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?

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

step-by-step explanation:

the original amount doesn't ''re just looking for the time it takes for 1/2 of the substance to we have

(1/2) =)     take the ln   of both sides

ln (1/2)   = )   and we can write

ln(1/2) = ) lne     and lne   = 1     so we have

ln(1/2) = -.1092t     divide both sides by -.1092

t = ln(1/2)/-.1092 = about 6.347 days = about 6.3 days (rounded)