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 16 minutes per visit Assume that time spent on the social networking site per visit is normaily distributed and that the standard deviation is 5 minutes Complete parts (a) through (d) below a. if you select a random sample of 36 sessions, what is the probability that the sample mean is between 15.5 and 16 5 minutes? (Round to three decimal places as needed) b. If you select a random sample of 36 sessiens, what is the probablity that the sample mean is between 15 and 16 minutes? (Round to three decimal places as needed) C. if you select a random sample of 144 sessions, what is the probability that the sample mean is between 15 5 and 16 5 minutes? (Round to three decimal places as needed.) d. Explain the diflerence in the results of (a) and (c) The sample size in (c) is greater than the sample size in (a), so the standard er or ofthe mean orthe standard deviation of the samping distribution in e s than in (a) As the standard error that includes the population mean will always values become more concentrated around the mean. Therefore, the probability that the sample mean will fall in a region when the sample size increases

Explanation / Answer

a) std error of mean =std deviaiton/(n)1/2 =5/(36)1/2 =0.8333

a) P(15.5<X<16.5)=P((15.5-16)/0.8333<Z<(16.5-16)/0.8333)=P(-0.6<Z<0.6)=0.7257-0.2742 =0.451

b)P(15<X<16)=P(-1.2<Z<0)=0.5-0.1151 =0.385

c)

std error of mean =std deviaiton/(n)1/2 =5/(144)1/2 =0.4167

P(15.5<X<16.5)=P((15.5-16)/0.4167<Z<(16.5-16)/0.4167)=P(-1.2<Z<1.2)=0.8849-0.1151 =0.770

d) .......... in(c) is smaller then in *(a). as the std error decreases values become..............

will always greater when the sample size increases,

please revert for any clarification required

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