Answer:
Using the normal probability distribution, with a capacity of 350, it is enough for all abused on 90.82% of nights.
274 shelters will be needed.
Step-by-step explanation:
Problems of normally distributed samples are 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)
The Z-score measures how many standard deviations the measure is from the mean. 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, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
![\mu = 250, \sigma = 75](https://tex.z-dn.net/?f=%5Cmu%20%3D%20250%2C%20%5Csigma%20%3D%2075)
If the city’s shelters have a capacity of 350, will that be enough places for abused women on 95% of all nights?
What is the percentile of 350?
This is the pvalue of Z when X = 350.
![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{250 - 150}{75}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B250%20-%20150%7D%7B75%7D)
![Z = 1.33](https://tex.z-dn.net/?f=Z%20%3D%201.33)
has a pvalue of 0.9082.
Using the normal probability distribution, with a capacity of 350, it is enough for all abused on 90.82% of nights.
If not, what number of shelter openings will be needed?
The 95th percentile, which is X when Z = 1.645. So
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![1.645 = \frac{X - 150}{75}](https://tex.z-dn.net/?f=1.645%20%3D%20%5Cfrac%7BX%20-%20150%7D%7B75%7D)
![X - 150 = 1.645*75](https://tex.z-dn.net/?f=X%20-%20150%20%3D%201.645%2A75)
![X = 274](https://tex.z-dn.net/?f=X%20%3D%20274)
274 shelters will be needed.