The time it takes for a statistics professor to mark a singl

The time it takes for a statistics professor to mark a single midterm test is normally distributed with a mean of 4.9 minutes and a standard deviation of 2.3 minutes. There are 64 students in the professor\'s class. What is the probability that he needs more than 5 hours to mark all of the midterm tests?

Suppose that the number of customers who enter a supermarket each hour is normally distributed with a mean of 610 and a standard deviation of 210. The supermarket is open 16 hours per day. What is the probability that the total number of customers who enter the supermarket in one day is greater than 11100? (Hint: Calculate the average hourly number of customers necessary to exceed 11100 in one 16-hour day.)

Solution

a)
Normal Distribution
Mean ( u ) =4.9
Standard Deviation ( sd )=2.3
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
P(X > 5) = (5-4.9)/2.3/ Sqrt ( 64 )
= 0.1/0.288= 0.3478
= P ( Z >0.3478) From Standard Normal Table
= 0.364                  
b)
Normal Distribution
Mean ( u ) =610
Standard Deviation ( sd )=210
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
Per hourly based people = 11100/16 = 693.75
P(X > 693.75) = (693.75-610)/210/ Sqrt ( 16 )
= 83.75/52.5= 1.5952
= P ( Z >1.5952) From Standard Normal Table
= 0.0553