Question

an elevator has a placard stating that the maximum capacity is 3600 lb - 25 passengers. so, 25 adult male passengers can have a mean weight of up to 3600/25 = 144 pounds. assume that weights of males are normally distributed with a mean of 185 lb and a standard deviation of 27 lb.
a. find the probability that 1 randomly selected adult male has a weight greater than 144 lb.
b. find the probability that a sample of 25 randomly selected adult males has a mean weight greater than 144 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 144 lb is 0.9357.
(round to four decimal places as needed.)
b. the probability that a sample of 25 randomly selected adult males has a mean weight greater than 144 lb is 1.0000.
(round to four decimal places as needed.)
c. does this elevator appear to be safe?
a. yes, because there is a good chance that 25 randomly selected people will not exceed the elevator capacity
b. no, because there is a good chance that 25 randomly selected adult male passengers will exceed the elevator capacity
c. no, because 25 randomly selected people will never be under the weight limit.
d. yes, because 25 randomly selected adult male passengers will always be under the weight limit.

Explanation:

Step1: Analyze part a

Given a normal distribution with mean \(\mu = 185\) and standard deviation \(\sigma=27\). For a single value \(x = 144\), the z - score is \(z=\frac{x-\mu}{\sigma}=\frac{144 - 185}{27}=\frac{-41}{27}\approx - 1.52\). Using the standard normal table or a calculator, \(P(X>144)=1 - P(X\leq144)\). Since \(P(Z\leq - 1.52)\) from the standard normal table is \(0.0643\), then \(P(X>144)=1 - 0.0643 = 0.9357\).

Step2: Analyze part b

For a sample of size \(n = 25\), the standard error of the mean is \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}=\frac{27}{\sqrt{25}}=\frac{27}{5} = 5.4\). The z - score for \(\bar{x}=144\) is \(z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}=\frac{144 - 185}{5.4}=\frac{-41}{5.4}\approx - 7.59\). Using the standard normal table or a calculator, \(P(\bar{X}>144)=1 - P(\bar{X}\leq144)\). Since \(P(Z\leq - 7.59)\approx0\), then \(P(\bar{X}>144)=1-0 = 1.0000\).

Step3: Analyze part c

Since the probability that the mean weight of 25 adult males is greater than 144 lb (the maximum mean weight for capacity) is very high (close to 1), there is a good chance that 25 randomly selected adult male passengers will exceed the elevator capacity.

Answer:

a. \(0.9357\)
b. \(1.0000\)
c. B. No, because there is a good chance that 25 randomly selected adult male passengers will exceed the elevator capacity.