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Tems11 [23]
3 years ago
5

An art history professor assigns letter grades on a test according to the following scheme. A: Top 13%13% of scores B: Scores be

low the top 13%13% and above the bottom 56%56% C: Scores below the top 44%44% and above the bottom 21%21% D: Scores below the top 79%79% and above the bottom 9%9% F: Bottom 9%9% of scores Scores on the test are normally distributed with a mean of 79.779.7 and a standard deviation of 8.48.4. Find the numerical limits for a B grade. Round your answers to the nearest whole number, if necessary.
Mathematics
1 answer:
amm18123 years ago
3 0

Answer:

The numerical limits for a B grade is between 81 and 89.

Step-by-step explanation:

Problems of normally distributed samples are solved using 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.

In this problem, we have that:

\mu = 79.7, \sigma = 8.4

B: Scores below the top 13% and above the bottom 56%

Below the top 13%:

Below the 100-13 = 87th percentile. So below the value of X when Z has a pvalue of 0.87. So below X when Z = 1.127. So

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

1.127 = \frac{X - 79.7}{8.4}

X - 79.7 = 8.4*1.127

X = 89

Above the bottom 56:

Above the 56th percentile, so above the value of X when Z has a pvalue of 0.56. So above X when Z = 0.15. So

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

0.15 = \frac{X - 79.7}{8.4}

X - 79.7 = 8.4*0.15

X = 81

The numerical limits for a B grade is between 81 and 89.

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9.

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