Question

An elevator has a placard stating that the maximum capacity is 19201920 lblong dash—1212 passengers.​ So, 1212 adult male passengers can have a mean weight of up to 1920 divided by 12 equals 160 pounds.1920/12=160 pounds. If the elevator is loaded with 1212 adult male​ passengers, find the probability that it is overloaded because they have a mean weight greater than 160160 lb.​ (Assume that weights of males are normally distributed with a mean of 163 lb163 lb and a standard deviation of 29 lb29 lb​.) Does this elevator appear to be​ safe?

Answer 1

This means that there is a 1-0.3594 = 0.6406 = 64.06% probability that the elevator is overloaded. This a good chance that the elevator's limit weight will be exceeded. So, this elevator does not appear to be safe.

Explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by

`Z = (X - \mu)/(\sigma)`

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

In this problem, we have that:

Assume that weights of males are normally distributed with a mean of 163 lb and a standard deviation of 29 lb.

This means that
`\mu = 163, \sigma = 29`.

We have a sample of 12 adults, and we want to calculate the zscore of THE SAMPLE'S AVERAGE so we need to find the standard deviation of the sample. This is

`s = (\sigma)/(√(12)) = 8.37`

Find the probability that it is overloaded because they have a mean weight greater than 160.

This is 1 subtracted by the pvalue of Z when
`X = 160`

`Z = (X - \mu)/(\sigma)`

`Z = (160 - 163)/(8.37)`

`Z = -0.36`

`Z = -0.36` has a pvalue of 0.3594.

This means that there is a 1-0.3594 = 0.6406 = 64.06% probability that the elevator is overloaded. This a good chance that the elevator's limit weight will be exceeded. So, this elevator does not appear to be safe.