You are asked to do a study of shelters for abused and battered women to determine the necessary 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 question posed by the city council.

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

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:

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:

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.

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

274 shelters will be needed.

$4

step-by-step explanation:

$30×5=$150

$150 minus $130=$20

$20÷5=$4.

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step-by-step explanation: