Question

In a large section of a statistics​ class, the points for the final exam are normally​ distributed, with a mean of 71 and a standard deviation of 7. Grades are assigned such that the top​ 10% receive​ A's, the next​ 20% received​ B's, the middle​ 40% receive​ C's, the next​ 20% receive​ D's, and the bottom​ 10% receive​ F's. Find the lowest score on the final exam that would qualify a student for an​ A, a​ B, a​ C, and a D.

Answer 1

Answer:

The lowest score on the final exam that would qualify a student for an​ A is 80.

The lowest score on the final exam that would qualify a student for a B is 74.68.

The lowest score on the final exam that would qualify a student for a C is 67.33.

The lowest score on the final exam that would qualify a student for a​ D is 62.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Mean of 71 and a standard deviation of 7.

This means that $\mu = 71, \sigma = 7$

Grades are assigned such that the top​ 10% receive​ A's, the next​ 20% received​ B's, the middle​ 40% receive​ C's, the next​ 20% receive​ D's, and the bottom​ 10% receive​ F's.

This means that:

90th percentile and above: A

70th percentile and below 90th: B

30th percentile to the 70th percentile: C

10th percentile to the 30th: D

Lowest score for an A:

Top 10% receive A, which means that the lowest score that would qualify a student for an A is the 100 - 10 = 90th percentile, which is X when Z has a pvalue of 0.9, so X when Z = 1.28.

$Z = \frac{X - \mu}{\sigma}$

$1.28 = \frac{X - 71}{7}$

$X - 71 = 7*1.28$

$X = 80$

The lowest score on the final exam that would qualify a student for an​ A is 80.

Lowest score for a B:

70th percentile, which is X when Z has a pvalue of 0.7, so X when Z = 0.525.

$Z = \frac{X - \mu}{\sigma}$

$0.525 = \frac{X - 71}{7}$

$X - 71 = 7*0.525$

$X = 74.68$

The lowest score on the final exam that would qualify a student for a B is 74.68.

Lowest score for a C:

30th percentile, which is X when Z has a pvalue of 0.3, so X when Z = -0.525.

$Z = \frac{X - \mu}{\sigma}$

$-0.525 = \frac{X - 71}{7}$

$X - 71 = 7*(-0.525)$

$X = 67.33$

The lowest score on the final exam that would qualify a student for a C is 67.33.

Lowest score for a D:

10th percentile, which is X when Z has a pvalue of 0.1, so X when Z = -1.28.

$Z = \frac{X - \mu}{\sigma}$

$-1.28 = \frac{X - 71}{7}$

$X - 71 = 7*(-1.28)$

$X = 62$

The lowest score on the final exam that would qualify a student for a​ D is 62.