Question

) e number of viewers ordering a particular pay-per-view program is normally distributed. Past history shows that 33.00% of the time fewer than 20,000 people order the program. Only ten percent of the time do more than 28,000 people order the program. What is the mean and standard deviation of the number of people ordering the program, respectively?

Answer 1

Answer:

The mean of the number of people ordering the program is 22,046.5 and the standard deviation is 4,651.16.

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.

Past history shows that 33.00% of the time fewer than 20,000 people order the program

This means that X = 20000 has a pvalue of 0.33. So when X = 20000, Z = -0.44.

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

$-0.44 = \frac{20000 - \mu}{\sigma}$

$20000 - \mu = -0.44\sigma$

$\mu = 20000 + 0.44\sigma$

Only ten percent of the time do more than 28,000 people order the program.

This means that X = 28000 has a pvalue of 1-0.1 = 0.9. So when X = 28000, Z = 1.28.

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

$1.28 = \frac{28000 - \mu}{\sigma}$

$28000 - \mu = 1.28\sigma$

$\mu = 28000 - 1.28\sigma$

Also

$\mu = 20000 + 0.44\sigma$

So

$20000 + 0.44\sigma = 28000 - 1.28\sigma$

$1.72\sigma = 8000$

$\sigma = \frac{8000}{1.72}$

$\sigma = 4651.16$

$\mu = 20000 + 0.44\sigma = 20000 + 0.44*4651.16 = 22046.5$

The mean of the number of people ordering the program is 22,046.5 and the standard deviation is 4,651.16.