Time spent using e-mail per session is normally distributed, with =11 minutes an

Question

Time spent using e-mail per session is normally distributed, with =11 minutes and =2 minutes. Complete parts (a) through (d).

If you select a random sample of 50 sessions, what is the probability that the sample mean is between 10.8 and 11.2 minutes? nothing (Round to three decimal places as needed.)

b. If you select a random sample of 50 sessions, what is the probability that the sample mean is between 10.5 and 11minutes? nothing (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 10.8 and 11.2 minutes? nothing (Round to three decimal places as needed.)

d. Explain the difference in the results of (a) and (c).

Choose the correct answer below. 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) is less OR greater than in (a). As the standard deviation decreases OR increases , values become less OR more concentrated around the mean. Therefore, the probability of a region that includes the mean will always increase OR decrease when the sample size increases.

Explanation / Answer

a. P(10.8<xbar<11.2)=P(10.8-11/(2/sqrt(50)<z<11.2-11/(2/sqrt(50))=P(-0.71<z<0.71)=0.5223

b. n=50, P(10.5<xbar<11)=P(-1.77<z<0)=0.4616

c. P(10.8<xbar<11.2) for n=100, we get P(-1<z<1)=0.6827

d. As we raise n we see probability increases.

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