Question

You are asked to do a study of shelters for abused and battered women to determine the nec- essary capacity in your city to provide housing for most of these women. After recording data for a whole year, you find that the mean number of women in shelters each night is 250, with a standard deviation of 75. Fortunately, the distribution of the number of women in the shelters each night is normal, so you can answer the following questions posed by the city council.
a. If the city's shelters have a capacity of 350, will that be enough places for abused womer on 95% of all nights? If not, what number of shelter openings will be needed?
b. The current capacity is only 220 openings, because some shelters have been closed. What is the percentage of nights that the number of abused women seeking shelter will exceed current capacity?

Answer 1

Answer:

a) 350 shelters is not enough, as 374 shelters are needed.

b) On 65.54% of the nights, the number of abused women seeking shelter will exceed current capacity

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

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

$Z = \frac{X - \mu}{\sigma}$

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 question, we have that:

$\mu = 250, \sigma = 75$

a. If the city's shelters have a capacity of 350, will that be enough places for abused womer on 95% of all nights? If not, what number of shelter openings will be needed?

We have to find the 95th percentile of the distribution, that is, X when Z has a pvalue of 0.95, so X when Z = 1.645.

$Z = \frac{X - \mu}{\sigma}$

$1.645 = \frac{X - 250}{75}$

$X - 250 = 1.645*75$

$X = 373.3$

Rounding up, 374

350 shelters is not enough, as 374 shelters are needed.

b. The current capacity is only 220 openings, because some shelters have been closed. What is the percentage of nights that the number of abused women seeking shelter will exceed current capacity?

This is 1 subtracted by the pvalue of Z when X = 220. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{220 - 250}{75}$

$Z = -0.4$

$Z = -0.4$ has a pvalue of 0.3446

1 - 0.3446 = 0.6554

On 65.54% of the nights, the number of abused women seeking shelter will exceed current capacity