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
(round to four decimal places as needed).

Explanation:

Step1: Calculate the z - score for part b

The formula for the z - score of a sample mean is \(z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\), where \(\bar{x} = 144\), \(\mu=185\), \(\sigma = 27\), and \(n = 25\).

$$ z=\frac{144 - 185}{\frac{27}{\sqrt{25}}}=\frac{- 41}{\frac{27}{5}}=\frac{-41}{5.4}\approx - 7.59 $$

Step2: Find the probability

We want to find \(P(\bar{X}>144)\). Since \(P(\bar{X}>144)=1 - P(\bar{X}\leq144)\), and for \(z=-7.59\), using the standard normal distribution table (or a calculator with a normal - distribution function), \(P(Z\leq - 7.59)\approx0\)

Answer:

The probability that a sample of 25 randomly selected adult males has a mean weight greater than 144 lb is \(1.0000\) (rounded to four decimal places).