Question

If Designing Manholes According to the website www.torchmate.com, “manhole covers must be a minimum of 22 in. in diameter, but can be as much as 60 in. in diameter.” Assume that a manhole is constructed to have a circular opening with a diameter of 22 in. Men have shoulder breadths that are normally distributed with a mean of 18.2 in. and a standard deviationof 1.0 in. (based on data from the National Health and Nutrition Examination Survey).

1. What percentage of men will fit into the manhole?
2. Assume that the Connecticut’s Eversource company employs 36 men who work in manholes. If 36 men are randomly selected, what is the probability that their mean shoulder breadth is less than 18.5 in.? Does this result suggest that money can be saved by making smaller manholes with a diameter of 18.5 in.? Why or why not?

Short answer

a. 99.99% of men would fit the manhole.

b. The manholes can have smaller diameters of 18.5 in as there are 96.41% sample mean manhole diameters that have an average diameter lesser than18.5 in.

Step-by-step solution

Given information

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

Men’s 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.

Describe the random variable

Let X be the width of shoulders for the men.

Then,

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

Compute the probability

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,

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

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.

Discuss the sample mean distribution

b.

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

Then,

$\overline{X}~N\left({\mu }_{\overline{X}},{{\sigma }_{\overline{X}}}^2\right)$

Where,

$$\begin{gathered} {\mu }_{\overline{X}}=18.2 \\ {\sigma }_{\overline{X}}=\frac{{\sigma }_X}{\sqrt{n}} \\ =\frac{1}{\sqrt{36}} \\ =0.1667 \end{gathered}$$

Determine the probabilities

The z-scores for 18.5 in is,

$$\begin{gathered} z=\frac{x-\mu }{\frac{\sigma }{\sqrt{n}}} \\ =\frac{18.5-18.2}{0.1667} \\ =1.7996 \\ \approx 1.80 \end{gathered}$$

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

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

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

Analyze the results

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