Question

A survey of Massachusetts high school students showed that the mean number of hours these students reported watching crime TV shows was 2.4 hours with standard deviation S=0.8 and that the distribution of this variable is normal. Using this information, find a. Find the z-score for a student who reported watching 2 hours of crime shows a week. b. Find the z-score for a student who reported watching 3.6 hours of crime shows a week. c. What percentage of students watch less than 3.6 hours of shows a week? d. What percentage of the sample watches between 2 hours and 3.6 hours of crime shows a week?

Answer 1

Answer:

a) $Z = -0.5$

b) $Z = 1.5$

c) 93.32% of students watch less than 3.6 hours of shows a week

d) 62.57% of the sample watches between 2 hours and 3.6 hours of crime shows a week

Step-by-step explanation:

Problems of normally distributed samples can be 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:

$\mu = 2.4, \sigma = 0.8$

a. Find the z-score for a student who reported watching 2 hours of crime shows a week.

This is Z when X = 2. So

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

$Z = \frac{2 - 2.4}{0.8}$

$Z = -0.5$

b. Find the z-score for a student who reported watching 3.6 hours of crime shows a week

X = 3.6. So

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

$Z = \frac{3.6 - 2.4}{0.8}$

$Z = 1.5$

c. What percentage of students watch less than 3.6 hours of shows a week?

pvalue of Z when X = 3.6.

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

$Z = \frac{3.6 - 2.4}{0.8}$

$Z = 1.5$

$Z = 1.5$ has a pvalue of 0.9332

93.32% of students watch less than 3.6 hours of shows a week

d. What percentage of the sample watches between 2 hours and 3.6 hours of crime shows a week?

pvalue of Z when X = 3.6 subtracted by the pvalue of Z when X = 2.

X = 3.6

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

$Z = \frac{3.6 - 2.4}{0.8}$

$Z = 1.5$

$Z = 1.5$ has a pvalue of 0.9332

X = 2

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

$Z = \frac{2 - 2.4}{0.8}$

$Z = -0.5$

$Z = -0.5$ has a pvalue of 0.3075

0.9332 - 0.3075 = 0.6257

62.57% of the sample watches between 2 hours and 3.6 hours of crime shows a week