91影视

Dimensions of manhole covers are between 22in to 60in.

Men鈥檚 shoulder width is normally distributed with mean 18.2in and standard deviation 1.0 in.

A manhole being studied has a diameter of 22 in.

Let X be the width of shoulders for the men.

Then,

$$\begin{gathered} X~N\left(渭,{蟽}^2\right) \\ ~N\left(18.2,{1.0}^2\right) \end{gathered}$$

a.

The men with shoulder width less than 22 in. would fit in the manhole.

The z-score associated to 22in on the distribution of X is,

$$\begin{gathered} z=\frac{x-渭}{蟽} \\ z=\frac{22-18.2}{1} \\ =3.8 \end{gathered}$$

The probability that a man would fit the manhole is,

$P\left(X<22\right)=P\left(Z<3.8\right)$

From the standard normal table, the left tailed area of 3.8 is obtained as the area under the curve, left of 3.8 which is 0.9999.

Therefore, 99.99% of men will fit into the manhole.

b.

Let $\stackrel{炉}{X}$ be the sample mean width of shoulders for 36 randomly selected men.

Then,

$\stackrel{炉}{X}~N\left({渭}_{\stackrel{炉}{X}},{{蟽}_{\stackrel{炉}{X}}}^2\right)$

Where,

$$\begin{gathered} {渭}_{\stackrel{炉}{X}}=18.2 \\ {蟽}_{\stackrel{炉}{X}}=\frac{{蟽}_X}{\sqrt{n}} \\ =\frac{1}{\sqrt{36}} \\ =0.1667 \end{gathered}$$

The z-scores for 18.5 in is,

$$\begin{gathered} z=\frac{x-渭}{\frac{蟽}{\sqrt{n}}} \\ =\frac{18.5-18.2}{0.1667} \\ =1.7996 \\ 鈮�1.80 \end{gathered}$$

The probability that the mean shoulder breadth is less than 18.5in. is,

$$\begin{gathered} P\left(\stackrel{炉}{X}<18.5\right)=P\left(Z<1.80\right) \\ =0.9641 \end{gathered}$$

Using the standard normal table, the required probability is 0.9641.

The result suggests that in a group of 36 men, the average breadth for shoulder would be less than 18.5in. for 96.41% of men who work on manholes.

Thus, it suggests that manholes can have a smaller manhole diameter of 18.5in