Question

If a normal distribution has a mean of 132 and a standard devistion of 20, what is the value that has a z score of 3.6

Answer 1

z = (x - μ)/σ

Substituting given values, you have

3.6 = (x - 132)/20

(20)(3.6) = x - 132 . . . . multiply by 20

72 + 132 = x . . . . . . . . .add 132

204 = x

The value that has a z-score of 3.6 is 204.