Chapter 4, Problem 2PT1

To determine

To find: The correct option for standard deviation

Answer to Problem 2PT1

Option e is correct.

Explanation of Solution

Given information:

Height of a child $=40\text{ inches}$

Percentile in which the child’s height is $=90\%$

Mean of heights of children of similar age $=38\text{ inches}$

The heights are normally distributed.

Formula used:

Z score

  $\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ \text{where}\\ \mu \text{ is the mean}\\ \sigma \text{ is the standard deviation}\end{array}$

Calculation:

A height at the $90th$ percentile would mean that $90\%$ of the heights are less than or equal to $40\text{ inches}$ .

  $\begin{array}{l}P\left(X\le 40\right)=0.90\\ P\left(z<\frac{40-38}{\sigma }\right)=0.90\end{array}$

Let X be the heights of children.

So, $X~N\left(38,{\sigma }^2\right)$

Reading from z tables,

  $P\left(z<1.2816\right)=0.90$

Comparing the values

  $\begin{array}{l}1.2816=\frac{40-38}{\sigma }\\ \sigma =\frac{40-38}{1.2816}\\ \sigma =1.56\end{array}$

Therefore, standard deviation of heights of all children of similar age is $1.56\text{ inches}$