Time spent using e-mail per session is normally distributed, with mu = 11 minute

Question

Time spent using e-mail per session is normally distributed, with mu = 11 minutes and sigma = 3 minutes. Complete parts (a) through (d). a. If you select a random sample of 25 sessions, what is the probability that the sample mean is between 10.8 and 11.2 minutes? b. If you select a random sample of 25 session, what is the probability that the sample mean is between 1.05 and 11 minutes? 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? d. Explain the difference in the results of (a) and (c) Choose the correct answer below.

Explanation / Answer

Mean is 11 and standard deviation is 3

For question part a and b n is given to be 25 and thus the standard error is 3/sqrt(25)=0.6

a) P(10.8<x<11.2)=P((10.8-11)/0.6<z<(11.2-11)/0.6)=P(-0.33<z<0.33) or 2*P(z<0.33)-1 from normal distribution table it is 2*0.6293-1=0.2586

b) P(10.5<x<11)=P((10.5-11)/0.6<z<0)=P(-0.833<z<0)=P(z<0)-(1-P(z<0.833))=0.5-(1-0.7967)=0.2967

c) here n=100 and thus standard error is 3/sqrt(100)=0.3

thus P(10.8<x<11.2)=P((10.8-11)/0.3<z<(11.2-11)/0.3)=P(-0.66<z<0.66) or 2*P(z<0.66)-1 from normal distribution table it is 2*0.7454-1=0.4908

d) less,decreases, more, increase

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