Question

Suppose that an airline uses a seat width of $16.5\mathrm{in}$. Assume men have hip widths that are normally distributed with a mean of $14.4\mathrm{in}$. and a standard deviation of $1\mathrm{in}$. Find the probability that if an individual man is randomly​ selected, his hip width will be greater than $16.5\mathrm{in}$. ​(Round to four decimal places as​ needed.)

Answer 1

Step-1: Find $z$ score:

Given mean is $\left(\mu \right)$ is$14.4\mathrm{in}$

Standard deviation, $\left(\sigma \right)$ is $1\mathrm{in}$.

Let $X$ be the hip width.

The distribution in the given question is normal distribution.

Therefore, standard normal distribution table $z-\mathrm{table}$ is a table used to know the percentage of values that are below $z$ score in the standard normal distribution.

$z_{\mathrm{score}}=\frac{X-\mu }{\sigma }\dots \left(1\right)$

where $X$ is given value, $\mu$ is mean and $\sigma$ is the standard deviation.

Substitute $X=16.5$, $\mu =14.4$ and $\sigma =1$ in equation $\left(1\right)$

$$\begin{gathered} z_{\mathrm{score}}=\frac{16.5-14.4}{1} \\ \Rightarrow z_{\mathrm{score}}=\frac{2.1}{1} \\ \therefore z_{\mathrm{score}}=2.1 \end{gathered}$$

Step-2: Find probability:

Probability that if an individual man is randomly​ selected, his hip width will be greater than $16.5\mathrm{in}$ is

$$\begin{gathered} P\left(X>16.5\right)=P\left(z>z_{score}\right) \\ \Rightarrow P\left(X>16.5\right)=P\left(z>2.1\right)\left[\because z_{score}=2.1\right] \\ \Rightarrow P\left(X>16.5\right)=1-P\left(z\le 2.1\right)\left[\because P\left(z>a\right)=1-P\left(z\le a\right)\right] \\ \Rightarrow P\left(X>16.5\right)=1-0.9821\left[\because \mathrm{From}\mathrm{standard}\mathrm{normal}\mathrm{z}-\mathrm{table},P\left(z\le 2.1\right)=0.9821\right] \\ \therefore P\left(X>16.5\right)=0.0179 \end{gathered}$$

Hence, probability that if an individual man is randomly​ selected, his hip width will be greater than $16.5\mathrm{in}$ is $0.0179\mathrm{in}.$