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?

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 and standard deviation , the zscore of a measure X is given by:

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:

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.

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

.632

step-by-step explanation:

6 * 1/10 = 3/5

3*0.01 = 0.03

3/5 + 0.03 = .63

2*0.002 = .002

.63 + 0.002 = .632

step-by-step explanation: the answer is 27. first multiply .32 by 25 which equals 8. then add 8 to 25 which gives you 33. then you subtract that from 6 since she gave 6 to her brother.

,, add !