In a large section of a statistics class, the points for the final exam are normally distributed, with a mean of 72 and a standard deviation of 9. Grades are to be assigned according to the following rule:

The top 10% receive A's.
The next 20% receive B's.
The middle 40% receive C's.
The next 20% received D's.
The bottom 10% receive F's.

What is the least amount of points that a student must score, on the final exam, in order to earn a C? (Round your answer to two decimal places.)

the answer is 72?

http://davidmlane.com/hyperstat/z_table.html

area = .4
mean = 72
SD = 9
.4 are between 71.5 and 72.5
so 71.5

No, the answer is not 72. To determine the least amount of points required to earn a C, we need to find the cutoff point below which 60% of the students fall. To do this, we can use the cumulative distribution function (CDF) of the normal distribution.

First, we need to find the z-score corresponding to the 60th percentile (or 0.60). Using a standard normal distribution table or a calculator, we find the z-score to be approximately 0.253.

Next, we can use the formula for z-score:

z = (X - μ) / σ

Rearranging the formula to solve for X:

X = z * σ + μ

Plugging in the values:

X = 0.253 * 9 + 72

Simplifying:

X ≈ 2.28 + 72

X ≈ 74.28

Therefore, the least amount of points that a student must score on the final exam to earn a C is approximately 74.28. Rounded to two decimal places, this would be 74.29.

To determine the least amount of points required to earn a C, we need to find the score below which 60% of the class falls. In this case, the middle 40% of students receive C's.

To calculate this, we'll use the concept of z-scores, which measure how many standard deviations a data point is from the mean in a normal distribution.

First, we need to find the z-score that corresponds to the 60th percentile. Since the top 10%, next 20%, and next 20% are already accounted for, we're interested in the 30th to 70th percentile range (40% in the middle). The z-score corresponding to the 30th percentile is -0.524, and the z-score corresponding to the 70th percentile is 0.524.

To find the corresponding raw score, we can use the formula:
x = mean + (z * standard deviation)

For the mean (μ) of 72 and standard deviation (σ) of 9, substituting the values:

For the 30th percentile (z = -0.524):
x = 72 + (-0.524 * 9)
x = 72 - 4.716
x ≈ 67.28

For the 70th percentile (z = 0.524):
x = 72 + (0.524 * 9)
x = 72 + 4.716
x ≈ 76.72

Therefore, the least amount of points a student must score on the final exam to earn a C is approximately 67.28. Rounded to two decimal places, this would be 67.28.