SOLUTION: The statistical report shows that the average amount of time spent by each customer per week on online shopping is 1 hour 20 minutes with a variance of 1225 minutes. Assume that th

Algebra ->  Probability-and-statistics -> SOLUTION: The statistical report shows that the average amount of time spent by each customer per week on online shopping is 1 hour 20 minutes with a variance of 1225 minutes. Assume that th      Log On


   

Question 1031204: The statistical report shows that the average amount of time spent by each customer per week on online shopping is 1 hour 20 minutes with a variance of 1225 minutes. Assume that the data are normally distributed.
b. Calculate the probability that 9 randomly chosen customers will have a mean amount of time spend on on-line shopping between 35 minutes and 65 minutes in the second week of December.
The answer is 0.0992, but I'm not sure how to solve it. The 9 randomly chosen part is stumping me.
EDIT: I've already tried using NormalCD for probability of spending between 35 and 65 minutes online which is 0.2348. Am I supposed to use Binomial and Normal Distribution to solve the problem?
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!

Probability-and-statistics/1031204 (2016-04-23 13:50:48): The statistical report shows that the average amount of time spent by each customer per week on online shopping is 1 hour 20 minutes with a variance of 1225 minutes. Assume that the data are normally distributed.
b. Calculate the probability that 9 randomly chosen customers will have a mean amount of time spend on on-line shopping between 35 minutes and 65 minutes in the second week of December.
z(35) = (35-80)/(35/3) = -3.8571
z(65) = (65-80)/(35/3) = -1.2857
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P(35< x-bar < 65) = P(-3.8571 < z < -1.2857) = 0.0992
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Cheers,
Stan H.
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