If the grades on a police entrance exam are normally distributed with a mean of 100 and a standard deviation of 15, what percentage of students received grades of less than 115?

Use z-scores.

z = (x - mean)/sd

x = 115
mean = 100
sd = 15

Use a z-table to find probability of less than 115 using the z-score calculated above. Convert to a percent.

I hope this will help.

To find the percentage of students who received grades of less than 115, we need to calculate the area under the normal distribution curve to the left of 115. This can be done by finding the cumulative distribution function (CDF) of the normal distribution.

First, let's convert the individual score of 115 into a z-score. The z-score measures the number of standard deviations a particular value is from the mean. It is calculated using the formula:

z = (x - μ) / σ

where:
z = z-score
x = individual score
μ = mean
σ = standard deviation

Plugging in the values:
z = (115 - 100) / 15
z = 1

Now, we can use a standard normal distribution table or a statistical software to find the cumulative probability corresponding to a z-score of 1. The cumulative probability will give us the percentage of students who received grades less than 115.

Looking up the z-score of 1 in a standard normal table or using statistical software, we can find that the cumulative probability is approximately 0.8413.

Therefore, approximately 84.13% of the students received grades less than 115.