// Numbas version: exam_results_page_options {"name": "Normal Distribution", "extensions": ["eukleides", "stats", "jsxgraph", "permutations", "geogebra", "random_person", "codewords", "polynomials", "chemistry", "quantities"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Normal Distribution", "tags": [], "metadata": {"description": "", "licence": "None specified"}, "statement": "

This test assesses your understanding of the normal distribution.

", "advice": "

Here are some tips on how to set out your work, as well as how to complete the questions. None of the answers given below have been rounded to what the question states. 

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Hint: If your Z-Score is a negative value, or the question is asking for $P(Z>z)$ then ensure you calculate $1 - \\phi(z)$. E.g. $P(Z<-0.32) =1-P(Z<0.32) = 1-0.6255 = 0.3745$ or $P(Z>0.32)=1-P(Z<0.32) =1-0.6255 = 0.3745$ or $P(Z>-0.32)=1-(1-P(Z<0.32)) = 1 - (1 - 0.6255) =1 - 0.3745 = 0.6255$.

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Question 1

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1a: $X\\sim N(\\var{Mu1}, \\var{SD1})$

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(i) $P(X<\\var{SM1}) =P(Z<\\frac{\\bar{X} - \\mu}{\\sigma})= P(Z<\\frac{\\var{SM1}-\\var{Mu1}}{\\var{SD1}})= P(Z<\\var{Z1}) = \\var{P1}$

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(ii) $P(X<\\var{SM2})=P(Z<\\frac{\\bar{X} - \\mu}{\\sigma})= P(Z<\\frac{\\var{SM2}-\\var{Mu1}}{\\var{SD1}})= P(Z<\\var{Z2}) = \\var{P2}$

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(iii) $P(X<\\var{SM3}) =P(Z<\\frac{\\bar{X} - \\mu}{\\sigma})= P(Z<\\frac{\\var{SM3}-\\var{Mu1}}{\\var{SD1}})= P(Z<\\var{Z3}) = \\var{P3}$

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1b: $X\\sim N(\\var{Mu2}, \\var{SD2})$

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(i) $P(X>\\var{SM4}) =P(Z>\\frac{\\bar{X} - \\mu}{\\sigma})= P(Z>\\frac{\\var{SM4}-\\var{Mu2}}{\\var{SD2}})= P(Z>\\var{Z4}) = 1-P(Z<\\var{Z4})=\\var{P4}$

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(ii) $P(X>\\var{SM5}) =P(Z>\\frac{\\bar{X} - \\mu}{\\sigma})= P(Z>\\frac{\\var{SM5}-\\var{Mu2}}{\\var{SD2}})= P(Z>\\var{Z5}) = 1-P(Z<\\var{Z5})=\\var{P5}$

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(iii) $P(X>\\var{SM6}) =P(Z>\\frac{\\bar{X} - \\mu}{\\sigma})= P(Z>\\frac{\\var{SM6}-\\var{Mu2}}{\\var{SD2}})= P(Z>\\var{Z6}) = 1-P(Z<\\var{Z6})= \\var{P6}$

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(iv) $P(X>\\var{SM7}) =P(Z>\\frac{\\bar{X} - \\mu}{\\sigma})= P(Z>\\frac{\\var{SM7}-\\var{Mu2}}{\\var{SD2}})= P(Z>\\var{Z7}) = 1-P(Z<\\var{Z7})=\\var{P7}$

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Question 2:

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2a: $X\\sim N(\\var{Mu3}, \\var{SD3})$

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(i) $P(X<\\var{SM8}) =P(Z<\\frac{\\bar{X} - \\mu}{\\sigma})= P(Z<\\frac{\\var{SM8}-\\var{Mu3}}{\\var{SD3}})= P(Z<\\var{Z8}) = \\var{P8}$

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(ii) $P(X<\\var{SM9}) =P(Z<\\frac{\\bar{X} - \\mu}{\\sigma})= P(Z<\\frac{\\var{SM9}-\\var{Mu3}}{\\var{SD3}})= P(Z<\\var{Z9}) = \\var{P9}$

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(iii) $P(\\var{SM10Lower}<X<\\var{SM10Upper}) =P(Z<\\frac{\\bar{X}_{upper} - \\mu}{\\sigma})-P(Z<\\frac{\\bar{X}_{lower} - \\mu}{\\sigma})= P(Z<\\frac{\\var{SM10Upper}-\\var{Mu3}}{\\var{SD3}})-P(Z<\\frac{\\var{SM10Lower}-\\var{Mu3}}{\\var{SD3}}) = P(Z<\\var{Z10Upper}) - P(Z<\\var{Z10Lower})=\\var{P10}$

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2b: $X\\sim N(\\var{Mu4}, \\var{SD4})$

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(i) $P(X>\\var{SM11}) = 1 - P(X<\\var{SM11}) =1 - P(Z<\\frac{\\bar{X} - \\mu}{\\sigma})= 1 - P(Z<\\frac{\\var{SM11}-\\var{Mu4}}{\\var{SD4}})= 1 - P(Z<\\var{Z11}) = 1 - \\var{1-P11} = \\var{P11}$

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(ii) $Z=\\frac{X - \\mu}{\\sigma} \\rightarrow X = Z \\times \\sigma + \\mu  = \\var{-Z12} \\times \\var{SD4} + \\var{Mu4} = \\var{SM12}$

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[Note: We use the negative version of the Z-score, as we want to know the weight exceeded by 95% of the bags, so the weight should be less than the mean of $\\var{Mu4}$kg].

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Question 3:

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3a: $X\\sim N(\\var{Mu5}, \\var{SD5}^2)$

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(i) Since the normal distribution is repersented by $X \\sim N(\\mu,\\sigma^2) \\rightarrow X\\sim N(\\var{Mu5}, \\var{SD5}^2)$

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(ii) $P(X<\\var{SM13}) =P(Z<\\frac{\\bar{X} - \\mu}{\\sigma})= P(Z<\\frac{\\var{SM13}-\\var{Mu5}}{\\var{SD5}})= P(Z<\\var{Z13}) = \\var{P13}$

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(iii) $P(X=\\var{SM13})=P(Z=\\frac{\\bar{X} - \\mu}{\\sigma/ \\sqrt{n}}) = P(Z= \\frac{\\var{SM13}-\\var{Mu5}}{\\var{SD5}/\\sqrt{\\var{n}}}) = P(Z=\\var{Z14}) = \\var{P14}$

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(a) Given that $X\\sim N(\\var{Mu1},\\var{SD1}^2)$. Find,

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(i) $P(X<\\var{SM1})$

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Answer:[[0]]

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(ii) $P(X<\\var{SM2})$

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Answer:[[1]]

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(iii) $P(X<\\var{SM3})$

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Answer:[[2]]

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(b) Given that $X\\sim N(\\var{Mu2},\\var{SD2}^2)$. Find,

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(i) $P(X>\\var{SM4})$ 

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Answer:[[3]]

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(ii) $P(X>\\var{SM5})$

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Answer:[[4]]

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(iii) $P(X>\\var{SM6})$

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Answer:[[5]]

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(iv) $P(X>\\var{SM7})$

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Answer:[[6]]

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"useCustomName": true, "customName": "Question 2", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

(a) The salary of a maths teacher is normally distributed with a mean of £{Mu3} and a standard deviation of £{SD3}.

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If a maths teacher is chosen at random, what is probability that they;

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(i) earn less than £{SM8}?  

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Answer:[[0]]

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(ii) earn more than £{SM9}?

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Answer:[[1]]

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(iii) earn between £{SM10Lower} and £{SM10Upper}?

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Answer:[[2]]

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(b) A packing plant fills bags with cement. The weight $X$kg of a bag of cement can be modelled by a normal distribution with a mean of {Mu4}kg and a standard deviation of {SD4}kg. Find,

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(i) $P(X>\\var{SM11})$

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Answer: [[3]]

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(ii) Find the weight that is exceeded by the {SigLev}% of the bags. 

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Answer: [[4]]

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A brewer has a machine which he claims fills cans of beer to a nominal volume which is set at 440ml and a standard deviation of {SD5}ml. Printed on the side of the can is e440ml. It is therefore an offence for the average contents of all such filled cans to be less than 440ml.

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(i) If $X$ = r.v 'weight of beer in one can', state the parameters of $X$ for the distribution.

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Answer: $X\\sim N$( [[0]], [[1]]$^2$)

\n

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(ii) What is the probability of a randomly selected can weighing less than 435ml?

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Answer: [[2]]

\n

\n

(iii) What is the probability of a random sample of $\\var{n}$ cans having a mean of 435ml?

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Answer: [[3]]

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