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The current seat width of airlines is 17 in.

Men鈥檚 hip breadths have a mean of 14.4 in. and a standard deviation of 1.0 in., and are known to be normally distributed.

Let X be the random variable for men鈥檚 hip breadths.

$$\begin{gathered} X~N\left(渭,{蟽}^2\right) \\ ~N\left(14.4,{1.0}^2\right) \end{gathered}$$

a.

The hip breadth of a randomly selected man is 17 in.

The associated z-score is as follows:

$$\begin{gathered} z=\frac{x-渭}{蟽} \\ =\frac{17-14.4}{1} \\ =2.6 \end{gathered}$$

The probability that man has a hip breadth greater than 17 in. is given as follows:

$$\begin{gathered} P\left(X>17\right)=P\left(Z>2.6\right) \\ =1-P\left(Z<2.6\right)...\left(1\right) \end{gathered}$$

The cumulative probability is obtained from cell intersection in the standard normal table with row 2.6 and column 0.00, which is 0.9953.

Substitute the value into equation (1).

$$\begin{gathered} P\left(X>17\right)=1-0.9953 \\ =0.0047 \end{gathered}$$

Thus, the probability that a man has a hip breadth larger than 17 in. is 0.0047.

b.

As per the central limit theorem, the sample size of seats in Boeing 737 is larger than 30 (122); therefore, the sample mean distribution would follow a normal distribution as shown below:

Define$\stackrel{炉}{X}$ as the distribution of sample means of size 122 (n).

Then,

$$\begin{gathered} \stackrel{炉}{X}~N\left({渭}_{\stackrel{炉}{X}},{{蟽}_{\stackrel{炉}{X}}}^2\right) \\ ~N\left(14.4,{\left(\frac{1}{\sqrt{122}}\right)}^2\right) \\ ~N\left(14.4,{0.0905}^2\right) \end{gathered}$$

The z-score corresponding to mean hip breadth of 17 in. is as follows:

$$\begin{gathered} z=\frac{\stackrel{炉}{x}-{渭}_{\stackrel{炉}{X}}}{{蟽}_{\stackrel{炉}{X}}} \\ =\frac{17-14.4}{0.0905} \\ =28.72 \end{gathered}$$

The probability that the mean breadth of 122 men is greater than 17 in. is expressed in the following manner:

$$\begin{gathered} P\left(\stackrel{炉}{X}>17\right)=P\left(Z>28.72\right) \\ =1-P\left(Z<28.72\right)...\left(1\right) \end{gathered}$$

From the standard normal table, the cumulative probability for values larger than 3.55 is0.9999.

$P\left(\stackrel{炉}{X}>17\right)=0.0001$

Thus, the probability that the mean breadth of 122 men is greater than 17 in. is almost 0.0001.

c.

As the seats are occupied by the individual, the result in part (a) is most relevant to be considered in designing seats. The mean of the men鈥檚 hip breadth involves a group of individuals.