SOLUTION: Suppose that the number of customers who enter a supermarket each hour is normally distributed with a mean of 540 and a standard deviation of 250. The supermarket is open 13 hours
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Question 1162233: Suppose that the number of customers who enter a supermarket each hour is normally distributed with a mean of 540 and a standard deviation of 250. The supermarket is open 13 hours per day. What is the probability that the total number of customers who enter the supermarket in one day is greater than 7500? (Hint: Calculate the average hourly number of customers necessary to exceed 7500 in one 13-hour day.) Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! i believe this is how it's done.
mean is 540
standard deviation is 250.
n = 13
7500 hours during the day / 13 = average of 576.9230769.
if the average exceeds that, then the total day will exceed 7500.
the standard deviation remains the same.
you're looking for the probability that the average number of customer per day exceeds 576.9230769.
find the z-score and then look up the probability in the normal distribution tables is what i think you need to do.
formula for z-score is z = (x-m)/s
z is the z-score
x is the raw score
m is the mean
s is the standard deviation.
formula becomes z = (576.9230769 - 540) / 250 = .1476923077
look up a z-score of 0.15 in the z-score table to find the area to the right of that, or look up the full z-score in a z-score calculator to find the area to the right of that.
in the z-score table, you will get area to the left of .15 is equal to .55962.
the area to the right of that z-score is equal to 1 minus that = .44038.
the probability of getting a z-score greater than that z-score is the area to the right.
if you use a calculator, you can get greater accuracy.
you would look for the area to the right of a z-score of .1476923077.
you would find the area to the left of that z-score = .558707 and the ara to the right of the z-score = .441293