Assume that the weights of individuals are independent and normally distributed with a mean of 160 pounds and a standard deviation of 30 pounds. Suppose that 25 people squeeze into an elevator that is designed to hold 4300 pounds.

(a) What is the probability that the load (total weight) exceeds the design limit?

(b) What design limit is exceeded by 25 occupants with probability 0.0001?

Step 1 of 2:

Given the individual’s weight is normally distributed with a mean is 160 and a standard deviation is 30.

Suppose that 25 people squeeze into an elevator with a design limit 4300.

Let the random ‘T’ denotes the total weight of 25 individuals.

The expected value of T is

E(T) = 25160

E(T) = 4000

Then the variance of the total weight of 25 people in the elevator is

V(T) =

V(T) =

V(T) =

V(T) = 150

Our goal is:

a). We need to find the probability that the load exceeds the design limit.

b). We need to find what design limit is exceeded by 25 occupants with probability 0.0001.

a). Now we need to find the probability that the load exceeds the design limit.

Then Z formula is

Z =

Now we have to calculate P(T>4300).

P(T>4300) = P

P(T>4300) = P

P(T>4300) = P

P(Z>2) = 1-P

Using area under the normal curve table,

P(Z>2) = 1-0.9772

P(Z>2) = 0.0228

Therefore, the probability that the load exceeds the design limit is 0.0228.