Answer:
a) 
b) 
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
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:

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



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



c. What percentage of students watch less than 3.6 hours of shows a week?
pvalue of Z when X = 3.6.



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



has a pvalue of 0.9332
X = 2



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