Chapter 2, Problem 2CRE

(a)

To determine

To find: the $z$ -score of P's height.

Answer to Problem 2CRE

The z- score is 1.20

Explanation of Solution

Given:

A z-score of 0 corresponds to a height of 170 centimeters

A z-score of 1 corresponds to a height of 177.5cm

P is 15 years old and $179\text{cm}$ tall.

Calculation:

To find the mean, use the fact that a z-score of 0 corresponds to a height of 170.

Then the resultis shown below:

  $\begin{array}{l}==\frac{x-\mu }{\sigma }\hfill \\ \Rightarrow 0-\frac{170-\mu }{\sigma }\hfill \\ \Rightarrow 0-170-\mu \hfill \\ \Rightarrow y=170\text{cm}\hfill \end{array}$

To find it he standard deviation, use the fact that a 2 -score of 1 corresponds to a height of 177.5

Thon the result is shown below.

  $\begin{array}{l}z=\frac{x-\mu }{\sigma }\hfill \\ \Rightarrow 1=\frac{177.5-170}{\sigma }\hfill \\ \Rightarrow 1=\frac{7.5}{\sigma }\hfill \\ \Rightarrow \sigma =7.5\text{cm}\hfill \end{array}$

Thus, the mean and standard deviation of the height distribution for 15 year old males are 170 cm and 76 cm respectively

The standardized height low Pate is shown below

  $z=\frac{x-\mu }{\sigma }=\frac{179-170}{7.5}=1.20$

P is somewhat taller than average for his age. His height is 1.20 standard deviations abovethe average male height for his age.

Conclusion:

Thus, the z- score is 1.20

(b)

To determine

To explain:the given statement.

Answer to Problem 2CRE

Here, $85\%$ of boys Ps age are shorter than P's height

Explanation of Solution

Given:

P's height puts him at the 85 th percentile among 15 -year-old males.

Calculation:

SinceP’s height puts him at the ${85}^{\text{th }}$ percentile among 15 -year-old males, $85\%$ of boys Ps age are shorter than P's height

Conclusion:

Therefore, $85\%$ of boys Ps age are shorter than P's height