Answer:
There is a 98.26% probability that it is overloaded.
This elevator does not appear to be safe.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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 170 lb, so ![\mu = 170](https://tex.z-dn.net/?f=%5Cmu%20%3D%20170)
They have a standard deviation of standard deviation of 29 lb, so
.
We have a sample of 8 adults, so we have to find the standard deviation of the sample to use in the place of
in the Z score formula.
![s = \frac{\sigma}{8} = \frac{29}{8} = 3.625](https://tex.z-dn.net/?f=s%20%3D%20%5Cfrac%7B%5Csigma%7D%7B8%7D%20%3D%20%5Cfrac%7B29%7D%7B8%7D%20%3D%203.625)
Find the probability that it is overloaded because they have a mean weight greater than 162 lb, so ![X = 162](https://tex.z-dn.net/?f=X%20%3D%20162)
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{162 - 170}{3.625}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B162%20-%20170%7D%7B3.625%7D)
![Z = -2.21](https://tex.z-dn.net/?f=Z%20%3D%20-2.21)
has a pvalue 0.0174.
So, there is a 1-0.0174 = 0.9826 = 98.26% probability that it is overloaded.
Any probability that is above 95% is considerer unusually high. So, this elevator does not appear to be safe, since there is a 98.26% probability that it is overloaded.