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 19 minutes per visit. Assume that time spent on the social networking site per visit is normally distributed and that the standard deviation is 7 minutes. Complete parts (a) through (d) below. a. If you select a random sample of 16 sessions, what is the probability that the sample mean is between 18.5 and 19.5 minutes? (Round to three decimal places as needed.) b. If you select a random sample of 16 sessions, what is the probability that the sample mean is between 18 and 19 minutes? (Round to three decimal places as needed.) c. If you select a random sample of 100 sessions, what is the probability that the sample mean is between 18.5 and 19.5 minutes? (Round to three decimal places as needed.) d. Explain the difference in the results of (a) and (c) The sample size in (c) is greater than the sample size in (a), so the standard error of the mean (or the standard deviation of the sampling distribution) in (c) isthan in (a). As the standard error around the mean. Therefore, the probability that the sample mean will fall in a region that includes the population mean will always values become more concentrated V when the sample size increases

Explanation / Answer

SolutionA:

mean=19 min

sd=7 min

P(18.5<xbar<19.5)

n=16

P(18.5-19/7/sqrt(16)<Z<19.5-19/7/sqrt(16)

P(-0.29<Z<0.29)

P(Z<0.29)-P(Z<-0.29)

=0.6141-(1-0.6141)

=0.2282

ANSWER:0.2282

Solutionb:

P(18<xbar<19)

P(18-19/7/sqrt(16)<Z<(19-19)/7/sqrt(16)

P(-0.57<z<0)

=P(Z<0)-P(Z<-0.57)

=0.2157

ANSWER:0.2157

Solutionc:

P(18.5<xbar<19.5)

n=100

P(18.5-19/7/sqrt(100)<Z<19.5-19/7/sqrt(100)

P(-0.71<Z<0.71)

=0.5222

ANSWER:0.5222

Solutiond:

7/sqrt(16)=1.75

7/sqrt(100)=7/10=0.7

sample size increases,standard error decreases.

standard deviation of sampling distribution of (c) is less than in (a)

decreases

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