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 22 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 36 sessions, what is the probability that the sample mean is between 21.5 and 22.5 minutes? b. If you select a random sample of 36 sessions, what is the probability that the sample mean is between 21 and 22 minutes? c. If you select a random sample of 100 sessions, what is the probability that the sample mean is between 21.5 and 22.5 minutes? 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) is than in (a). As the standard

Explanation / Answer

for normal distribution z score =(X-mean)/std deviation

a) for sample size n=36; std error of mean =std deviation/(n)1/2 =7/6=1.1667

P(21.5<X<22.5)=P((21.5-22)/1.1667<Z<(22.5-22)/1.1667)=P(-0.4286<Z<0.4286)=0.6659-0.3341=0.331

b)

P(21<X<22)=P((21-22)/1.1667<Z<(22-22)/1.1667)=P(-0.8571<Z<0)=0.5-0.1957=0.304

c)

for sample size n=100; std error of mean =std deviation/(n)1/2 =7/10=0.7

P(21.5<X<22.5)=P((21.5-22)/0.7<Z<(22.5-22)/0.7)=P(-0.7143<Z<0.7143)=0.7625-0.2375=0.525

d)please provide full sentence of part D so that I can help

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