Question

9. The distribution of CHE 315 Exam I scores is nearly normal with a mean of 72.6 points, and a standard deviation of 4.78 points. The test score of the top 5% of students in the class is ___________ and higher

Answer 1

Solution :

Given that,  

mean = = 72.6

standard deviation = = 4.78

Using standard normal table ,

P(Z > z) = 5%

1 - P(Z < z) = 0.05

P(Z < z) = 1 - 0.05

P(Z < 1.65) = 0.95

z = 1.65

Using z-score formula,

x = z * +

x = 1.65 * 4.78 + 72.6 = 80.49

The test score of the top 5% of students in the class is 80.49 and higher