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. A. The professor has informed us that 7.93 percent ofher students received grades of A. What is the minimum scoreneeded to receive a grade of A?B. Students who made 57.93 or lower on the exam failedthe course. What percent of students failed the course?C. If 69.5% of the students received grades of C or better, what is the minimum score of those who received C's?

88.51

0.085343

67.39

Step-by-step explanation:

Given a normal distribution :

Mean (m) = 73

Standard deviation (s) = 11

Percentage A's = 7.93% = 0.0793

Minimum score needed to receive A :

Using the Z formula

Zscore corresponding to 0.0793 = 1.41

P(x ≥ x) :

Z = (x - m) / s

1.41 = (x - 73) / 11

11* 1.41 = x - 73

15.51 = x - 73

x = 15. 51 + 73

x = 88.51

B. Students who made 57.93 or lower on the exam failedthe course. What percent of students failed the course?

P(x ≤ 57.93)

Z = (x - m) / s

Z = (57.93 - 73) / 11

Z = - 15.07 / 11

Z = - 1.37

P( Z ≤ - 1.37) = 0.085343 ( Z probability calculator)

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

Minimum score of those who received C

Zscore corresponding to 0.695 = -0.51 (Z probability calculator)

Z = (x - m) / s

-0.51 = (x - 73) / 11

11* - 0.51= x - 73

-5.61 = x - 73

x = - 5.61 + 73

x = 67.39

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