Question

An adolescent adjustment scale has a mean of 50, and a standard deviation of 21. what score on the adjustment scale would put a student in the top 10% of the distribution? state your conclusion in words.

Answer 1

Solution :

Given that,

mean = = 50

standard deviation = = 21

Using standard normal table,

P( Z > z) = 10%

P(Z > z) = 0.10

1 - P( Z < z) = 0.10

P(Z < z) = 1 - 0.90

P(Z < z) = 0.90

z = 1.28

Using z-score formula,

x = z * +

x = 1.28 * 21 + 50

x = 76.88

x = 76.88