(percentage,2)}\\%\\]

Use this link to find some resources which will help you revise this topic.

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[[0]]%

Compound percentage change.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

The value of a car is initially {StartingPrice}. If the value decreases by {dec}%, and then increases by {inc}%, what is the final value?

Give your answer correct to two decimal places.

", "advice": "

There is a {dec}% decrease in price. This means that price after the decrease will be {100-dec}% of the old price.

\\[\\frac{\\var{100-dec}}{100} \\times \\var{StartingPrice} = \\var{(100-dec)/100*StartingPrice}\\]

Then there is a {inc}% increase in price. This means the final price will be {100+inc}% of the price after the decrease.

\\[\\frac{\\var{100+inc}}{100} \\times \\var{(100-dec)/100*StartingPrice} = £\\var{dpformat((100+inc)/100*(100-dec)/100*StartingPrice,2)}\\]

Use this link to find some resources which will help you revise this topic.

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£[[0]]

Find a percentage of an amount.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

What is {x}% of {y}?

", "advice": "

Taking {x}% of {y} is calculated by multiplying,

\\[\\frac{\\var{x}}{100}\\times\\var{y}.\\]

For this question we can calculate this by noticing that 10% of {y} is {y*0.1} and then since $\\var{x}\\%=\\var{x/10}\\times10\\%$ we can calculate {x}% of {y} as,
\\[\\var{x/10}\\times10\\%\\times \\var{y}=\\var{x/10}\\times\\var{y*0.1}=\\var{x/100*y}.\\]

Use this link to find some resources which will help you revise this topic.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"x": {"name": "x", "group": "Ungrouped variables", "definition": "random(10..90 #10)", "description": "", "templateType": "anything", "can_override": false}, "y": {"name": "y", "group": "Ungrouped variables", "definition": "random(10.. 100 # 10)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["x", "y"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "y*(x/100)", "maxValue": "y*(x/100)", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "NF11 One number as a percentage of another", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Don Shearman", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/680/"}, {"name": "Adelle Colbourn", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2083/"}, {"name": "Ruth Hand", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3228/"}, {"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Ben McGovern", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4872/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}, {"name": "Megan Oliver", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23526/"}], "tags": [], "metadata": {"description": "

Given the number of international students enrolled on a course of $n$ students, calculate the percentage of 'home' students.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

{num_students} of the {class_size} students enrolled on a course are international students. What percentage are 'home' students?

", "advice": "

First work out the number of students who are not international. In this case it is {class_size} - {num_students} = {class_size-num_students} students.

Then write this as a fraction out of {class_size}. $ \\frac{\\var{class_size-num_students}} {\\var{class_size}} $

Then convert this to a percentage. You should put this fraction into your calculator and then multiply by 100:

$ \\frac{\\var{class_size-num_students}} {\\var{class_size}} \\times 100 = \\var{(class_size-num_students)/class_size*100}\\%$

Use this link to find resources to help you revise how to calculate percentages.

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The number of students in the class who do speak a language other than English.

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", "minMarks": 0, "maxMarks": 0, "shuffleChoices": true, "displayType": "radiogroup", "displayColumns": 0, "showCellAnswerState": true, "choices": ["$\\var{class_size-num_students}\\%$", "$\\var{num_students*100/class_size}\\%$", "$\\var{num_students}\\%$", "$\\var{(class_size-num_students)/class_size*100}\\%$"], "matrix": [0, 0, 0, "1"], "distractors": ["Have you converted this to a percentage? Click on Reveal Answer and scroll down for Advice regarding this question.", "How many students do NOT speak a language other than English at home? Click on Reveal Answer and scroll down for Advice regarding this question.", "How many students do NOT speak a language other than English at home? Then convert this to a percentage. Click on Reveal Answer and scroll down for Advice regarding this question.", "Well done!"]}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "NF12 Simplify (cancel down) Fractions", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Simon Thomas", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3148/"}, {"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "Calculating the LCM and HCF of numbers by using prime factorisation.", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Express the fraction below in its simplest form:

\\[\\frac{\\var{x}}{\\var{y}}\\]

", "advice": "

To simplify a fraction we need to divide both numbers by their common factors.

We can write $\\var{x}$ and $\\var{y}$ as a product of prime factors as follows:

$\\var{x}=\\var{show_factors(x)}$

$\\var{y}=\\var{show_factors(y)}$.

So to fully simplify the fraction we need to divide both $\\var{x}$ and $\\var{y}$ by

\\[\\var{show_factors(hcf_xy)}.\\]

This gives us the fraction

\\[\\frac{\\var{x/hcf_xy}}{\\var{y/hcf_xy}}\\]

Use this link to find some resources which will help you revise this topic.

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Manipulate fractions in order to add and subtract them. The difficulty escalates through the inclusion of a whole integer and a decimal, which both need to be converted into a fraction before the addition/subtraction can take place.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Evaluate the following addition, giving the fraction in its simplest form.

", "advice": "

$\\displaystyle\\frac{\\var{a_coprime}}{\\var{b_coprime}}+\\frac{\\var{c_coprime}}{\\var{d_coprime}}$

To add or subtract fractions, we need to have a common denominator on both fractions.

To get a common denominator, we need to find the lowest common multiple of the two denominators.

The lowest common multiple of $\\var{b_coprime}$ and $\\var{d_coprime}$ is $\\var{lcm}.$

This will be the new denominator, and we need to multiply each fraction individually to ensure we get this denominator.

For $\\displaystyle\\frac{\\var{a_coprime}}{\\var{b_coprime}}$, we need to multiply the fraction by $\\displaystyle\\frac{\\var{lcm_b}}{\\var{lcm_b}}$ to give $\\displaystyle\\frac{\\var{alcm_b}}{\\var{lcm}}.$

For $\\displaystyle\\frac{\\var{c_coprime}}{\\var{d_coprime}}$, we need to multiply the fraction by $\\displaystyle\\frac{\\var{lcm_d}}{\\var{lcm_d}}$ to give $\\displaystyle\\frac{\\var{clcm_d}}{\\var{lcm}}.$

Now that we have each fraction in terms of a common denominator, we can now add the fractions together.

$\\displaystyle\\frac{\\var{alcm_b}}{\\var{lcm}}+\\frac{\\var{clcm_d}}{\\var{lcm}}=\\frac{(\\var{alcm_b}+\\var{clcm_d})}{\\var{lcm}}=\\frac{\\var{alcmclcm}}{\\var{lcm}}.$

From this, we can try to simplify the result down by finding the greatest common divisor of the numerator and denominator and dividing the whole fraction by this amount.

The greatest common divisor of $\\var{alcmclcm}$ and $\\var{lcm}$ is $\\var{gcd}.$

Simplifying using this value gives a final answer of $\\displaystyle\\frac{\\var{num}}{\\var{denom}}.$

Therefore, the expression cannot be simplified further, and $\\displaystyle\\frac{\\var{num}}{\\var{denom}}$ is the final answer.

Find out more about this topic using our resource

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PART A answer for the denominator of part a

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PART A variable a - random number between 1 and 5.

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PART A variable d - random number between 5 and 15.

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PART A variable c times variable b

", "templateType": "anything", "can_override": false}, "gcd_ab": {"name": "gcd_ab", "group": "Part a", "definition": "gcd(a,b)", "description": "

PART A simplification of fractions in the question.

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PART A lcm of b and d, divided by b

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PART A answer for the numerator input

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PART A variable b - random number between 5 and 10 and not the same value as d.

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PART A variable a times variable d

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PART A lcm of b and d, divided by d

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PART A variable c - random number between 1 and 5.

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PART A variable c times the lcm of b and d, divided by d

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PART A greatest common divisor of the variables alcmclcm and lcm

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PART A variable a times the lcm of b and d, divided by b

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PART A

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PART A

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PART A

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PART A lowest common multiple of variable b_coprime and variable d_coprime.

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PART A

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$\\displaystyle\\frac{\\var{a_coprime}}{\\var{b_coprime}}+\\frac{\\var{c_coprime}}{\\var{d_coprime}}=$ [[0]] [[1]]

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Evaluate the following additions and subtractions, giving each fraction in its simplest form. Write the numerator (the top number) as negative if the fraction is negative.

", "advice": "

$\\displaystyle\\frac{\\var{f_coprime}}{\\var{g_coprime}}-\\frac{\\var{h_coprime}}{\\var{j_coprime}}+2.$

The two fractions can be individually multiplied to achieve a common denominator of the lowest common multiple, $\\var{lcm2}.$

$\\displaystyle\\frac{\\var{f_coprime}}{\\var{g_coprime}}$ becomes $\\displaystyle\\frac{\\var{flcm2_g}}{\\var{lcm2}}$ and $\\displaystyle\\frac{\\var{h_coprime}}{\\var{j_coprime}}$ becomes $\\displaystyle\\frac{\\var{hlcm2_j}}{\\var{lcm2}}.$

We can now subtract the second fraction from the first.

$\\displaystyle\\frac{\\var{flcm2_g}}{\\var{lcm2}}-\\frac{\\var{hlcm2_j}}{\\var{lcm2}}=\\frac{\\var{flcmhlcm}}{\\var{lcm2}}.$

Find out more about this topic using our resource.

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PART B

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PART B

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PART B gcd of first fraction num and denom

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PART B

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PART B

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PART B

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PART B

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PART B

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PART B

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PART B

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PART B g_coprime

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PART B

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PART B

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PART B

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PART B

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PART B

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$\\displaystyle\\frac{\\var{f_coprime}}{\\var{g_coprime}}-\\frac{\\var{h_coprime}}{\\var{j_coprime}}=$ [[0]] [[1]]

", "gaps": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": "2", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "flcmhlcm", "maxValue": "flcmhlcm", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}, {"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "lcm2", "maxValue": "lcm2", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "NF16 Multiply Fractions", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Lauren Richards", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1589/"}, {"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}, {"name": "Oliver Spenceley", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23557/"}], "tags": ["improper fractions", "mixed numbers", "multiplication of fractions", "multiplying fractions", "squared fraction", "taxonomy"], "metadata": {"description": "

Several problems involving the multiplication of fractions, with increasingly difficult examples, including a mixed fraction and a squared fraction. The final part is a word problem.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Evaluate the following multiplication, giving the answer in its simplest form.

", "advice": "

To multiply $\\displaystyle\\frac{\\var{a_coprime}}{\\var{c_coprime}}\\times\\frac{\\var{b_coprime}}{\\var{d_coprime}}$, address the numerators and denominators separately.

Multiply the numerators across both fractions.

$\\var{a_coprime}\\times\\var{b_coprime}=\\var{ab}$,

and then multiply the denominators across both fractions.

$\\var{c_coprime}\\times\\var{d_coprime}=\\var{cd}$.

The values of the multiplied numerators and denominators will be the numerator and denominator of the new fraction: $\\displaystyle\\frac{\\var{ab}}{\\var{cd}}$.

This answer may need simplifying down, and to do this, find the greatest common divisor in both the numerator and denominator and divide by this number.

The greatest common divisor of $\\var{ab}$ and $\\var{cd}$ is $\\var{gcd}$.

By using $\\var{gcd}$ to cancel down the fraction, the final answer is $\\displaystyle\\simplify{{ab}/{cd}}$.

Use this link to find some resources which will help you revise this topic.

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Random number between 1 and 20

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Random number between 1 and 20.

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Variable c times variable d.

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Random number from 1 to 12.

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Random number from 1 to 12.

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Numerator of the improper fraction converted from a mixed number.

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Variable f times variable h

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Random number between 1 and 4 - integer part of the mixed number.

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Random number from 1 to 12.

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PART A

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Denominator of new fraction.

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Random number between 1 and 20.

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Numerator of gap 0

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gcd of the numerator of the improper fraction

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Random number from 1 to 12.

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Random number between 1 and 20

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Variable a times variable b

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$\\displaystyle\\frac{\\var{a_coprime}}{\\var{c_coprime}}\\times\\frac{\\var{b_coprime}}{\\var{d_coprime}}$ = [[0]] [[1]]

Several problems involving dividing fractions, with increasingly difficult examples, including mixed numbers and complex fractions.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Evaluate the following sums involving division of fractions. Simplify your answers where possible.

", "advice": "

When faced with dividing fractions, it much easier to switch one of the fractions around and multiply them together instead of divide them.

\\[ \\left( \\frac{\\var{f_coprime}}{\\var{g_coprime}}\\div\\frac{\\var{h_coprime}}{\\var{j_coprime}} \\right) \\equiv \\left( \\frac{\\var{f_coprime}}{\\var{g_coprime}}\\times\\frac{\\var{j_coprime}}{\\var{h_coprime}} \\right) = \\frac{\\var{fj}}{\\var{gh}} \\]

Then, simplify by finding the highest common divisor in the numerator and denominator which in this case is $\\var{gcd1}$.

This gives a final answer of $\\displaystyle\\simplify{{fj}/{gh}}$.

Use this link to find some resources which will help you revise this topic

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variable f4 times h4.

Used in part c)

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PART C

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Random number but not the same number as variable g4.

Used in part c.

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PART C

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PART A

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PART A

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PART B

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greatest common divisor of variables f1j1 and g1h1.

Used in part b).

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PART B

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PART B

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greatest common denominator for part c.

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Random number between 2 and 20 and not the same value as variable h1.

Used in part b).

", "templateType": "anything", "can_override": false}, "g1h1": {"name": "g1h1", "group": "Ungrouped variables", "definition": "g1_coprime*h1_coprime", "description": "

variable g1 times h1.

Used in part b).

", "templateType": "anything", "can_override": false}, "f": {"name": "f", "group": "part a", "definition": "random(2..10)", "description": "

Random number between 2 and 10.

Used in part a).

", "templateType": "anything", "can_override": false}, "f4": {"name": "f4", "group": "Ungrouped variables", "definition": "random(1..3)", "description": "

Random number.

Used in part c).

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Random number between 2 and 20.

Used in part b)

", "templateType": "anything", "can_override": false}, "g3": {"name": "g3", "group": "Ungrouped variables", "definition": "random(1..3)", "description": "

Random number.

Used in part c).

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variable f3 times h3.

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Random number from 2 to 10.

Used in part a).

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variable g times variable h.

Used in part a).

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PART A

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Unsimplified denominator of part c.

", "templateType": "anything", "can_override": false}, "j": {"name": "j", "group": "part a", "definition": "random(h..12 except h)", "description": "

Random number between 2 and 10 and not the same value as h.

Used in part a).

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variable f1 times j1.

Used in part b).

", "templateType": "anything", "can_override": false}, "h4_coprime": {"name": "h4_coprime", "group": "Ungrouped variables", "definition": "h4/gcd(g4,h4)", "description": "

PART C

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Random number between 2 and 30 and not the same value as variable f1.

Used in part b).

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variable f times variable j.

Used in part a).

", "templateType": "anything", "can_override": false}, "f3": {"name": "f3", "group": "Ungrouped variables", "definition": "random(1 .. 3#1)", "description": "

Random number between 2 and 6.

Used in part c).

", "templateType": "randrange", "can_override": false}, "f1_coprime": {"name": "f1_coprime", "group": "Ungrouped variables", "definition": "f1/gcd(f1,g1)", "description": "

PART B

", "templateType": "anything", "can_override": false}, "h3": {"name": "h3", "group": "Ungrouped variables", "definition": "random(5..8)", "description": "

Random number and not the same value as variable g3.

Used in part c).

", "templateType": "anything", "can_override": false}, "gcd1": {"name": "gcd1", "group": "part a", "definition": "gcd(fj,gh)", "description": "

greatest common divisor of variable fj and gh.

Used in part a).

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PART C

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PART A

", "templateType": "anything", "can_override": false}, "g4": {"name": "g4", "group": "Ungrouped variables", "definition": "random(1..5)", "description": "

Random number.

Used in part c).

", "templateType": "anything", "can_override": false}, "h1": {"name": "h1", "group": "Ungrouped variables", "definition": "random(2..10)", "description": "

Random number between 2 and 20.

Used in part b).

", "templateType": "anything", "can_override": false}, "num": {"name": "num", "group": "Ungrouped variables", "definition": "h4_coprime*(f3h3+g3_coprime)", "description": "

numerator of the improper fraction in part c. Unsimplified.

", "templateType": "anything", "can_override": false}, "g": {"name": "g", "group": "part a", "definition": "random(2..10)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["f1", "g1", "f1_coprime", "g1_coprime", "h1", "j1", "h1_coprime", "j1_coprime", "f1j1", "g1h1", "gcd2", "f3", "g3", "h3", "g3_coprime", "h3_coprime", "f4", "g4", "h4", "g4_coprime", "h4_coprime", "f3h3", "f4h4", "num", "denom", "gcd3"], "variable_groups": [{"name": "part a", "variables": ["g", "f", "f_coprime", "g_coprime", "h", "j", "h_coprime", "j_coprime", "fj", "gh", "gcd1"]}], "functions": {}, "preamble": {"js": "", "css": "fraction {\n display: inline-block;\n vertical-align: middle;\n}\nfraction > numerator, fraction > denominator {\n float: left;\n width: 100%;\n text-align: center;\n line-height: 2.5em;\n}\nfraction > numerator {\n border-bottom: 1px solid;\n padding-bottom: 5px;\n}\nfraction > denominator {\n padding-top: 5px;\n}\nfraction input {\n line-height: 1em;\n}\n\nfraction .part {\n margin: 0;\n}\n\n.table-responsive, .fractiontable {\n display:inline-block;\n}\n.fractiontable {\n padding: 0; \n border: 0;\n}\n\n.fractiontable .tddenom \n{\n text-align: center;\n}\n\n.fractiontable .tdnum \n{\n border-bottom: 1px solid black; \n text-align: center;\n}\n\n\n.fractiontable tr {\n height: 3em;\n}\n"}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

$\\displaystyle\\frac{\\var{f_coprime}}{\\var{g_coprime}}\\div\\frac{\\var{h_coprime}}{\\var{j_coprime}}=$ [[0]] [[1]]

Dividing amounts in ratios

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

The ratio of ethanol to water is {a}:{b} for an experiment. If I have {volWater}ml of water, how much ethanol do I need?

", "advice": "

If there is a ratio of {a}:{b} for ethanol:water then that means for every {b}ml of water we need {a}ml of ethanol.

In our experiment there is {volwater}ml of water so to find the amount of ethanol we divide by {b} and then multiply by {a}.

\\[\\var{volwater}\\text{ml}\\times\\frac{\\var{a}}{\\var{b}}=\\var{volwater*a/b}\\text{ml}\\]

Use this link to find some resources which will help you revise this topic.

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[[0]]ml

This question is about identifying what types of charts or visual representations of data you can use for different data sets.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

This question is about recognising what types of charts or visual representations of data you can use with what types of data sets.

", "advice": "

There are many different types of visual representations of data and sometimes there will be a choice of what you use.

Start by looking at these resources to build up your understanding of data display options and methods.

The table shows different names of charts on the left hand side and different descriptions of data sets along the top.

Pair up each description with the chart that would be most suitable.

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This question is about correctly interpreting pie charts.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

{geogebra_applet{\"https://www.geogebra.org/calculator/pmvumdrv\",[C: C,M: M]}}

The Pie Chart above shows the responses to a question asked by someone trying to plan a social event for their workplace. It shows answers given to the question \"Where would you like to go for a staff social?\" with the options \"Meal\", \"Cinema\" and \"Games Cafe\".

", "advice": "

A Pie chart of this type can only be used to make statements about the proportions of data in each category and does not provide information about the actual frequencies.

For more information on Pie Charts follow this link.

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From the following comments which can you say are definitely true, definitely false and which do you not have enough information to know?

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Reading a value from a simple bar chart.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "", "advice": "

The bar heights give the values of the spend.

Each company has two bars, the left one for last year (in red) and the right one for this year (in purple).
Isolate last years spend by looking at the the bars on the right side, and choose the tallest bar, corresponding to the highest value.

Use this link to find some resources which will help you revise this topic.

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new y values after the transformation

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the (random) original y values which relate to the x values

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vertical shift

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horizontal shift

", "templateType": "anything", "can_override": false}, "fakeanswer4": {"name": "fakeanswer4", "group": "Ungrouped variables", "definition": "random([yo9,yo7,yo8] except fakeanswer3)", "description": "", "templateType": "anything", "can_override": false}, "dd": {"name": "dd", "group": "Ungrouped variables", "definition": "random(0.7..1.3#0.01 except 1 except aa except bb except cc)", "description": "", "templateType": "anything", "can_override": false}, "percent": {"name": "percent", "group": "Ungrouped variables", "definition": "random(5..15#0.1 except 5 except 6 except 7 except 8 except 9 except 10 except 11 except 12 except 13 except 14 except 15)", "description": "", "templateType": "anything", "can_override": false}, "yo0": {"name": "yo0", "group": "Ungrouped variables", "definition": "random(20..40#1)", "description": "", "templateType": "anything", "can_override": false}, "xn": {"name": "xn", "group": "Ungrouped variables", "definition": "map((x-hsh)/hsc,x,xo)", "description": "

new transformed x values

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original x values

", "templateType": "anything", "can_override": false}, "yo4": {"name": "yo4", "group": "Ungrouped variables", "definition": "random(20..40#1 except yo1 except yo0 except yo2 except yo3)", "description": "", "templateType": "anything", "can_override": false}, "fakeanswer2": {"name": "fakeanswer2", "group": "Ungrouped variables", "definition": "random([yo1,yo0,yo2,yo3,yo4] except answer except fakeanswer1)", "description": "", "templateType": "anything", "can_override": false}, "f": {"name": "f", "group": "Ungrouped variables", "definition": "random(1.1..1.3#0.01 except a except b except c except d except e)", "description": "", "templateType": "anything", "can_override": false}, "yo51": {"name": "yo51", "group": "Ungrouped variables", "definition": "eee*yo5", "description": "", "templateType": "anything", "can_override": false}, "fakeanswer3": {"name": "fakeanswer3", "group": "Ungrouped variables", "definition": "random([yo6,yo7,yo8])", "description": "", "templateType": "anything", "can_override": false}, "increase": {"name": "increase", "group": "Ungrouped variables", "definition": "random(10..40#5)", "description": "", "templateType": "anything", "can_override": false}, "yo6": {"name": "yo6", "group": "Ungrouped variables", "definition": "random(41..70#1 except yo5)", "description": "", "templateType": "anything", "can_override": false}, "hsc": {"name": "hsc", "group": "Ungrouped variables", "definition": "if(selector='hsc',random(-2,-1,-0.5,0.5,2),1)", "description": "", "templateType": "anything", "can_override": false}, "yo41": {"name": "yo41", "group": "Ungrouped variables", "definition": "d*yo4", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["selector", "vsh", "hsh", "vsc", "hsc", "yo", "yn", "xo", "xn", "yo0", "yo1", "yo2", "yo3", "yo4", "maxx", "yo6", "yo7", "yo8", "yo9", "yo41", "yo5", "yo51", "a", "b", "c", "d", "eee", "f", "answer", "fakeanswer1", "fakeanswer2", "fakeanswer3", "fakeanswer4", "aa", "bb", "cc", "dd", "percent", "students", "yearvector", "ii", "year", "answervector", "increase"], "variable_groups": [], "functions": {}, "preamble": {"js": "function dragpoint_board() {\n var scope = question.scope;\n\n JXG.Options.text.display = 'internal';\n \n var yo0 = scope.variables.yo0.value;\n var yo1 = scope.variables.yo1.value;\n var yo2 = scope.variables.yo2.value;\n var yo3 = scope.variables.yo3.value;\n var yo4 = scope.variables.yo4.value;\n var yo5 = scope.variables.yo5.value;\n var yo6 = scope.variables.yo6.value;\n var yo7 = scope.variables.yo7.value; \n var yo8 = scope.variables.yo8.value;\n var yo9 = scope.variables.yo9.value; \n \n var div = Numbas.extensions.jsxgraph.makeBoard('550px','550px',{boundingBox:[-0.8,82,16,-8], axis:false, grid:true});\n \n question.display.html.querySelector('#dragpoint').append(div);\n \n var board = div.board;\n \n// board.suspendUpdate(); \n\n \n var dataArr = [yo0,yo5,0,yo1,yo6,0,yo2,yo7,0,yo3,yo8,0,yo4,yo9]; \n \n var xaxis = board.create('axis', [[0, 0], [12, 0]], {withLabel: true, name: \"Bank\", label: {offset: [250,-30]}});\n \n xaxis.removeAllTicks(); \n \n board.create('axis', [[0, 0], [0, 10]], {hideTicks:true, withLabel: false, name: \"\", label: {offset: [-110,300]}});\n \n var pop0 = board.create('point', [1.5,0],{name:'Morgan',fixed:true,size:0,color:'black',face:'diamond', label:{offset:[-20,-8]}});\n var pop1 = board.create('point',[4.5,0],{name:'Strome',fixed:true,size:0,color:'black',face:'diamond', label:{offset:[-20,-8]}});\n var pop2 = board.create('point',[7.5,0],{name:'Bentley',fixed:true,size:0,color:'black', face:'diamond', label:{offset:[-15,-8]}});\n var pop3 = board.create('point',[10.5,0],{name:'Sand',fixed:true,size:0,color:'black', face:'diamond', label:{offset:[-15,-8]}});\n var pop4 = board.create('point',[13.5,0],{name:'Karchen',fixed:true,size:0,color:'black', face:'diamond', label:{offset:[-15,-8]}});\n var leg1 = board.create('point',[12,75],{name:'last year',fixed:true,size:6,color:'#DA2228', face:'square', label:{offset:[9,0]}});\n var leg2 = board.create('point',[12,72],{name:'this year',fixed:true,size:6,color:'#6F1B75', face:'square', label:{offset:[9,0]}});\n \n \n// var chart = board.createElement('chart', dataArr, \n // {chartStyle:'bar', fillOpacity:1, width:1,\n // colorArray:['#8E1B77','#8E1B77','Red','Red','blue','red','blue','red','red','blue', 'red','blue','red','red'], shadow:false});\n \n//var chart = board.createElement('chart', dataArr, \n // {chartStyle:'bar', width:1,fillOpacity:1, fillColor:'red', shadow:false}); \n \n \n var a = board.create('chart', [[1,2,3],[yo0,yo5,0]], {chartStyle:'bar',colors:['#DA2228','#6F1B75','#6F1B75'],width:1,fillOpacity:1});\n var b = board.create('chart', [[4,5,6],[yo1,yo6,0]], {chartStyle:'bar',width:1,colors:['#DA2228','#6F1B75','#6F1B75'],fillOpacity:1});\n var c = board.create('chart', [[7,8,9],[yo2,yo7,0]], {chartStyle:'bar',width:1,colors:['#DA2228','#6F1B75','#6F1B75'],fillOpacity:1});\n var d = board.create('chart', [[10,11,12],[yo3,yo8,0]], {chartStyle:'bar',width:1,colors:['#DA2228','#6F1B75','#6F1B75'],fillOpacity:1});\n var e = board.create('chart', [[13,14],[yo4,yo9]], {chartStyle:'bar',width:1,colors:['#DA2228','#6F1B75'],fillOpacity:1});\n \n board.unsuspendUpdate();\n \n var txt1 = board.create('text',[-0.3,30, 'Investment \u00a3(m)'], {fontColor:'black', fontSize:14, rotate:90});\n \n // var txt = board.create('text',[0.5,75, 'Investment (m)'], {fontSize:14, rotate:90});\n \n // var txt1 = board.create('text',[8,76, 'red bars represent 2010'], {fontColor:'red', fontSize:14, rotate:90});\n \n // var txt2 = board.create('text',[8,73, 'blue bars represents 2011'], {fontSize:14, rotate:90});\n\n // var myColors = new Array('red', 'blue', 'white','red', 'blue', 'white','red', 'blue', 'white','red', 'blue', 'white','red', 'blue');\n \n \n \n //board.unsuspendUpdate();\n\n // Rotate text around the lower left corner (-2,-1) by 30 degrees.\n // var tRot = board.create('transform', [90.0*Math.PI/180.0, -1,40], {type:'rotate'}); \n // tRot.bindTo(txt);\n // board.update();\n\n \n//var chart2 = board.createElement('chart', dataArr, {chartStyle:'line,point'});\n//chart2[0].setProperty('strokeColor:black','strokeWidth:2','shadow:true');\n//for(var i=0; i<11;i++) {\n // chart2[1][i].setProperty({strokeColor:'black',fillColor:'white',face:'[]', size:4, strokeWidth:2});\n//}\n//board.unsuspendUpdate(); \n \n //board.unsuspendUpdate();\n\n}\n\nquestion.signals.on('HTMLAttached',function() {\n dragpoint_board();\n});", "css": "table#values th {\n background: none;\n text-align: center;\n}"}, "parts": [{"type": "1_n_2", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Banking Sector: IT Infrastructure Spending

What was the maximum spend by a single company last year?

", "minMarks": 0, "maxMarks": 0, "shuffleChoices": true, "displayType": "radiogroup", "displayColumns": 0, "showBlankOption": true, "showCellAnswerState": true, "choices": ["

£{answer} m

", "

£{fakeanswer1} m

", "

£{fakeanswer2} m

", "

£{fakeanswer3} m

", "

£{fakeanswer4} m

"], "matrix": ["1", 0, 0, 0, 0], "distractors": ["", "", "", "", ""]}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "SD04 Interpret a Box Plot", "extensions": ["geogebra"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Lauren Desoysa", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21504/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}, {"name": "Michael Pan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23528/"}], "tags": [], "metadata": {"description": "

Interpreting the elements of a box plot

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

The diagram below shows a box plot of some data.

{geogebra_applet{\"https://www.geogebra.org/m/aj2hcbhg\",[lv: lv,lq: lq,m: m,uq: uq,hv: hv]}}

", "advice": "

A boxplot (also known as a box-and-whisker diagram or plot) is a convenient way of graphically depicting groups of numerical data through their five-number summaries: the smallest observation (sample minimum), lower quartile (Q1), median (Q2), upper quartile (Q3), and largest observation (sample maximum). A boxplot may also indicate which observations, if any, might be considered outliers.

For more information on box plots follow this link.

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Which of these statements are true and which are false?

", "minMarks": 0, "maxMarks": 0, "minAnswers": 0, "maxAnswers": 0, "shuffleChoices": true, "shuffleAnswers": false, "displayType": "radiogroup", "warningType": "none", "showCellAnswerState": true, "markingMethod": "sum ticked cells", "choices": ["The range of the data is $\\var{range}$.", "The Interquarttile range of the data is larger than the range of the data.", "You can calculate the mean of the data from this Box plot.", "

The median of the data is $\\var{m}$.

", "The mode of the data is $\\var{lv-3}$."], "matrix": [["1", 0], [0, "1"], [0, "1"], ["1", 0], [0, "1"]], "layout": {"type": "all", "expression": ""}, "answers": ["True.", "False."]}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "SD05 Interpret contingency table", "extensions": ["stats"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Chris Graham", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/369/"}, {"name": "Stanislav Duris", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1590/"}, {"name": "Ruth Hand", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3228/"}, {"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Upuli Wickramaarachchi", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23527/"}], "tags": [], "metadata": {"description": "

Calculate an intersection probability given a two way table.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "", "advice": "

a) Each row and column must sum to the 'total'.

b) Look for the column containing '$\\var{q1a}$' and the row containing '$\\var{q1b}$'. The entry where they intersect, $\\var{q1*total}$, is the value we are interested in.

Since we require a probability, this is $\\var{q1*total}$ out of $\\var{total}$, i.e.

\\[ \\frac{\\var{q1*total}}{\\var{total}} \\]

Use this link to find some resources which will help you revise this topic

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$\\var{total}$ items are sampled. Complete the table.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

	$\\var{nB}$	not $\\var{nB}$	TOTAL
$\\var{nA}$	[[0]]	$\\var{AnB'}$	$\\var{A}$
not $\\var{nA}$	$\\var{notAnB}$	[[1]]	[[2]]
TOTAL	[[3]]	[[4]]	$\\var{total}$

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If one item is picked at random, use the table to calculate the probability that the item is '{q1a}' and '{q1b}'.

Give your answer as a fraction, or a decimal correct to 2dp.

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Reading a value from a histogram.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

The histogram shows information about distances run on a Sunday by some randomly asked people in a park.

", "advice": "

To calculate the frequencies using the following formula:

frequency = class width $\\times$ frequency density

Hence our table becomes:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

Distance, $km$	Frequency
$0 < t \\leq 10$	$10 \\times \\var{yo0} = \\var{freq0}$
$10 < t \\leq 15$	$5 \\times \\var{yo1} = \\var{freq1}$
$15 < t \\leq 20$	$5 \\times \\var{yo2} = \\var{freq2}$
$20 < t \\leq 30$	$10 \\times \\var{yo3} = \\var{freq3}$

Hence, to find the total number of people that ran that day:

$\\var{freq0} + \\var{freq1} + \\var{freq2} + \\var{freq3} = \\var{total}.$

The frequency of runners that ran less than $10km$ is found by calculating the frequency from the first bar on the histogram:

class width $\\times$ frequency density $= 10 \\times \\var{yo0} = \\var{freq0}$.

The frequency of runners that ran between $15km$ and $20km$ is found by calculating the frequency from the third bar on the histogram:

class width $\\times$ frequency density $= 5 \\times \\var{yo2} = \\var{freq2}$.

To estimate how many runners ran more than $25km$ we need again need to use the frequency = class width \\times frequency density formula.

Here the class width is $5$ because we are looking for the frequency of runners that ran between $25km$ and $30km$ from the college.

Frequency $= 5 \\times \\var{yo3} = \\var{5*yo3}.$

If this number is a decimal we round up to get $\\var{m25}$.

As in part d),to estimate how many runners ran less than $7km$ we use the frequency = class width \\times frequency density formula.

Here the class width is $7$ because we are looking for the frequency of runners that ran between 0 and 7 km.

Frequency $= 7 \\times \\var{yo0} = \\var{7*yo0}.$

If this number is a decimal we round up to get $\\var{l7}$.

To estimate how many runners ran between $5km$ and $12.5km$ we use a method similar to that in part d) and e) but, this time, we need to use information from both the first and second bar on the histogram.

Let's first calculate how many runners ran between $5km$ and $10km$:

frequency $=$ class width $\\times$ frequency density $= 5 \\times \\var{yo0} = \\var{5*yo0}$.

Now we need to calculate how many runners ran between $10km$ and $12.5km$:

frequency $=$ class width $\\times$ frequency density $= 2.5 \\times \\var{yo1} = \\var{2.5*yo1}$.

Putting this together the number of runners that ran between $5km$ and $12.5km$ is $\\var{5*yo0} + \\var{2.5*yo1} = \\var{5*yo0 + 2.5*yo1}.$

If this number is a decimal we round up to get $\\var{ef}.$

The number of runners that ran between $10km$ and $14km$ is $4 \\times \\var{yo1} = \\var{4*yo1}.$

If this number is a decimal we round up to get $\\var{eh}.$

The number of runners that ran further than $18km$ is:

$2 \\times \\var{yo2} + 10 \\times \\var{yo3} = \\var{2*yo2} + \\var{freq3} = \\var{2*yo2+freq3}.$

If this number is a decimal we round up to get $\\var{ej}.$

Did you get a decimal answer? Were you surprised to see a whole number answer when you got a decimal on one of the estimate questions? Look at the context of the question, you cannot have $0.5$ of a student so we round our answers up to the next whole number!

Use this link to find some resources which will help you revise this topic.

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Frequency Density

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Frequency Density

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Frequency Density

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Frequency Density

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Sum of frequencies

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Frequency

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Frequency

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Frequency

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Frequency

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How many people ran more than 25km

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How many runners ran less than 7km

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Number of people running between 5 & 12.5 km

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Number of runners running between 10 & 14km

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Number of runners running more than 18km

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How many people were asked about the distance they ran that day?

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How many runners ran less than 10 km that day?

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How many people ran between 10 and 15 km that day?

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Estimate how many runners ran more than 25 km that day?

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Estimate how many runners ran less than 7 km that day?

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Estimate how many people ran between 5 km and 12.5 km.

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Francesca recorded the number of customers in a supermarket every two hours.

She began at 9 am and finished at 7 pm.

The line graph below shows her results.

{geogebra_applet{\"https://www.geogebra.org/classic/s4w7nmga\",[y1:y1,y3:y3,y4:y4,y6:y6]}}

", "advice": "

a) You want to find the point on the graph with the greatest frequency. You can see that this is at 1pm when there were $\\var{y3}$ cars in the car park.

b) We find 11am on the $x$-axis and look vertically upwards until we find the point. From here we go horiztonally across to the $y$-axis to read the frequency. We can see that at 11am there were $\\var{y2}$ cars in the car park.

c) We want to find $\\var{number}$ on the $y$-axis and then look horiztonally across until we find the point. From here we move down to the $x$-axis to see at which time there were $\\var{number}$ cars in the car park. We can see this occured at $\\var{answerc}\\var{time}$.

d) We must find 6pm on the $x$-axis. This isn't marked on like the other times but we know it sits half way between 5pm and 7pm. From here we look vertically upwards until we meet the red line on our graph. Notice we don't have a point for this time, hence why this is an estimate. From here we move hortizontally across to the $y$-axis to find the frequency. At 6pm we estimate that there were $\\var{y56}$ cars in the car park.

e) We must find the frequency of cars at 1pm and at 3pm using the same steps as in part b. At 3pm there were $\\var{y4}$ cars in the car park and we subtract this from $\\var{y3}$ which is the number of cars in the car park at 1pm. Hence, $\\var{y3}-\\var{y4}=\\var{y3-y4}$.

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Number of customers in supermarket at 7pm.

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Number of customers in supermarket at 5pm.

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Number of customers in supermarket at 1pm.

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Number of customers in supermarket at 11am.

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Number of customers in supermarket at 9am

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Number of customers in supermarket at 6pm.

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Randomly select which number of customers user needs to identify associated time for.

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Associated time to randomly selected frequency variable (number)

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For printing am or pm depending on time selelected - to appear in advice only

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When were the most customers in the supermarker?

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How many customers were in the supermarket at 11 am?

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At what time were there {number} customers in the supermarket?

If your answer was 12pm you would just write 12 in the box.

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Estimate the number of customers in the supermarket at 6pm.

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How many less customers were there in the supermarket at 3pm than 1pm?

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Decide whether each of the described sets of data is drawn from a discrete or continuous distribution.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Decide whether the following data sets are discrete or continuous.

", "advice": "

Data can either be discrete or continuous.

Discrete data consists of integers, units of measurement cannot be split. Only certain values have a meaning.
Continuous data can be shown on a number line, all points on this line have a meaning. In between two measurements there's always a valid \"middle\" measurement. So for example, between 156.3cm and 156.4cm there's 156.35cm.

a)

Height is a continuous variable. For example, 180.3cm and 180.4cm have a valid midpoint 180.35cm.Weight is a continuous variable. For example, 54.5kg and 54.6kg have a valid midpoint 54.55kg.Time is a continuous variable. For example, 54.2s and 54.3s have a valid midpoint 54.25s.Temperature is a continuous variable, it can take any value between -273.15°C (absolute zero) and positive infinity. For example, 25°C and 26°C have a valid midpoint 25.5°C. Hence, this data is continuous.

b)

The number of Stage 1 students will always be an integer. You cannot split one student into two, for example value 19.5 students does not make sense. Therefore, this is a discrete set of data.The result of rolling 3 dice can take values of integers from 3 up to 18. For example, values 3 and 4 do not have any valid middle measurement. Therefore, this is a discrete set of data.Shoe sizes are a discrete set of data. For example, sizes 39 and 40 mean something while the middle value 39.5 does not.The number of chocolate bars sold on Monday will always be an integer. There is no middle measurement between 1 and 2 bars sold. You cannot buy a half of a bar. Therefore, this is a discrete set of data.The number of movies downloaded will always be an integer. You can either download a movie successfully or unsuccessfuly, so this is a discrete set of data. It is impossible to split 0 and 1 movies downloaded into 0.5. The number of cinema tickets sold will always be a whole number. There is no middle measurement between 1 and 2 tickets sold. You simply cannot buy half of a ticket. Therefore, this is a discrete set of data.

c)

The number of Stage 1 students will always be an integer. You cannot split one student into two, for example value 19.5 students does not make sense. Therefore, this is a discrete set of data.The result of rolling 3 dice can take values of integers from 3 up to 18. For example, values 3 and 4 do not have any valid middle measurement. Therefore, this is a discrete set of data.Shoe sizes are a discrete set of data. For example, sizes 39 and 40 mean something while the middle value 39.5 does not..The number of chocolate bars sold on Monday will always be an integer. There is no middle measurement between 1 and 2 bars sold. You cannot buy a half of a bar. Therefore, this is a discrete set of data.The number of movies downloaded will always be an integer. You can either download a movie successfully or unsuccessfuly, so this is a discrete set of data. It is impossible to split 0 and 1 movies downloaded into 0.5.The number of cinema tickets sold will always be a whole number. There is no middle measurement between 1 and 2 tickets sold. You simply cannot buy half of a ticket. Therefore, this is a discrete set of data.

d)

e)

When we round continuous variables to the nearest integer, this data becomes discrete, as there are no valid middle measurements between the integers. Therefore, the weight of a dog to the nearest kgthe height of Olympic medalists to the nearest cmthe time taken to run 10km to the nearest min is discrete and not continuous.

f)

The number of Stage 1 students will always be an integer. You cannot split one student into two, for example value 19.5 students does not make sense. Therefore, this is a discrete set of data.The result of rolling 3 dice can take values of integers from 3 up to 18. For example, values 3 and 4 do not have any valid middle measurement. Therefore, this is a discrete set of data.Shoe sizes are a discrete set of data. For example, sizes 39 and 40 mean something while the middle value 39.5 does not.The number of chocolate bars sold on Monday will always be an integer. There is no middle measurement between 1 and 2 bars sold. You cannot buy half of a bar of chocolate. Therefore, this is a discrete set of data.The number of movies downloaded will always be an integer. You can either download a movie successfully or unsuccessfuly, so this is a discrete set of data. It is impossible to split 0 and 1 movies downloaded into 0.5.The number of cinema tickets sold will always be a whole number. There is no middle measurement between 1 and 2 tickets sold. You simply cannot buy half of a ticket. Therefore, this is a discrete set of data.

Use this link to find some resources which will help you revise this topic

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"rand3": {"name": "rand3", "group": "Ungrouped variables", "definition": "random(0..5 except rand except rand2)", "description": "", "templateType": "anything", "can_override": false}, "cont": {"name": "cont", "group": "Ungrouped variables", "definition": "[\"The height of Newcastle University students.\", \"The weight of Olympic medalists.\", \"The time taken to brush teeth.\", \"The maximum daily temperature.\"]", "description": "", "templateType": "anything", "can_override": false}, "disc": {"name": "disc", "group": "Ungrouped variables", "definition": "[\"The number of Stage 1 students.\", \"The result of rolling 3 dice.\", \"Shoe sizes.\", \"The number of chocolate bars sold on Monday.\", \"The number of movies downloaded.\", \"The number of cinema tickets sold.\"]", "description": "", "templateType": "anything", "can_override": false}, "trick": {"name": "trick", "group": "Ungrouped variables", "definition": "[\"The weight of a dog to the nearest kg.\", \"The height of Olympic medalists to the nearest cm.\", \"The time taken to run 10km to the nearest min.\"]", "description": "", "templateType": "anything", "can_override": false}, "ranc2": {"name": "ranc2", "group": "Ungrouped variables", "definition": "random(0..3 except ranc)", "description": "", "templateType": "anything", "can_override": false}, "rant": {"name": "rant", "group": "Ungrouped variables", "definition": "random(0..2)", "description": "", "templateType": "anything", "can_override": false}, "rand": {"name": "rand", "group": "Ungrouped variables", "definition": "random(0..5)", "description": "", "templateType": "anything", "can_override": false}, "rand2": {"name": "rand2", "group": "Ungrouped variables", "definition": "random(0..5 except rand)", "description": "", "templateType": "anything", "can_override": false}, "ranc": {"name": "ranc", "group": "Ungrouped variables", "definition": "random(0..3)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["disc", "cont", "trick", "ranc", "rant", "ranc2", "rand2", "rand3", "rand"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "1_n_2", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

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{disc[rand]}

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Discrete

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{disc[rand2]}

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Discrete

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Continuous

{cont[ranc2]}

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Discrete

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Continuous

{trick[rant]}

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Discrete

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{disc[rand3]}

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"], "matrix": ["1", 0], "distractors": ["", ""]}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "SM01 Calculate Range", "extensions": ["stats"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Chris Graham", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/369/"}, {"name": "Stanislav Duris", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1590/"}, {"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Lauren Desoysa", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21504/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}, {"name": "Michael Pan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23528/"}], "tags": ["mean", "measures of average and spread", "median", "mode", "range", "taxonomy"], "metadata": {"description": "

This question provides a list of data to the student. They are asked to find the \"range\".

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

A random sample of 20 residents from Newcastle were asked about the number of times they went to see a play at the theatre last year.

Here is the list of their answers:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\var{a[0]}$	$\\var{a[1]}$	$\\var{a[2]}$	$\\var{a[3]}$	$\\var{a[4]}$	$\\var{a[5]}$	$\\var{a[6]}$	$\\var{a[7]}$	$\\var{a[8]}$	$\\var{a[9]}$
$\\var{a[10]}$	$\\var{a[11]}$	$\\var{a[12]}$	$\\var{a[13]}$	$\\var{a[14]}$	$\\var{a[15]}$	$\\var{a[16]}$	$\\var{a[17]}$	$\\var{a[18]}$	$\\var{a[19]}$

", "advice": "

Range is the difference between the highest and the lowest value in the data.

To find this, we subtract the lowest value from the highest value:

\\[ \\var{max(a)} - \\var{min(a)} = \\var{range} \\text{.}\\]

Use this link to find some resources which will help you revise this topic.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"a2": {"name": "a2", "group": "Ungrouped variables", "definition": "repeat(random(1..9), 20)", "description": "

Option 2 for the list. Only used if there is only one mode and option 1 was not used.

Option 1 for the list. Only used if there is only one mode.

", "templateType": "anything", "can_override": false}, "a_s": {"name": "a_s", "group": "final list", "definition": "sort(a)", "description": "

Sorted list.

", "templateType": "anything", "can_override": false}, "modea2": {"name": "modea2", "group": "Ungrouped variables", "definition": "mode(a2)", "description": "", "templateType": "anything", "can_override": false}, "a3": {"name": "a3", "group": "Ungrouped variables", "definition": "shuffle([ random(0..1),\n 2, \n random(4..6),\n random(0..3 except 2), \n random(0..3 except 2),\n random(4..6),\n 2,\n 2,\n random(4..6),\n random(7..8),\n random(0..3 except 2 except 1), \n random(4..6),\n 2,\n random(1..3 except 2), \n random(7..8),\n 2,\n random(7..8),\n random(4..6), \n random(0..3 except 2), \n 2\n])", "description": "

Option 3 for the list. Ensures there is only one mode (2) while still randomising the data.

", "templateType": "anything", "can_override": false}, "modetimes": {"name": "modetimes", "group": "final list", "definition": "map(\nlen(filter(x=j,x,a)),\nj, 0..8)", "description": "

The vector of number of times of each value in the data.

The final list.

", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["modea1", "modea2", "a1", "a2", "a3"], "variable_groups": [{"name": "final list", "variables": ["a", "a_s", "range", "modetimes"]}], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Find the range.

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Calculating the Mean from a basic list of integers.

", "licence": "None specified"}, "statement": "

Calculate the Mean from a list

", "advice": "

The MEAN is the sum, divided by the number of values summed i.e.

$\\frac{\\var{list[0]} + \\var{list[1]} + \\var{list[2]} + \\var{list[3]} + \\var{list[4]}}{5}$

use your calculator to find

mean = $\\var{mean}$.

Use this link to find some resources which will help you revise this topic.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"list": {"name": "list", "group": "Ungrouped variables", "definition": "repeat(random(0..20), 5)", "description": "", "templateType": "anything", "can_override": false}, "mean": {"name": "mean", "group": "Ungrouped variables", "definition": "mean(list)", "description": "", "templateType": "anything", "can_override": false}, "median": {"name": "median", "group": "Ungrouped variables", "definition": "median(list)", "description": "", "templateType": "anything", "can_override": false}, "order": {"name": "order", "group": "Ungrouped variables", "definition": "sort(list)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["list", "mean", "median", "order"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Given a list of numbers:

{list}

Calculate the mean: [[0]]

This question provides a list of data to the student. They are asked to find the \"mode\".

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

A random sample of 20 residents from Newcastle were asked about the number of times they went to see a play at the theatre last year.

Here is the list of their answers:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\var{a[0]}$	$\\var{a[1]}$	$\\var{a[2]}$	$\\var{a[3]}$	$\\var{a[4]}$	$\\var{a[5]}$	$\\var{a[6]}$	$\\var{a[7]}$	$\\var{a[8]}$	$\\var{a[9]}$
$\\var{a[10]}$	$\\var{a[11]}$	$\\var{a[12]}$	$\\var{a[13]}$	$\\var{a[14]}$	$\\var{a[15]}$	$\\var{a[16]}$	$\\var{a[17]}$	$\\var{a[18]}$	$\\var{a[19]}$

", "advice": "

The mode is the value that occurs the most often in the data.

To find a mode, we can look at our sorted list:

$\\var{a_s[0]}, \\var{a_s[1]}, \\var{a_s[2]}, \\var{a_s[3]}, \\var{a_s[4]}, \\var{a_s[5]}, \\var{a_s[6]}, \\var{a_s[7]}, \\var{a_s[8]}, \\var{a_s[9]}, \\var{a_s[10]}, \\var{a_s[11]}, \\var{a_s[12]}, \\var{a_s[13]}, \\var{a_s[14]}, \\var{a_s[15]}, \\var{a_s[16]}, \\var{a_s[17]}, \\var{a_s[18]}, \\var{a_s[19]}$.

We notice that $\\var{mode1}$ occurs the most ($\\var{modetimes[mode1]}$ times) so $\\var{mode1}$ is the mode.

Use this link to find some resources which will help you revise this topic.

Option 2 for the list. Only used if there is only one mode and option 1 was not used.

Option 1 for the list. Only used if there is only one mode.

", "templateType": "anything", "can_override": false}, "a_s": {"name": "a_s", "group": "final list", "definition": "sort(a)", "description": "

Sorted list.

Option 3 for the list. Ensures there is only one mode (2) while still randomising the data.

", "templateType": "anything", "can_override": false}, "modetimes": {"name": "modetimes", "group": "final list", "definition": "map(\nlen(filter(x=j,x,a)),\nj, 0..8)", "description": "

The vector of number of times of each value in the data.

", "templateType": "anything", "can_override": false}, "mode1": {"name": "mode1", "group": "final list", "definition": "mode[0]", "description": "

Mode as a value.

", "templateType": "anything", "can_override": false}, "mode": {"name": "mode", "group": "final list", "definition": "mode(a)", "description": "

Mode as a vector.

", "templateType": "anything", "can_override": false}, "a": {"name": "a", "group": "final list", "definition": "if(len(modea1) = 1, a1, if(len(modea2) = 1, a2, a3))", "description": "

The final list.

", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["modea1", "modea2", "a1", "a2", "a3"], "variable_groups": [{"name": "final list", "variables": ["a", "a_s", "mode", "mode1", "modetimes"]}], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Find the mode.

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This question provides a list of data to the student. They are asked to find the \"median\".

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

A random sample of 20 residents from Newcastle were asked about the number of times they went to see a play at the theatre last year.

Here is the list of their answers:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\var{a[0]}$	$\\var{a[1]}$	$\\var{a[2]}$	$\\var{a[3]}$	$\\var{a[4]}$	$\\var{a[5]}$	$\\var{a[6]}$	$\\var{a[7]}$	$\\var{a[8]}$	$\\var{a[9]}$
$\\var{a[10]}$	$\\var{a[11]}$	$\\var{a[12]}$	$\\var{a[13]}$	$\\var{a[14]}$	$\\var{a[15]}$	$\\var{a[16]}$	$\\var{a[17]}$	$\\var{a[18]}$	$\\var{a[19]}$

", "advice": "

The median is the middle value. We need to sort the list in order:

\\[ \\var{a_s[0]}, \\quad \\var{a_s[1]}, \\quad \\var{a_s[2]}, \\quad \\var{a_s[3]}, \\quad \\var{a_s[4]}, \\quad \\var{a_s[5]}, \\quad \\var{a_s[6]}, \\quad \\var{a_s[7]}, \\quad \\var{a_s[8]}, \\quad \\var{a_s[9]}, \\quad \\var{a_s[10]}, \\quad \\var{a_s[11]}, \\quad \\var{a_s[12]}, \\quad \\var{a_s[13]}, \\quad \\var{a_s[14]}, \\quad \\var{a_s[15]}, \\quad \\var{a_s[16]}, \\quad \\var{a_s[17]}, \\quad \\var{a_s[18]}, \\quad \\var{a_s[19]} \\]

There is an even number of responses, so there are two numbers in the middle (10^th and 11^th place). To find the median, we need to find the mean of these two numbers $\\var{a_s[9]}$ and $\\var{a_s[10]}$:

\\begin{align}
\\frac{\\var{a_s[9]} + \\var{a_s[10]}}{2} &= \\frac{\\var{a_s[9] + a_s[10]}}{2} \\\\
&= \\var{median} \\text{.}
\\end{align}

Use this link to find some resources which will help you revise this topic.

Option 2 for the list. Only used if there is only one mode and option 1 was not used.

", "templateType": "anything", "can_override": false}, "modea1": {"name": "modea1", "group": "Ungrouped variables", "definition": "mode(a1)", "description": "", "templateType": "anything", "can_override": false}, "median": {"name": "median", "group": "final list", "definition": "median(a)", "description": "", "templateType": "anything", "can_override": false}, "a1": {"name": "a1", "group": "Ungrouped variables", "definition": "repeat(random(0..8), 20)", "description": "

Option 1 for the list. Only used if there is only one mode.

", "templateType": "anything", "can_override": false}, "a_s": {"name": "a_s", "group": "final list", "definition": "sort(a)", "description": "

Sorted list.

Option 3 for the list. Ensures there is only one mode (2) while still randomising the data.

", "templateType": "anything", "can_override": false}, "mean": {"name": "mean", "group": "final list", "definition": "mean(a)", "description": "", "templateType": "anything", "can_override": false}, "modetimes": {"name": "modetimes", "group": "final list", "definition": "map(\nlen(filter(x=j,x,a)),\nj, 0..8)", "description": "

The vector of number of times of each value in the data.

Mode as a value.

", "templateType": "anything", "can_override": false}, "mode": {"name": "mode", "group": "final list", "definition": "mode(a)", "description": "

Mode as a vector.

", "templateType": "anything", "can_override": false}, "a": {"name": "a", "group": "final list", "definition": "if(len(modea1) = 1, a1, if(len(modea2) = 1, a2, a3))", "description": "

The final list.

", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["modea1", "modea2", "a1", "a2", "a3"], "variable_groups": [{"name": "final list", "variables": ["a", "a_s", "mean", "median", "mode", "mode1", "range", "modetimes"]}], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Find the median.

", "minValue": "median", "maxValue": "median", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "SM05 Calculate the mean (Frequency table) - With a calculator", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/496/"}, {"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Marta Emmett", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/11961/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}, {"name": "Megan Oliver", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23526/"}], "tags": [], "metadata": {"description": "", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Using a calculator, work out the mean for each of these frequency tables, give your answers to two decimal places.

", "advice": "

To calculate the mean shoe size we need to add a third column to our table where we multiply the grade by the number (frequency) of students who are that shoe size.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

Shoe Size	Frequency	*Shoe Size Frequency**
3	{p1}	3 * {p1} = {3*p1}
4	{p2}	4 * {p2} = {4*p2}
5	{p3}	5 * {p3} = {5*p3}
6	{p4}	6 * {p4} = {6*p4}
7	{p5}	7 * {p5} = {7*p5}
8	{p6}	8 * {p6} = {8*p6}

Now we find the total sum of that third column:

$\\var{3*p1} + \\var{4*p2} + \\var{5*p3} + \\var{6*p4} + \\var{7*p5} + \\var{8*p6} = \\var{3*p1 + 4*p2 + 5*p3 + 6*p4 + 7*p5 + 8*p6}.$

To find the mean shoe size we must divide this total by the total number of students:

$\\frac{\\var{3*p1 + 4*p2 + 5*p3 + 6*p4 + 7*p5 + 8*p6}}{\\var{p1}+\\var{p2}+\\var{p3}+\\var{p4}+\\var{p5}+\\var{p6}} = \\frac{\\var{3*p1 + 4*p2 + 5*p3 + 6*p4 + 7*p5 + 8*p6}}{\\var{sum1}} = \\var{mean1a}.$

The question asks us for our answer to two decimal places so the last thing we need to do is round. Hence, the mean is $\\var{mean1}$.

You can use this same method to answer the other parts of this question!

Use this link to find some resources which will help you revise this topic.

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The mean - part a

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Frequency 1 - part a

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Frequency 2 - part a

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Frequency 3 - part a

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Frequency 4 - part a

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Frequency 5 - part a

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Frequency 6 - part a

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Mean rounded to two decimal places - part a

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Sum of frequencies - part a

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Sum of frequencies - part b

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Frequency 1 - part b

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Frequency 2 - part b

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Frequency 3 - part b

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Frequency 4 - part b

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The mean - part b

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Mean rounded to two decimal places - part b

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Frequency 1 - part c

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Frequency 2 - part c

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Frequency 3 - part c

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Frequency 4 - part c

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Frequency 5 - part c

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Frequency 6 - part c

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Mean rounded to two decimal places - part c

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Sum of frequencies - part c

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The mean - part c

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The table below shows the distribution of shoe sizes amongst {sum1} students at a school in Sheffield.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

Shoe size	Frequency
3	{p1}
4	{p2}
5	{p3}
6	{p4}
7	{p5}
8	{p6}

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The table below shows the distribution of number of pets that {sum2} randomly asked people have.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

Number of Pets	Frequency
0	{p1b}
1	{p2b}
2	{p3b}
3	{p4b}

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The table below shows the distribution of the number of drinks per order at a coffee shop in Manchester.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

Number of Drinks	Frequency
0	{p1c}
1	{p2c}
2	{p3c}
3	{p4c}
4	{p5c}
5	{p6c}

", "minValue": "mean3", "maxValue": "mean3", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "SM06 Choosing the appropriate average", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Lauren Desoysa", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21504/"}, {"name": "Michael Pan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23528/"}], "tags": [], "metadata": {"description": "

This question asks the student to choose the appropriate measure of average and spread for a data with outliers.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Which of the following measures would you choose if you were dealing with data which includes outliers? Select one measure of average and one measure of spread.

", "advice": "

The median is a more appropriate measure of average when your data contains outliers because outliers do not affect the median.

The interquartile range is the best measure of variability for skewed distributions or data sets with outliers. Because it’s based on values that come from the middle half of the distribution, it’s unlikely to be influenced by outliers.

Use this link to find some resources which will help you revise this topic.

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Identifying measures of spread or location (average)

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Match each of the following with what they measure.

", "advice": "

The mean is a measure of location or central tendancy. It is calcuated by summing all of the data values and dividing by the number of values.

The median is a measure of location or central tendancy. It is the middle value of an ordered data set.

The inter-quartile range is a measure of spread. The interquartile range is the difference between upper and lower quartiles.The lower quartile, or first quartile (Q1), is the value under which 25% of data points are found when they are arranged in increasing order. The upper quartile, or third quartile (Q3), is the value under which 75% of data points are found when arranged in increasing order. The inter-quartile range therefore gives us an idea of the middle 50% of the ordered data set.

The standard deviation is a measure of spread. It measures the dispersion of a data set relative to its mean.

The variance is a measure spread because it is the square of the standard deviation.

A p-value the probability that a particular statistical measure, such as the mean or standard deviation, of an assumed probability distribution will be greater than or equal to (or less than or equal to in some instances) observed results. A p-value is used to determine statistical significance, not measures of spread or location.

Use this link to find some resources which will help you revise this topic.

", "rulesets": {"std": ["all", "fractionNumbers"]}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": [], "variable_groups": [], "functions": {}, "preamble": {"js": "function dragpoint_board() {\n var scope = question.scope;\n\n JXG.Options.text.display = 'internal';\n \n var yo0 = scope.variables.yo0.value;\n var yo1 = scope.variables.yo1.value;\n var yo2 = scope.variables.yo2.value;\n var yo3 = scope.variables.yo3.value;\n var yo4 = scope.variables.yo4.value;\n var yo5 = scope.variables.yo5.value;\n var yo6 = scope.variables.yo6.value;\n var yo7 = scope.variables.yo7.value; \n var yo8 = scope.variables.yo8.value;\n var yo9 = scope.variables.yo9.value; \n \n var div = Numbas.extensions.jsxgraph.makeBoard('550px','550px',{boundingBox:[-0.8,82,16,-8], axis:false, grid:true});\n \n $(question.display.html).find('#dragpoint').append(div);\n \n var board = div.board;\n \nboard.suspendUpdate(); \n\n \n var dataArr = [yo0,yo5,0,yo1,yo6,0,yo2,yo7,0,yo3,yo8,0,yo4,yo9]; \n \n var xaxis = board.create('axis', [[0, 0], [12, 0]], {withLabel: true, name: \"Bank\", label: {offset: [250,-30]}});\n \n xaxis.removeAllTicks(); \n \n board.create('axis', [[0, 0], [0, 10]], {hideTicks:true, withLabel: false, name: \"\", label: {offset: [-110,300]}});\n \n var pop0 = board.create('point', [1.5,0],{name:'Morgan',fixed:true,size:0,color:'black',face:'diamond', label:{offset:[-20,-8]}});\n var pop1 = board.create('point',[4.5,0],{name:'Strome',fixed:true,size:0,color:'black',face:'diamond', label:{offset:[-20,-8]}});\n var pop2 = board.create('point',[7.5,0],{name:'Bentley',fixed:true,size:0,color:'black', face:'diamond', label:{offset:[-15,-8]}});\n var pop3 = board.create('point',[10.5,0],{name:'Sand',fixed:true,size:0,color:'black', face:'diamond', label:{offset:[-15,-8]}});\n var pop4 = board.create('point',[13.5,0],{name:'Karchen',fixed:true,size:0,color:'black', face:'diamond', label:{offset:[-15,-8]}});\n\n var leg1 = board.create('point',[12,75],{name:'last year',fixed:true,size:6,color:'#DA2228', face:'square', label:{offset:[9,0]}});\n var leg2 = board.create('point',[12,72],{name:'this year',fixed:true,size:6,color:'#6F1B75', face:'square', label:{offset:[9,0]}});\n \n \n// var chart = board.createElement('chart', dataArr, \n // {chartStyle:'bar', fillOpacity:1, width:1,\n // colorArray:['#8E1B77','#8E1B77','Red','Red','blue','red','blue','red','red','blue', 'red','blue','red','red'], shadow:false});\n \n//var chart = board.createElement('chart', dataArr, \n // {chartStyle:'bar', width:1,fillOpacity:1, fillColor:'red', shadow:false}); \n \n \n var a = board.create('chart', [[1,2,3],[yo0,yo5,0]], {chartStyle:'bar',colors:['#DA2228','#6F1B75','#6F1B75'],width:1,fillOpacity:1});\n var b = board.create('chart', [[4,5,6],[yo1,yo6,0]], {chartStyle:'bar',width:1,colors:['#DA2228','#6F1B75','#6F1B75'],fillOpacity:1});\n var c = board.create('chart', [[7,8,9],[yo2,yo7,0]], {chartStyle:'bar',width:1,colors:['#DA2228','#6F1B75','#6F1B75'],fillOpacity:1});\n var d = board.create('chart', [[10,11,12],[yo3,yo8,0]], {chartStyle:'bar',width:1,colors:['#DA2228','#6F1B75','#6F1B75'],fillOpacity:1});\n var e = board.create('chart', [[13,14],[yo4,yo9]], {chartStyle:'bar',width:1,colors:['#DA2228','#6F1B75'],fillOpacity:1});\n \n board.unsuspendUpdate();\n \n var txt1 = board.create('text',[-0.3,30, 'Investment \u00a3(m)'], {fontColor:'black', fontSize:14, rotate:90});\n \n // var txt = board.create('text',[0.5,75, 'Investment (m)'], {fontSize:14, rotate:90});\n \n // var txt1 = board.create('text',[8,76, 'red bars represent 2010'], {fontColor:'red', fontSize:14, rotate:90});\n \n // var txt2 = board.create('text',[8,73, 'blue bars represents 2011'], {fontSize:14, rotate:90});\n\n // var myColors = new Array('red', 'blue', 'white','red', 'blue', 'white','red', 'blue', 'white','red', 'blue', 'white','red', 'blue');\n \n \n \n //board.unsuspendUpdate();\n\n // Rotate text around the lower left corner (-2,-1) by 30 degrees.\n // var tRot = board.create('transform', [90.0*Math.PI/180.0, -1,40], {type:'rotate'}); \n // tRot.bindTo(txt);\n // board.update();\n\n \n//var chart2 = board.createElement('chart', dataArr, {chartStyle:'line,point'});\n//chart2[0].setProperty('strokeColor:black','strokeWidth:2','shadow:true');\n//for(var i=0; i<11;i++) {\n // chart2[1][i].setProperty({strokeColor:'black',fillColor:'white',face:'[]', size:4, strokeWidth:2});\n//}\n//board.unsuspendUpdate(); \n \n //board.unsuspendUpdate();\n\n}\n\nquestion.signals.on('HTMLAttached',function() {\n dragpoint_board();\n});", "css": "table#values th {\n background: none;\n text-align: center;\n}"}, "parts": [{"type": "m_n_x", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minMarks": 0, "maxMarks": 0, "minAnswers": 0, "maxAnswers": 0, "shuffleChoices": true, "shuffleAnswers": true, "displayType": "radiogroup", "warningType": "none", "showCellAnswerState": true, "markingMethod": "sum ticked cells", "choices": ["Variance", "Mean", "Median", "Inter-quartile range", "P-value", "Standard deviation"], "matrix": [["1", 0, 0], [0, "1", 0], [0, "1", 0], ["1", 0, 0], [0, 0, "1"], ["1", 0, 0]], "layout": {"type": "all", "expression": ""}, "answers": ["Measure of Spread", "Measure of location (average)", "Neither measure of location nor measure of spread"]}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "SM08 Calculate Range (Decimal)", "extensions": ["stats"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Chris Graham", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/369/"}, {"name": "Stanislav Duris", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1590/"}, {"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Lauren Desoysa", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21504/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}, {"name": "Megan Oliver", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23526/"}, {"name": "Michael Pan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23528/"}], "tags": ["mean", "measures of average and spread", "median", "mode", "range", "taxonomy"], "metadata": {"description": "

This question provides a list of data to the student. They are asked to find the \"range\".

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Here is a list of 20 random numbers:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\var{a[0]}$	$\\var{a[1]}$	$\\var{a[2]}$	$\\var{a[3]}$	$\\var{a[4]}$	$\\var{a[5]}$	$\\var{a[6]}$	$\\var{a[7]}$	$\\var{a[8]}$	$\\var{a[9]}$
$\\var{a[10]}$	$\\var{a[11]}$	$\\var{a[12]}$	$\\var{a[13]}$	$\\var{a[14]}$	$\\var{a[15]}$	$\\var{a[16]}$	$\\var{a[17]}$	$\\var{a[18]}$	$\\var{a[19]}$

", "advice": "

Range is the difference between the highest and the lowest value in the data.

To find this, we subtract the lowest value from the highest value:

\\[ \\var{max(a)} - \\var{min(a)} = \\var{range} \\text{.}\\]

Use this link to find some resources which will help you revise this topic.

Option 1 for the list. Only used if there is only one mode.

", "templateType": "anything", "can_override": false}, "a_s": {"name": "a_s", "group": "final list", "definition": "sort(a)", "description": "

Sorted list.

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Find the range.

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Here are the ages of 7 people:

{a1}, {a2}, {a3}, {a4}, {a5}, {a6}, {a7}.

", "advice": "

First we should order our list of ages from smallest to largest:

$\\var{list[0]}, \\var{list[1]}, \\var{list[2]}, \\var{list[3]}, \\var{list[4]}, \\var{list[5]}, \\var{list[6]}$.

To find the range of ages we must subtract the smallest age from the largest age,

$\\var{max(list)} - \\var{min(list)} = \\var{range}$.

Hence, the range is $\\var{range}$.

To calculate the interquartile range we subtract the lower quartile from the upper quartile.

To calculate the lower quartile:

Since we have an odd number of ages we had one to the total and divide by 4,

$\\frac{7+1}{4} = \\frac{8}{4} = 2.$

So we are looking for the $2^{nd}$ value in our list which is $\\var{lq}$.

Hence the lower quartile is $\\var{lq}$.

To find the upper quartile:

We still add one to the number of ages because this number is odd and then we find $75$% of it,

$\\frac{3\\times(7+1)}{4} = \\frac{3\\times8}{4} = \\frac{24}{4} = 6.$

So we are looking for the $6^{th}$ value in our list which is $\\var{uq}$.

Hence the lower quartile is $\\var{uq}$.

Now we can calculate the interquartile range by subtracting the lower quartile from the upper quartile,

$\\var{uq}-\\var{lq} = \\var{iqr}.$

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Age 1

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Age 3

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Age 4

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Age 6

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Age 7

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Ages in ascending order

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Upper quartile

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Work out the range of ages.

Work out the interquartile range of the ages.

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Tina needs to take a train to work.

The table below shows information about how long the train is delayed for on each day.

The policy of the train company is that if your train is delayed for longer than 30 minutes, you will recieve a 50% discount on your train ticket.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

Delay Time	Frequency
0 < t $\\leq$ 10	{f1}
10 < t $\\leq$ 20	{f2}
20 < t $\\leq$ 30	{f3}
30 < t $\\leq$ 40	{f4}
40 < t $\\leq$ 50	{f5}

", "advice": "

To find the estimated mean we must work out the midpoint of the delay time intervals and multiply this by the frequency to create the $fx$ column on the table.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

Delay Time	Frequency	Midpoint	Frequency $\\times$ Midpoint, $fx$
0 < t $\\leq$ 10	{f1}	5	= 5$\\times\\var{f1} = \\var{f1m}$
10 < t $\\leq$ 20	{f2}	15	= 15$\\times\\var{f2} = \\var{f2m}$
20 < t $\\leq$ 30	{f3}	25	= 25$\\times\\var{f3} = \\var{f3m}$
30 < t $\\leq$ 40	{f4}	35	= 35$\\times\\var{f4} = \\var{f4m}$
40 < t $\\leq$ 50	{f5}	45	= 45$\\times\\var{f5} = \\var{f5m}$

Now to find the mean we calculate the sum of the $fx$ column and divide it by the sum of the frequencies:

Estimated Mean = $\\frac{\\var{f1m}+\\var{f2m}+\\var{f3m}+\\var{f4m}+\\var{f5m}}{5+15+25+35+45} = \\frac{\\var{freqmid}}{\\var{sum}} = \\var{meana}.$

Hence the estimated mean delay time is $\\var{mean}$ to two decimal places.

To work out what percentage of delays that were over 30 minutes we first need to calculate the number of days where the train was delayed for over 30 minutes.

The number of delays that over 30 minutes is $\\var{f4}+\\var{f5} = \\var{f4+f5}$.

We divide this by the total number of deliveries to find the fraction of delays that were over 30 minutes and then multiply this number by 100 to find the percentage:

$\\frac{\\var{f4+f5}}{\\var{f1}+\\var{f2}+\\var{f3}+\\var{f4}+\\var{f5}} \\times 100\\% = \\frac{\\var{f4+f5}}{\\var{sum}} \\times 100\\% = \\var{percentagea} \\times 100\\%$.

Hence, the percentage of delays that were over 30 minutes is $\\var{percentage}\\%$ to two decimal places.

The percentage of delays that take over 40 minutes is:

$\\frac{\\var{f5}}{\\var{f1}+\\var{f2}+\\var{f3}+\\var{f4}+\\var{f5}} \\times 100\\% = \\frac{\\var{f5}}{\\var{sum}} \\times 100\\% = \\var{percent40} \\times 100\\%$, to one decimal place.

So, only $\\var{percent40}\\%$ of the time, delays were longer than 40 minutes so the manager would probably wanto the policy changed to 40 minutes.

Use this link to find some resources which will help you revise this topic.

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Percentage over 30 mins

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Using a calculator, calculate an estimate for the mean delay time. Give your answer to two decimal places.

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What percentage of delays were over 30 minutes? Give your answer to one decimal place.

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The train company propose that they should only offer the 50% discount if the train is delayed for over 40 minutes.

If you were the manager of the train company would you agree? Type yes or no into the box and think about the reasons for your answer.

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This frequence table stores data about the lengths of some plants:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

Length ($x$ cm)	Frequency
10 < $x$ $\\leq$ 20	{f1}
20 < $x$ $\\leq$ 30	{f2}
30 < $x$ $\\leq$ 40	{f3}
40 < $x$ $\\leq$ 50	{f4}

", "advice": "

To find the median we must find the middle value. To do this we sum the frequencies and divide the result by 2:

$\\frac{\\var{f1}+\\var{f2}+\\var{f3}+\\var{f4}}{2} = \\frac{\\var{total}}{2} = \\var{middle1}$.

Hence we must find value number $\\var{middle1}$, we can do this by totalling the cumulative frequency in the table below.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

\n Length $x$ cm \n	Frequency	Cumulative Frequency
1. 10 < $x$ $\\leq$ 20	{f1}	$\\var{f1}$
2. 20 < $x$ $\\leq$ 30	{f2}	$\\var{f1}+\\var{f2}=\\var{cf2}$
3. 30 < $x$ $\\leq$ 40	{f3}	$\\var{f1}+\\var{f2} +\\var{f3} =\\var{cf3}$
4. 40 < $x$ $\\leq$ 50	{f4}	$\\var{f1}+\\var{f2} +\\var{f3} + \\var{f4} =\\var{total}$

You can see value number $\\var{middle1}$ lies in interval number $\\var{interval1}$ hence this is the median class interval for the time taken.

We can estimate the median by using interpolation:

$$
\\text{Estimate of median} = \\text{class start value} + \\frac{\\text{position in class}}{\\text{frequency in class}}\\times \\text{class width}
$$

In this case this is $\\var{lower1}+\\frac{\\var{middle1}-\\var{cf11}}{\\var{fint1}}\\times 10 =\\var{lower1}+\\var{interpolation1}=\\var{median1}.$

(Note: that sometimes it is convenient to use $(\\var{total} + 1)/2 = \\var{middle2}$ as the way to work out the median position. The distinction does not really matter as this is an estimated value and often the decision is made based on which is the most convenient to calculate). This question has been written so that both answers will be marked correctly.

To calculate the $i^{th}$ quartile ($Q_i$) we must use the following formula:

$Q_i = l + \\frac{\\frac{iN}{4}-F}{f}\\times h,$

where:

$l$ = lower limit of the interval in which $Q_i$ lies;

$N$ = Total number of observations;

$F$ = Cumulative frequency of class previous to the $i^{th}$ quartile class;

$f$ = Frequency of $i^{th}$ quartile class.

Since we want to calculate the lower quartile, $i=1$.

If we calculate $\\frac{iN}{4} = \\frac{\\var{total}}{4} = \\var{quarter}$, we can see in the table from part a) that the lower quartile will be in interval number $\\var{intervallq}$. This means that $F = \\var{fprev}$ and $f = \\var{frequency}$.

Hence,

$Q_1 = \\var{lowerlq} + \\frac{\\var{quarter} - \\var{Fprev}}{\\var{frequency}} \\times 10 = \\var{lowerlq} + \\frac{\\var{quarter-Fprev}}{\\var{frequency}} \\times 10 = \\var{lowerlq} + \\var{((quarter-Fprev)/frequency)*10} = \\var{lq}.$

So, The lower quartile is $\\var{lqround}$ to two decimal places.

To calculate the upper quartile we use the same formula as in part b) but this time $i=3$. So, $\\frac{iN}{4} = \\frac{\\var{3*total}}{4} = \\var{3*quarter}$, we can see in the table from part a) that the lower quartile will be in interval number $\\var{intervaluq}$. This means that $F = \\var{ufprev}$ and $f = \\var{ufrequency}$.

Hence,

$Q_3 = \\var{upperlq} + \\frac{\\var{3*quarter} - \\var{uFprev}}{\\var{ufrequency}} \\times 10 = \\var{upperlq} + \\frac{\\var{3*quarter-uFprev}}{\\var{ufrequency}} \\times 10 = \\var{upperlq} + \\var{((3*quarter-uFprev)/ufrequency)*10} = \\var{uq}.$

So, The lower quartile is $\\var{uqround}$ to two decimal places.

To calculate the interquartile range we subtract the lower quartile from the upper quartile.

Hence,

$IQR = Q_3 - Q_1 = \\var{uq}-\\var{lq}=\\var{iqr}$.

So the interquartile range rounded to two decimal places is $\\var{iqrround}$.

Use this link to find some resources which will help you revise this topic.

Frequency 1

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Frequency 2

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Frequency 4

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Lower bound of interval which median sits

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Interval which lower quartile sits in

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Frequency of class which lower quartile sits in

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Lower quartile rounded to 2 decimal places

", "templateType": "anything", "can_override": false}, "uqround": {"name": "uqround", "group": "Part c", "definition": "precround(uq,2)", "description": "

Upper quartile rounded to 2 decimal places

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Upper quartile

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cumulative frequency of class previous to the one which the upper quartile sits in

", "templateType": "anything", "can_override": false}, "ufrequency": {"name": "ufrequency", "group": "Part c", "definition": "if(intervaluq=1,f1,if(intervaluq=2,f2,if(intervaluq=3,f3,if(intervaluq=4,f4,0))))", "description": "

Frequency of class upper quartile sits in

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class which upper quartile sits in

", "templateType": "anything", "can_override": false}, "iqr": {"name": "iqr", "group": "Part d", "definition": "rational(uq-lq)", "description": "

Interquartile range

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width of class which lower quartile sits in

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width of class upper quartile sits in

", "templateType": "anything", "can_override": false}, "iqrround": {"name": "iqrround", "group": "Part d", "definition": "precround(iqr,2)", "description": "

Round interquartle range to two decimal places

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Calculate the lower quartile, give your answer to 2 decimal places.

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Calculate an estimate for the upper quartile using linear interpolation, give your answer to two decimal places.

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Estimate the interquartile range, give your answer to two decimal places.

", "minValue": "iqrround", "maxValue": "iqrround", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "SN01 Correlation", "extensions": ["jsxgraph"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Richard Miles", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/882/"}, {"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Lauren Desoysa", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21504/"}, {"name": "Upuli Wickramaarachchi", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23527/"}], "tags": [], "metadata": {"description": "

Tests understanding of scatter plots and related concepts.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

The scatter plot below shows the relationship between an employee’s height in centimetres and how long it takes them to walk to work in minutes.

\n\n\n\n\n\n\n\n\n\n\n\n

time (mins)	{drawgraph()}
	height (cm)

", "advice": "

The graph shows that there is a positive correlation between a person's height and how long it takes them to walk to work.

A postive correlation is a relationship between two variables where both variables move in the same diection.

This tells us that as a person's height increases, the time it takes to walk to work increases.

Use this link to find some resources which will help you revise this topic

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p6y

", "templateType": "anything", "can_override": false}, "p6x": {"name": "p6x", "group": "Points", "definition": "random(176..185 except p4x)", "description": "", "templateType": "anything", "can_override": false}, "p4y": {"name": "p4y", "group": "Points", "definition": "random(36..45)", "description": "", "templateType": "anything", "can_override": false}, "p4x": {"name": "p4x", "group": "Points", "definition": "random(176..185)", "description": "", "templateType": "anything", "can_override": false}, "p2y": {"name": "p2y", "group": "Points", "definition": "random(16..25)", "description": "", "templateType": "anything", "can_override": false}, "p2x": {"name": "p2x", "group": "Points", "definition": "random(156..165)", "description": "", "templateType": "anything", "can_override": false}, "sumxx": {"name": "sumxx", "group": "Regression variables", "definition": "p1x^2+p2x^2+p3x^2+p4x^2+p5x^2+p6x^2", "description": "", "templateType": "anything", "can_override": false}, "sumxy": {"name": "sumxy", "group": "Regression variables", "definition": "p1x*p1y+p2x*p2y+p3x*p3y+p4x*p4y+p5x*p5y+p6x*p6y", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": [], "variable_groups": [{"name": "Graph Limits", "variables": ["minx", "maxx", "miny", "maxy"]}, {"name": "Points", "variables": ["p1x", "p1y", "p2x", "p2y", "p3x", "p3y", "p4x", "p4y", "p5x", "p5y", "p6x", "p6y"]}, {"name": "Calculation variables", "variables": ["tallest", "timemax", "timemin", "timediff"]}, {"name": "Regression variables", "variables": ["sumx", "sumy", "sumxy", "sumxx", "slope", "yintercept", "regy1", "regy2", "roundedslope"]}], "functions": {"drawgraph": {"parameters": [], "type": "html", "language": "javascript", "definition": " var miny = Numbas.jme.unwrapValue(scope.variables.miny);\n var maxy = Numbas.jme.unwrapValue(scope.variables.maxy);\n var minx = Numbas.jme.unwrapValue(scope.variables.minx);\n var maxx = Numbas.jme.unwrapValue(scope.variables.maxx);\n var regy1 = Numbas.jme.unwrapValue(scope.variables.regy1);\n var regy2 = Numbas.jme.unwrapValue(scope.variables.regy2);\n\n var p1x = Numbas.jme.unwrapValue(scope.variables.p1x);\n var p1y = Numbas.jme.unwrapValue(scope.variables.p1y);\n var p2x = Numbas.jme.unwrapValue(scope.variables.p2x);\n var p2y= Numbas.jme.unwrapValue(scope.variables.p2y);\n var p3x = Numbas.jme.unwrapValue(scope.variables.p3x);\n var p3y= Numbas.jme.unwrapValue(scope.variables.p3y);\n var p4x = Numbas.jme.unwrapValue(scope.variables.p4x);\n var p4y= Numbas.jme.unwrapValue(scope.variables.p4y);\n var p5x = Numbas.jme.unwrapValue(scope.variables.p5x);\n var p5y= Numbas.jme.unwrapValue(scope.variables.p5y);\n var p6x = Numbas.jme.unwrapValue(scope.variables.p6x);\n var p6y= Numbas.jme.unwrapValue(scope.variables.p6y);\n \n var div = Numbas.extensions.jsxgraph.makeBoard('400px','400px',\n {boundingBox:[minx,maxy,maxx,miny],\n axis:false,\n showNavigation:false,\n grid:true});\n var brd = div.board; \n var xaxis=brd.createElement('axis', [[minx,0],[maxx,0]]);\n var yaxis=brd.createElement('axis', [[minx+5,miny],[minx+5,maxy]]);\n var li1=brd.create('line',[[minx,regy1],[maxx,regy2]],{fixed:true,withLabel:false});\n var pt1=brd.create('point',[p1x,p1y],{visible:true,withLabel:false}); \n var pt2=brd.create('point',[p2x,p2y],{visible:true,withLabel:false}); \n var pt3=brd.create('point',[p3x,p3y],{visible:true,withLabel:false}); \n var pt4=brd.create('point',[p4x,p4y],{visible:true,withLabel:false}); \n var pt5=brd.create('point',[p5x,p5y],{visible:true,withLabel:false}); \n var pt6=brd.create('point',[p6x,p6y],{visible:true,withLabel:false}); \nreturn div;\n "}}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "1_n_2", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Mark the statement that best describes what this scatter plot shows.

", "minMarks": 0, "maxMarks": 0, "shuffleChoices": true, "displayType": "radiogroup", "displayColumns": 0, "showCellAnswerState": true, "choices": ["

In general, there is a positive correlation between a person's height and how long it takes them to walk to work.

", "

In general, there is a negative correlation between a person's height and how long it takes them to walk to work.

", "

In general, there is a no correlation between a person's height and how long it takes them to walk to work.

"], "matrix": ["1", 0, 0], "distractors": ["", "", ""]}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "SP01 Probability - \"sample space\"", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Ruth Hand", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3228/"}, {"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}], "tags": [], "metadata": {"description": "

Calculate probability of selecting coloured counters from a bag.

", "licence": "None specified"}, "statement": "

A bag contains:

$\\var{srn}$ small, red tokens,
$\\var{sbn}$ small, blue tokens,
$\\var{brn}$ large, red tokens, and
$\\var{bbn}$ large, blue tokens.

", "advice": "

part a)

A probability is a fraction. You can give your answer as a fraction, decimal or percentage as these are all equivalent.

The formula for probability is:

\\[ P(A) = \\frac{\\text{number of possibilities for A}}{\\text{number of total possible outcomes}} \\]

For this question the total possible outcomes are $\\var{srn}+\\var{sbn}+\\var{brn}+\\var{bbn} = \\var{total}$.

Therefore

\\[ P(\\text{A large red token}) = \\frac{\\var{brn}}{\\var{total}} = \\var[fractionnumbers]{brn/total}\\]

part b)

For this question we need to know the total number of small tokens, i.e. $\\var{srn}+\\var{sbn} = \\var{srn+sbn}$.

Therefore

\\[ P(\\text{A small token}) = \\frac{\\var{srn+sbn}}{\\var{total}} = \\var[fractionnumbers]{(srn+sbn)/total}\\]

Use this link to find some resources which will help you revise this topic.

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You take a token at random.

What is the probability that it is a large, red token?

Give your answer as a fraction, or a decimal correct to 2dp.

You take a token at random.

What is the probability that it is a small token?

Give your answer as a fraction, or a decimal correct to 2dp.

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Predicting the probability of an unbiased coin landing on heads based on the results of previous throws.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "", "advice": "

When we flip an unbiased coin there are two possible events that we could measure: the coin lands on heads or the coin lands on tails.

Each toss of the coin is independent; if we flip a coin once and it lands on heads then the next time we flip the coin it is still equally likely to land on either heads or tails.

It doesn't matter what the coin landed on previously as this outcome does not affect the outcome of the next flip of the coin.

Even when we flip an unbiased coin $\\var{no_flips}$ times and it lands on heads each time; the next time we flip the coin, it is still equally likely to land on either heads or tails.

So the probability that the coin lands on heads the next time that the coin is flipped is still $\\displaystyle\\frac{1}{2}$.

Use this link to find some resources which will help you revise this topic.

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Number of flips of the coin

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An unbiased coin is flipped $\\var{no_flips}$ times. Given that the coin landed on heads each time, what is the probability of the coin landing on heads the next time it is flipped?

", "minValue": "1/2", "maxValue": "1/2", "correctAnswerFraction": true, "allowFractions": true, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "SP03 Calculating probability from a Contingency Table", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Megan Oliver", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23526/"}], "tags": [], "metadata": {"description": "", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Complete the two way table.

", "advice": "\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

	Football	Rugby	Tennis	Total
Year 7	$\\var{f7}$	$\\var{r7}$	$\\var{total7}-(\\var{f7}+\\var{r7}) = \\var{total7}-\\var{f7+r7} = \\var{t7}$	$\\var{total7}$
Year 8	$\\var{totalf}-\\var{f7} = \\var{f8}$	$\\var{r8}$	$\\var{t8}$	$\\var{total}-\\var{total7} = \\var{total8}$
Total	$\\var{totalf}$	$\\var{r7}+\\var{r8}=\\var{totalr}$	$\\var{t7}+\\var{t8}=\\var{totalt}$	$\\var{total}$

Use this link to find some resources which will help you revise this topic.

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Thank you for completing the Skills Audit for Maths and Stats. Hopefully it has been useful in directing you to resources and services that can support your studies. The Skills Audit for Maths and Stats will remain open to you throughout the academic year and you can always revisit it again later.

For any further information or questions please contact mash@sheffield.ac.uk

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