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

Question 1037984: 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 spent on on-line shopping between 35 minutes and 65 minutes in the second week of December.
I saw that this question was answered before but it had no explanation on it the answer is 0.0992
(https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.1031204.html) but I have never learned to do it this way in class
Mean is 80
Standard Deviation is 35
So far I have P( x ≤ 65 ) - P( X ≤ 35 )
= 0.33411757 - 0.09927139
= 0.23974252
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
Prob x <=65 is a z of (65-80)/35/3=-1.2857
prob of x <=35 is a z of (35-80)/35/3= -3.857.
This is a sample of 9, and the mean of samples of 9 follows a normal distribution as well with the same mean, but the standard deviation is different. It is the standard deviation (35), divided by the sqrt of the sample size (9). That's where the 3 comes from. It is much less likely for a group of 9 people to have a mean between 35 and 65 minutes than it is for an individual to have it.
The probability of x <65 is 0.0992.
The probability of x <35 is 0.00006, or essentially near zero. In other words, it is essentially the probability of the mean being <65 minutes.

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