According to a social media blog, time spent on a certain social networking webs

Question

According to a social media blog, time spent on a certain social networking website has a mean of 20 minutes per visit. Assume that time spent on the social networking site per visit is normally distributed and that the standard deviation is 33 minutes. Complete parts (a) through (d) below.

A)a. If you select a random sample of 25 sessions, what is the probability that the sample mean is between 19.5 and 20.5 minutes?

b. If you select a random sample of 25 sessions, what is the probability that the sample mean is between 19 and 20 minutes?

c. If you select a random sample of 100 sessions, what is the probability that the sample mean is between 19.5 and 20.5 minutes?

Explanation / Answer

a) Z = (x-Mean)/(Sigma/sqrt(n)

Z1 = (20.5-20)/(33/sqrt(25)) = 0.075

Z2 = -0.075

Probability =P(z<0.075) - P(z<-0.075) = 0.530171-0.469829=0.060342

2)

Z1 = (21-20)/(33/sqrt(25)) = 0.1515

Z2 = -0.1515

Probability =P(z<0.1515) - P(z<-0.1515) = 0.560209-0.439791= 0.1204

3)

Z1 = (20.5-20)/(33/sqrt(100)) = 0.1515

Z2 = -0.1515

Probability =P(z<0.1515) - P(z<-0.1515) = 0.560209-0.439791= 0.1204

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