Question

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?

Short answer

The probability of the elevator being overloaded is 0.7019.

The elevator does not appear to be safe, because we have 70.19% chance of the elevator to be overloaded.

Step-by-step solution

Given information

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.

Elevator capacity based on 25% safety factor

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.

Step 3:Describe the distribution of weights

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

Find the z-score

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.

Determine probability using technology

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.

Describe if the elevator is safe

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.