Question

Suppose the scores of students on an exam are Normally distributed with a mean of 303 and a standard deviation of 39. Then approximately 99.7% of the exam scores lie between the numbers and such that the mean is halfway between these two integers. (You are not to use Rcmdr for this question.)

Answer #1

Given that,

mean = = 303

standard deviation = = 39

middle 99.7% of score is

P(-z < Z < z) = 0.997

P(Z < z) - P(Z < -z) = 0.997

2 P(Z < z) - 1 = 0.997

2 P(Z < z) = 1 + 0.997 = 1.997

P(Z < z) =1.997 / 2 = 0.9985

P(Z < 2.96) = 0.9985

z ± 2.96 using z table

Using z-score formula

x= z * +

x= ± 2.96*39+303

x= 187.56 , 418.44

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