Question #175167

An elevator in the hospital has a weight limit of 1650 pounds. Suppose an average weight of

the people who use this elevator is 140 pounds with a standard deviation of 25 pounds.

Assuming that the distribution of weights is approximately normal and a random sample of

11 people are selected, what is the probability that the random sample will exceed the weight

limit?


The formula we are given is z=-μ/(σ/√n)



1
Expert's answer
2021-03-25T18:48:39-0400

Solution:

Let XX be the random variable.

Given, μ=140;σ=25\mu=140 ; \sigma=25

And an average weight for a sample of 11 would cause the total weight to exceed the 1650 pounds weight limit.

x=165011=150{\overline{x}}=\frac{1650}{11}=150

We need to find P(X>150)P(X>150).

Now, z=xμσ=15014025=0.4z=\frac{{\overline{x}}-\mu}{\sigma}=\frac{150-140}{25}=0.4

P(Z>z)=P(Z>0.4)=0.34450.35P(Z>z)=P(Z>0.4)=0.3445 \approx 0.35

Thus, required probability is 0.35.


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