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Suppose that the heights, in inches, of 22-year-old women are normally distributed with mean 64.5 inches and standard deviation 5 inches.

a)

What proportion of such women are less than 61.5 inches tall?

Proportion =  Expected answer: % (to 1 decimal place).

This feedback is on your last submitted answer. Submit your changed answer to get updated feedback.

b)

What proportion of such women are more than 71.5 inches tall?

Proportion =  Expected answer: % (to 1 decimal place).

This feedback is on your last submitted answer. Submit your changed answer to get updated feedback.

Advice

a)

Converting to N(0,1)

P(X<61.5)=P(Z<61.5−64.55)=P(Z<−0.6)=1−P(Z<0.6)=1−0.7257468822≈0.274

Hence the proportion is 27.4% to 1 decimal place.

b)

The proportion of women more than 71.5 inches tall is given by finding the probability:

P(X>71.5=1−P(X<71.5)=1−P(Z<71.5−64.55)=1−0.9192433408≈0.081

Hence the proportion is 8.1% to 1 decimal place.
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The electricity consumption, $X$, of a frozen foods warehouse each week in the summer months is normally distributed with mean 750kWh and standard deviation 60kWh.

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

 

Find the probability that in a particular week the electricity consumption is less than 710 kWh:

Probability = Expected answer:  (to 4 decimal places)

Find the probability that in a particular week the electricity consumption is greater than 800 kWh:

Probability = Expected answer:  (to 4 decimal places)

This feedback is on your last submitted answer. Submit your changed answer to get updated feedback.

Advice

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

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

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

$\simplify[all,!collectNumbers]{P(X > {upper}) = P(Z > ({upper} -{m}) / {s}) = P(Z > {upper-m}/{s}) = 1 -P(Z < {upper-m}/{s})} = 1-\var{p1} = \var{precround(1 -p1,4)}$ to 4 decimal places.
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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, $X \sim \operatorname{N}(\var{m},\var{s^2})$.

a)

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

Probability = Expected answer: (to 3 decimal places).

This feedback is on your last submitted answer. Submit your changed answer to get updated feedback.

b)

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)

$\var{percentile}$% percentile = Expected answer: (round to the nearest whole number).

This feedback is on your last submitted answer. Submit your changed answer to get updated feedback.

Advice

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

a)

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

b)

We need to find $x$ such that:

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

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

 
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