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 190 lb and a standard deviation of 32 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?

a. the probability that 1 randomly selected adult male has a weight greater than 142 lb is 0.9332.
(round to four decimal places as needed.)

b. the probability that a sample of 26 randomly selected adult males has a mean weight greater than 142 lb is
(round to four decimal places as needed.)

Explanation:

Step1: Calculate the z - score for the sample mean

The formula for the z - score of a sample mean is \(z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\), where \(\bar{x} = 142\), \(\mu=190\), \(\sigma = 32\), and \(n = 26\).
Substitute the values into the formula:
\(z=\frac{142 - 190}{\frac{32}{\sqrt{26}}}\)
First, calculate the denominator \(\frac{32}{\sqrt{26}}\approx\frac{32}{5.099}\approx6.275\)
Then, calculate the numerator \(142-190=-48\)
So, \(z=\frac{- 48}{6.275}\approx - 7.65\)

Step2: Find the probability

We want to find \(P(\bar{X}>142)\). Since \(P(\bar{X}>142)=1 - P(\bar{X}\leq142)\)
Looking up the z - score of \(z=-7.65\) in the standard normal distribution table (or using a calculator with normal distribution functions), \(P(Z\leq - 7.65)\approx0\)
So, \(P(\bar{X}>142)=1 - 0=1\)

Answer:

\(1.0000\)