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 21 minutes per visit. Assume that time spent on the social networking site per visit is normally distributed and that the standard deviation is 4 minutes. Complete parts (a) through (d) below. a. If you select a random sample of 25 sessions, what is the probability that the sample mean is between 20.5 and 21.5 minutes? (Round to three decimal places as needed.) b. If you select a random sample of 25 sessions, what is the probability that the sample mean is between 20 and 21 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 20.5 and 21.5 minutes? (Round to three decimal places as needed.)

Explanation / Answer

Mean is 21 and sd is 4 , z is calculated as (x-mean)/(sd/sqrt(N))

a) P(20.5<xbar<21.5) =P((20.5-21)/(4/sqrt(25))<z<((21.5-21)/(4/sqrt(25))) =P(-0.625<z<0.625) or P(z<0.625)-P(z<-0.625) or 2*P(z<0.625)-1 , from the normal distribution table we get p=0.7357 for z=0.625 (from the table) , thus answer is 2*0.7357-1 =0.4714

b)P(20<xbar<21) =P((20-21)/(4/sqrt(25))<z<((21-21)/(4/sqrt(25)))=P(-1.25<z<0) or P(z<0)-P(z<-1.25) =0.5-(1-0.8944) =0.3944

c) P(20.5<xbar<21.5) =P((20.5-21)/(4/sqrt(144))<z<((21.5-21)/(4/sqrt(144))) =P(-1.5<z<1.5) or 2*P(z<0.5)-1 =2*0.6915-1 =0.383

d) less, decreases, decrease

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