Answer the question(s) based on the following situation: 150 students in a math class take the final exam. The scores on the exam have an approximately normal distribution with center ì = 65 and standard deviation ó = 10.The third quartile of the scores on the exam is approximately

Z = (x - mean)/standard deviation

In a table labeled something like "areas under normal distribution," look for Z score at 75th percentile (third quartile). Insert values into the above equation and solve.

To find the third quartile of the scores on the exam, we need to calculate the z-score for the desired quartile and use it to find the corresponding value in the standard normal distribution.

First, let's calculate the z-score for the third quartile. The third quartile is equivalent to the 75th percentile, so we need to find the z-score that separates the top 25% of the distribution.

To calculate the z-score, we can use the formula:

z = (x - μ) / σ

Where:
x = the desired quartile (unknown)
μ = the mean (center) of the distribution, 65
σ = the standard deviation, 10

Since the third quartile is to the right of the mean, we're interested in the positive z-score. Using a standard normal distribution table or calculator, we can find that the z-score for the 75th percentile is approximately 0.675.

Now, we can solve for x in the z-score formula:

0.675 = (x - 65) / 10

Multiplying both sides of the equation by 10, we get:

6.75 = x - 65

Adding 65 to both sides of the equation, we find:

x = 65 + 6.75

x ≈ 71.75

Therefore, the third quartile of the scores on the exam is approximately 71.75.