Question

An elevator has a stated maximum capacity of 12 people or 2004 pounds. If 12 people have weights with a mean greater than (2004/12) = 167 pounds, the capacity will be exceeded. Assume that weights of men are normally distributed with a mean of 182.9 pounds and a standard deviation of 40.8 pounds. Show your work and round your answers to FOUR decimal places.a. Compute the probability that a randomly selected man will have a weight greater than 167 pounds.b. Compute the probability that 12 randomly selected men will have a mean weight that is greater than 167 pounds.c. Does the elevator appear to have the correct weight limit? Why or why not?

Answer 1

Answer:

a) 0.6517; b) 0.9115; c) No

Step-by-step explanation:

For part a, we will use the formula for a z score of an individual:

$z=\frac{X-\mu}{\sigma}\\\\=\frac{167-182.9}{40.8}\\\\=\frac{-15.9}{40.8}\approx -0.39$

Using a z table, we see that the area under the curve to the left of this value is 0.3483.  However, we want the probability greater than this, which is the area to the right of this value under the curve; this means we subtract from 1:

1-0.3483 = 0.6517

For part b, we will use the formula for a z score of the mean of a sample:

$z=\frac{\bar{X}-\mu}{\sigma \div \sqrt{n}}\\\\=\frac{167-182.9}{40.8\div \sqrt{12}}\\\\=\frac{-15.9}{40.8\div 3.4641}\\\\=\frac{-15.9}{11.778}\approx -1.35$

Using a z table, we see that the area under the curve to the left of this value is 0.0885.  This means the area under the curve to the right of this value is

1-0.0885 = 0.9115

For part c,

The fact that the probability that any 12 men on the elevator will have a mean weight greater than 167, putting their total weight above 2004 pounds, is 91% means the elevator does not have the appropriate limit.  There is a high chance the maximum will be exceeded.