Question

A professor at a local university noted that the exam grades of her students were normally distributed with a mean of 73 and a standard deviation of 11. If 69.5 percent of the students received grades of C or better, what is the minimum score of those who received C's?

Answer 1

Answer:

The minimum score of those who received C's is 67.39.

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:

$\mu = 73, \sigma = 11$

If 69.5 percent of the students received grades of C or better, what is the minimum score of those who received C's?

This is X when Z has a pvalue of 1-0.695 = 0.305. So it is X when Z = -0.51.

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

$-0.51 = \frac{X - 73}{11}$

$X - 73 = -0.51*11$

$X = 67.39$

The minimum score of those who received C's is 67.39.