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1. Converting to $\\operatorname{N}(0,1)$

\n

$\\simplify[all,!collectNumbers]{P(X < {lower}) = P(Z < ({lower} -{m}) / {s}) = P(Z < {lower-m}/{s}) = 1 -P(Z < {m-lower}/{s})} = 1 -\\var{p} = \\var{precround(1 -p,4)}$ to 4 decimal places.

\n

2. Converting to $\\operatorname{N}(0,1)$

\n

$\\simplify[all,!collectNumbers]{P(X > {upper}) = P(Z > ({upper} -{m}) / {s}) = P(Z > {upper-m}/{s}) = 1 -P(Z < {upper-m}/{s})} = 1-\\var{p1} = \\var{precround(1 -p1,4)}$ to 4 decimal places.

", "rulesets": {}, "parts": [{"prompt": "

Find the probability that in a particular week the {amount} is less than {lower} {units1}:

\n

Probability = [[0]] (to 4 decimal places)

\n

Find the probability that in a particular week the {amount} is greater than {upper} {units1}:

\n

Probability = [[1]] (to 4 decimal places)

", "marks": 0, "gaps": [{"precisionType": "dp", "precisionMessage": "You have not given your answer to the correct precision.", "allowFractions": false, "scripts": {}, "precision": "4", "maxValue": "prob1+tol", "minValue": "prob1-tol", "strictPrecision": false, "correctAnswerFraction": false, "showCorrectAnswer": true, "precisionPartialCredit": 0, "marks": 1, "type": "numberentry", "showPrecisionHint": false}, {"precisionType": "dp", "precisionMessage": "You have not given your answer to the correct precision.", "allowFractions": false, "scripts": {}, "precision": "4", "maxValue": "prob2+tol", "minValue": "prob2-tol", "strictPrecision": true, "correctAnswerFraction": false, "showCorrectAnswer": true, "precisionPartialCredit": 0, "marks": 1, "type": "numberentry", "showPrecisionHint": false}], "showCorrectAnswer": true, "scripts": {}, "type": "gapfill"}], "statement": "

The {amount}, $X$, of {stuff} is normally distributed with mean {m}{units1} and standard deviation {s}{units1}.

\n

i.e. \\[X \\sim \\operatorname{N}(\\var{m},\\var{s}^2)\\]

\n

 

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1/1/2012:

\n \t\t \t\t

Can be configured to other applications using the string variables suppplied. Included tag sc.

\n \t\t \n \t\t", "description": "

Given a random variable $X$  normally distributed as $\\operatorname{N}(m,\\sigma^2)$ find probabilities $P(X \\gt a),\\; a \\gt m;\\;\\;P(X \\lt b),\\;b \\lt m$.

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a)

\n

Converting to $\\operatorname{N}(0,1)$

\n

\\[\\begin{eqnarray*}P(X \\lt \\var{lower})&=&P\\left(Z \\lt \\frac{\\var{lower}-\\var{m}}{\\var{s}}\\right)\\\\&=&P(Z\\lt \\var{z})=1-P(Z \\lt \\var{-z})=1-\\var{normalcdf(-z,0,1)}\\\\&\\approx&\\var{prop/100}\\end{eqnarray*}\\]

\n

Hence the proportion is $\\var{prop}$% to 1 decimal place.

\n

b)

\n

The proportion of {mw} more than $\\var{upper1}$ {units1} tall is given by finding the probability:

\n

\\begin{align}
\\operatorname{P}(X>\\var{upper1} &= 1 - \\operatorname{P}(X<\\var{upper1})\\\\
&= 1 - \\operatorname{P}(Z<\\dfrac{\\var{upper1}-\\var{m}}{\\var{s}})\\\\
&= 1 - \\var{w2}\\\\
&\\approx \\var{prop1/100}
\\end{align}

\n

Hence the proportion is $\\var{prop1}$% to 1 decimal place.

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What proportion of such {mw} are less than $\\var{lower}$ {units1} tall?

\n

Proportion = [[0]]% (to 1 decimal place).

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What proportion of such {mw} are more than $\\var{upper1}$ {units1} tall?

\n

Proportion = [[0]]% (to 1 decimal place).

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Suppose that the heights, in {units1}, of $\\var{age}$-year-old {mw} are normally distributed with mean $\\var{m}$ {units1} and standard deviation $\\var{s}$ {units1}.

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7/02/2013:

\n

Finished first draft. Tag sc included.

", "description": "

Given a normal distribution $X \\sim N(m,\\sigma^2)$ find $P(X \\lt a),\\; a \\lt m$ and find $P(X\\gt b), \\; b\\gt m$.

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Converting to $\\operatorname{N}(0,1)$:

\n

a)

\n

\\[\\begin{eqnarray*}P(\\var{lower1} \\lt X \\lt \\var{upper1})&=&P(X \\lt \\var{upper1})- P(X \\lt \\var{lower1})\\\\&=&P\\left(Z \\lt \\frac{\\var{upper1}-\\var{m}}{\\var{s}}\\right)-P\\left(Z \\lt \\frac{\\var{lower1}-\\var{m}}{\\var{s}}\\right)\\\\&=&P(Z \\lt \\var{zu})-P(Z \\lt \\var{zl})=\\var{precround(u,4)}-\\var{precround(l,4)},\\\\&=&\\var{p}\\end{eqnarray*}\\]  (to 3 decimal places.)

\n

b)

\n

We need to find $x$ such that:

\n

\\[\\begin{eqnarray*}P(X \\lt x) &\\le& \\var{percentile/100}\\\\ \\Rightarrow P\\left(Z \\le \\frac{x-\\var{m}}{\\var{s}}\\right) &\\le& \\var{percentile/100}\\\\ \\Rightarrow \\frac{x-\\var{m}}{\\var{s}} &\\le& \\Phi^{-1}(\\var{percentile/100})=\\var{precround(normalinv(percentile/100,0,1),4)}\\\\ \\Rightarrow x &\\le&\\var{precround(normalinv(percentile/100,0,1),4)}\\times\\var{s}+\\var{m}=\\var{perX}\\end{eqnarray*}\\] to the nearest whole number.

\n

Hence $X=\\var{perX}$ gives the $\\var{percentile}$% percentile.

\n

 

", "rulesets": {}, "parts": [{"prompt": "

What is the probability that {athing} has {this} between $\\var{lower1}$ and $\\var{upper1}$?

\n

Probability = [[0]] (to 3 decimal places).

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What value of $X$ is equivalent to the $\\var{percentile}$% percentile? (i.e. $\\var{percentile}$% of medium sized factories in Europe that have a workcount below this value)

\n

$\\var{percentile}$% percentile = [[0]] (round to the nearest whole number).

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Let $X$ be the {something}. It is believed that $X$ has a normal distribution, with mean $\\var{m}$ and standard deviation $\\var{s}$. That is, $X \\sim \\operatorname{N}(\\var{m},\\var{s^2})$.

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7/02/2013:

\n

Added scenario tag sc. Have to use $\\operatorname{normalinv}(p/100,m,\\sigma)$ for a given percentile $p$% where $p$ ranges from 40 to 99. Might be better to choose standard values for p.

", "description": "

Given normal distribution $\\operatorname{N}(m,\\sigma^2)$ find $P(a \\lt X \\lt b),\\; a \\lt m,\\;b \\gt m$ and also find the value of $X$ corresponding to a given percentile $p$%. 

", "licence": "None specified"}, "type": "question", "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}], "pickingStrategy": "all-ordered"}], "allQuestions": true, "pickQuestions": 0, "navigation": {"onleave": {"action": "none", "message": ""}, "browse": true, "reverse": true, "preventleave": true, "showfrontpage": true, "allowregen": true, "showresultspage": "never"}, "timing": {"timeout": {"action": "none", "message": ""}, "timedwarning": {"action": "none", "message": ""}, "allowPause": true}, "metadata": {"notes": "", "description": "

A resource for Business students to practise calculating Normal probabilities and finding $Z$-values. 

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