The army reports that the distribution of head circumference among male soldiers is approximately Normal with mean of $22.8$inches and standard deviation $1.1$inches.

(a) A male soldier whose head circumference is $23.9$-inches would be at what percentile? Show your method clearly.

(b) The army’s helmet supplier regularly stocks helmets that fit male soldiers with head circumferences between $20$and $26$inches. Anyone with a head circumference outside that interval requires a customized helmet order. What percent of male soldiers require custom helmets? Show your work, including a well-labeled sketch of a Normal curve.

(c) Find the interquartile range for the distribution of head circumference among male soldiers. Show your method clearly.

Given in the question that,$\mu =22.8\text{inches and}\sigma =1.1\text{inches .}$

According to definitions, the $z$-score for a soldier with a head circumference of $23.9$inches is as follows:

localid="1649926606875" $$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{23.9-22.8}{1.1} \\ =1 \end{gathered}$$

We find that the proportion related to $z=1$scores is $0.8413$using ordinary normal tables.

This signifies that the head circumference of this soldier is in the ${84}^{th}$percentile.

This is depicted in the graph below.

Given in the question that, the army’s helmet supplier regularly stocks helmets that fit male soldiers with head circumferences between $20$and$26$ inches.

Anyone with a head circumference outside of that range, according to the report, requires a tailored helmet order.

According to definitions, the $z$score for a soldier with a $20$-inch head circumference is as follows:

localid="1649926624263" $$\begin{gathered} z=\frac{20-22.8}{1.1} \\ =-2.55 \end{gathered}$$

Let's find the $z-$score that corresponds to the soldier whose head circumference is $26$inches :

localid="1649926628791" $$\begin{gathered} z=\frac{26-22.8}{1.1} \\ =2.91 \end{gathered}$$

Using standard normal tables, we observe the proportion corresponding to the score$z=-2.55$as $0.0054$and the proportion corresponding to the score $z=2.91$as$0.9982.$Thus the area above $2.91$is $1-0.9982=0.0018$. This is shown in the below graph.

$0.9982.$

This means, that the area in both tails is:

localid="1649402393448" $0.0054+0.0018=0.0072\%$

The army reports that the distribution of head circumference among male soldiers is approximately Normal with mean of $22.8$inches and standard deviation $1.1$inches.

The quartiles of a Standard Normal distribution are $-0.67$and $0.67$. To find the quartiles of the head circumference distribution we solve the following equations for $x$

localid="1649926639197" $$\begin{gathered} -0.67=\frac{x-228}{1.1} \\ x=22.063 \end{gathered}$$

localid="1649926644659" $$\begin{gathered} 0.67=\frac{x-228}{1.1} \\ x=23.537 \end{gathered}$$

This shows that$Q_i=22.063\text{and}{Q}_i=23.537\text{.}$

So

localid="1649926656947" $$\begin{gathered} IQR=Q_i-Q_i \\ =23.537-22.063 \\ =1.174 \end{gathered}$$$=23.537-22.063$