205k views
1 vote
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?

User Arrel
by
7.7k points

1 Answer

1 vote

Answer:

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.

User Artyer
by
8.7k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.