Example 2 referred to an elevator with a maximum capacity of 4000 lb. When rating elevators, it is common to use a 25% safety factor, so the elevator should actuallybe able to carry a load that is 25% greater than the stated limit. The maximum capacity of 4000 lb becomes 5000 lb after it is increased by 25%, so 27 adult male passengers can have a mean weight of up to 185 lb. If the elevator is loaded with 27 adult male passengers, find the probability that it is overloaded because they have a mean weight greater than 185 lb. (As in Example 2, assume that weights of males are normally distributed with a mean of 189 lb and a standard deviation of 39 lb.) Does this elevator appear to be safe?

The maximumcapacity of the elevator is 4000 lb.

The maximum capacity increased to 5000 lb due to 25% safety factor.

Correspondingly, the mean weight of 27 male passengers is desired to be 185 lb.

Consider the maximum capacity 4000lb which increased by 25%.

$$\begin{gathered} \text{Increase}=4000\times 25\% \\ =4000\times \frac{25}{100} \\ =1000 \end{gathered}$$

Thus, the elevator capacity increased to 5000 lb.

Let X be the weight of male passengers in the elevator.

Refer to example 2 for the distribution of weights as normal.

$$\begin{gathered} X~N\left(\mu ,{\sigma }^2\right) \\ ~N\left(189,{39}^2\right) \end{gathered}$$

Define $\overline{X}$ as the sample mean distribution of 27 male passengers.

The distribution of sample mean would follow the normal distribution as the population is normal, with mean and standard deviation as,

$$\begin{gathered} {\mu }_{\overline{X}}=\mu \\ {\sigma }_{\overline{X}}=\frac{\sigma }{\sqrt{n}} \\ =\frac{39}{\sqrt{27}} \\ =7.5055 \end{gathered}$$

Thus, $\overline{X}~N\left(189,{7.5055}^2\right)$ .

The z-score associated with the mean weight of 185 on sample mean distribution,

$$\begin{gathered} z=\frac{\overline{x}-{\mu }_{\overline{X}}}{{\sigma }_{\overline{X}}} \\ =\frac{185-189}{7.5055} \\ =-0.5329 \end{gathered}$$

The probability that elevator is overloaded as the mean weight exceeds 185 lb when the elevator has 27 male passengers is expressed as,

$$\begin{gathered} P\left(\overline{X}>185\right)=P\left(Z>-0.5329\right) \\ =1-P\left(Z<-0.5329\right)...\left(1\right) \end{gathered}$$

Where, the probability is the right tailed area of z-score -0.53, under the standard normal curve.

From the standard normal table, the left tailed probability for z-score -0.53is computed as 0.2981.

Substitute value in equation (1),

$$\begin{gathered} P\left(\overline{X}>185\right)=1-0.2981 \\ =0.7019 \end{gathered}$$

Thus, the probability that the elevator is overloaded when there are 27 adult males in the elevator is 0.7019.

The elevator does not appear to be safe sincein spite of a safety factor of 25%, if the elevator is loaded with 27 adult males, there are 70.19% chances that elevator is overloaded.

As the chances of being overloaded is quite high, the elevator is not safe.