## Question 84:

Asked on February 19, 2008

1

## Answer:

No answer provided yet.Using the properties of the

standard normal curve, we know that

95% of the values are below 1.65 standard deviations above the mean, measured in

z-scores.
The average number of women in shelters is 250. One standard deviation above this is (250+75) or 325 containing 84% of the nights. For 95% of the nights we need to add 1.65 standard deviations to the mean (1.65*75 = 123.8 added to 250 = 373.75).

- So to answer the first question a 350 capacity will not be enough for 95% of the nights.
- The answer to the second question is they would need a capacity of 374 (rounded up to the whole woman).
- For the third question we need to calculate the z-score, which is the data-point minus the mean divided by the standard deviation or (220-250)/75 = a z-score of -.4. When a z-score is negative the percentage of area is less than 50%, so we already know less than half the nights will have enough shelter capacity. Using the Z-score to Percentile calculator for one-sided area provides us with around 34.5% of nights will have capacity or 65.5% of nights will not have enough capacity.