Question

an elevator has a placard stating that the maximum capacity is 4000 lb - 27 passengers. so, 27 adult male passengers can have a mean weight of up to 4000/27 = 148 pounds. assume that weights of males are normally distributed with a mean of 188 lb and a standard deviation of 20 lb. a. find the probability that 1 randomly - selected adult male has a weight greater than 148 lb. b. find the probability that a sample of 27 randomly selected adult males has a mean weight greater than 148 lb. c. what do you conclude about the safety of this elevator? a. the probability that 1 randomly selected adult male has a weight greater than 148 lb is (round to four decimal places as needed.)

Explanation:

Step1: Calculate z - score for single - male case

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $x = 148$, $\mu=188$, and $\sigma = 20$.
$z=\frac{148 - 188}{20}=\frac{-40}{20}=-2$

Step2: Find the probability for single - male case

We want $P(X>148)$. Since $P(X>148)=1 - P(X\leq148)$, and from the standard normal table, $P(Z\leq - 2)=0.0228$. So $P(X>148)=1 - 0.0228 = 0.9772$.

Step3: Calculate z - score for sample - mean case

The formula for the z - score of the sample mean is $z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$, where $\bar{x}=148$, $\mu = 188$, $\sigma=20$, and $n = 27$.
$z=\frac{148-188}{\frac{20}{\sqrt{27}}}=\frac{-40}{\frac{20}{\sqrt{27}}}\approx\frac{-40}{3.849}\approx - 10.4$. From the standard - normal table, $P(Z\leq - 10.4)\approx0$. So $P(\bar{X}>148)=1 - P(\bar{X}\leq148)=1-0 = 1$.

Step4: Conclusion about elevator safety

The probability that a single male's weight exceeds the maximum average weight allowed per passenger is relatively high ($0.9772$). The probability that the mean weight of 27 male passengers exceeds the maximum average weight allowed per passenger is almost $1$. This indicates that the elevator is not safe for 27 adult male passengers on average.

Answer:

a. $0.9772$
b. $1$
c. The elevator is not safe for 27 adult male passengers on average.