Mathematics, 09.02.2021 04:30 Miguel9825

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.

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Answer from: GodlyGamer8239

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.

Answer from: Quest

Your cleary not see the answer is 25

Answer from: Quest

Jim has run 5/6 of a mile so far. i hope that answers your question! good luck : )

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