Question

an elevator has a placard stating that the maximum capacity is 3700 lb—26 passengers. so, 26 adult male passengers can have a mean weight of up to 3700/26 = 142 pounds. assume that weights of males are normally distributed with a mean of 180 lb and a standard deviation of 26 lb.
a. find the probability that 1 randomly selected adult male has a weight greater than 142 lb.
b. find the probability that a sample of 26 randomly selected adult males has a mean weight greater than 142 lb.
c. what do you conclude about the safety of this elevator?

Explanation:

Step1: Calculate z - score for part a

For a single - value \(x\) from a normal distribution \(N(\mu,\sigma)\), the z - score is \(z=\frac{x - \mu}{\sigma}\). Here, \(\mu = 180\), \(\sigma=26\), and \(x = 142\).

$$z=\frac{142 - 180}{26}=\frac{- 38}{26}\approx - 1.46$$

Using the standard normal table (or a calculator with a normal - distribution function, e.g., \(P(Z\lt z)\) for \(z=-1.46\)), \(P(X\gt142)=1 - P(X\lt142)\). From the standard normal table, \(P(Z\lt - 1.46)=0.0721\). So \(P(X\gt142)=1 - 0.0721 = 0.9279\)

Step2: Calculate z - score for part b

For the sampling distribution of the sample mean \(\bar{X}\) with sample size \(n = 26\), the mean of the sampling distribution is \(\mu_{\bar{X}}=\mu = 180\) and the standard deviation is \(\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}=\frac{26}{\sqrt{26}}\approx5.099\).
The z - score is \(z=\frac{\bar{x}-\mu_{\bar{X}}}{\sigma_{\bar{X}}}\), where \(\bar{x} = 142\)

$$z=\frac{142 - 180}{5.099}=\frac{-38}{5.099}\approx - 7.45$$

Using the standard normal table (or a calculator with a normal - distribution function), \(P(\bar{X}\gt142)=1 - P(\bar{X}\lt142)\). Since for \(z=-7.45\), \(P(Z\lt - 7.45)\approx0\), so \(P(\bar{X}\gt142)\approx1\)

Step3: Conclusion for part c

The elevator's placard is based on the assumption that the mean weight of 26 passengers is \(142\) lb. But from part b, the probability that the mean weight of 26 randomly selected adult males is greater than \(142\) lb is approximately \(1\). This means that, on average, groups of 26 adult males will exceed the weight limit. So the elevator is not safe for its intended use (carrying 26 adult males) as the probability of exceeding the weight limit is very high.

Answer:

a. The probability that 1 randomly selected adult male has a weight greater than \(142\) lb is approximately \(0.9279\)
b. The probability that a sample of \(26\) randomly selected adult males has a mean weight greater than \(142\) lb is approximately \(1\)
c. The elevator is not safe for its intended use (carrying 26 adult males) as the probability of the mean weight of 26 adult males exceeding \(142\) lb is very high.