Question

A psychology professor assigns letter grades on a test according to the following scheme. A: Top 10% scores, B: Scores below 10% and above the bottom 63%, C: Scores below the top 37% and above the bottom 20%, D: Scores below the top 80% and above the bottom 7%, F: Bottom 7 % of scores Scores on the test are normally distributed with a mean of 71.5 and a standard deviation of 9.5. Find the numerical limits for a D grade. Round your answers to the nearest whole number if necessary.

Answer 1

Answer:

The numerical limits for a D grade is between 57 and 64.

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.

In this question:

$\mu = 71.5, \sigma = 9.5$

D: Scores below the top 80% and above the bottom 7%

Between the 7th and the 100 - 80 = 20th percentile.

7th percentile:

X when Z has a pvalue of 0.07. So X when Z = -1.475.

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

$-1.475 = \frac{X - 71.5}{9.5}$

$X - 71.5 = -1.475*9.5$

$X = 57.49$

So 57

20th percentile:

X when Z has a pvalue of 0.2. So X when Z = -0.84.

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

$-0.84 = \frac{X - 71.5}{9.5}$

$X - 71.5 = -0.84*9.5$

$X = 63.52$

So 64

The numerical limits for a D grade is between 57 and 64.