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 150. The supermarket is open 14 hours per day. What is the probability that the total number of customers who enter the supermarket in one day is greater than 7200? (Hint: Calculate the average hourly number of customers necessary to exceed 7200 in one 14-hour day.)

You need an average of 514.3 per hour to have 7200 in 14 hours. So, use a normal distribution to find the probability of getting 514 or more per hour, when the mean is 540 and the std. dev. is 150.

I get a value of 56.8%, using a computation tool at
http://psych.colorado.edu/~mcclella/java/normal/accurateNormal.html

To find the probability that the total number of customers who enter the supermarket in one day is greater than 7200, we first need to calculate the average hourly number of customers necessary to exceed 7200 in a 14-hour day.

We can calculate the average hourly number of customers by dividing the total number of customers (7200) by the number of hours the supermarket is open (14):

Average hourly number of customers = Total number of customers / Number of hours
= 7200 / 14
≈ 514.29 (rounded to two decimal places)

Now, we need to find the probability that the number of customers who enter the supermarket each hour is greater than the average hourly number of customers (514.29).

To do this, we need to standardize the average hourly number of customers by converting it to a z-score using the formula:

z = (x - μ) / σ

Where:
x = Average hourly number of customers (514.29)
μ = Mean (540)
σ = Standard deviation (150)

z = (514.29 - 540) / 150
≈ -0.1694 (rounded to four decimal places)

Next, we need to find the probability of the desired event by calculating the area under the normal curve to the right of the z-score.

The probability can be looked up in a standard normal distribution table, or it can be calculated using software or calculators.

Using a standard normal distribution table, the probability of z being greater than -0.1694 is approximately 0.5659 (from the table or calculator).

Therefore, the probability that the total number of customers who enter the supermarket in one day is greater than 7200 is approximately 0.5659 or 56.59%.

To find the probability that the total number of customers who enter the supermarket in one day is greater than 7200, we need to calculate the average hourly number of customers necessary to exceed 7200 in a 14-hour day.

Since the number of customers who enter the supermarket each hour is normally distributed with a mean of 540 and a standard deviation of 150, we can use the properties of the normal distribution to calculate the average hourly number of customers.

The total number of customers in one day is equal to the product of the average hourly number of customers and the number of hours the supermarket is open. In this case, the average hourly number of customers needed to exceed 7200 in a 14-hour day can be calculated as follows:

Average hourly number of customers = (7200 / 14) = 514.29

Now, we need to calculate the probability that the number of customers who enter the supermarket each hour is greater than the average hourly number of customers.

To do this, we need to standardize the average hourly number of customers using the formula:

Z = (X - μ) / σ

where X is the value we want to standardize, μ is the mean of the distribution, and σ is the standard deviation of the distribution.

In this case, X = 514.29, μ = 540, and σ = 150.

Z = (514.29 - 540) / 150 = -0.17

Now, we can use a standard normal distribution table or calculator to find the probability that the number of customers who enter the supermarket each hour is greater than the average hourly number of customers.

Looking up the corresponding area under the standard normal distribution curve for Z = -0.17, we find that the probability is approximately 0.4325.

Therefore, the probability that the total number of customers who enter the supermarket in one day is greater than 7200 is 0.4325 or 43.25%.