Question

The demand for a daily newspaper at a newsstand at a busy intersection is known to be normally distributed with a mean of 150 and a standard deviation of 25. How many newspapers should the newsstand operator order to ensure that he runs short on no more than 20% of days

Answer 1

Answer:

171 newspapers.

Step-by-step explanation:

Problems of normally distributed samples are solved using 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 problem, we have that:

$\mu = 150, \sigma = 25$

How many newspapers should the newsstand operator order to ensure that he runs short on no more than 20% of days

The number of newspapers must be on the 100-20 = 80th percentile. So this value if X when Z has a pvalue of 0.8. So X when Z = 0.84.

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

$0.84 = \frac{X - 150}{25}$

$X - 150 = 0.84*25$

$X = 171$

So 171 newspapers.