Question

Assume that student are to be sperated into a group with IQ scores in the bottom 30%, a second group with scores in the middle 40%, and a third group with scores in the top 30%. The Wechsler Adult Intelligence Scale yields an IQ score obtained through a test, and the scores are normally distributed with a mean of 100 and a standard deviation of 15. Find the IQ score that seperates the three groups.

Answer 1

Solution :

mean = = 100

standard deviation = = 15

Using standard normal table,

a ) First Group

P(Z < z) = 30%

P(Z < z) = 0.30

P(Z < - 0.52) = 0.30

z = - 0.52

Using z-score formula,

x = z * +

x = - 0.52*15 + 100

x = 92.2

b ) Second group

Using standard normal table,

P(-z < Z < z) = 40%
P(Z < z) - P(Z < z) = 0.40
2P(Z < z) - 1 = 0.40
2P(Z < z ) = 1 + 0.40
2P(Z < z) = 1.4
P(Z < z) = 1.40 / 2
P(Z < z) = 0.70
z = 0.52 znd z = - 0.52

Using z-score formula,

x = z * +

x = 0.52 * 0.15 + 100

= 107.8

x = 107.8

x = z * +

x = - 0.52 * 15 + 100

= 92.2

x = 92.2

c ) Tird group

Using standard normal table,

P(Z > z) = 30%

1 - P(Z < z) = 0.30

P(Z < z) = 1 - 0.30 = 0.70

P(Z < 0.52) = 0.70

z = 0.52

Using z-score formula,

x = z * +

x = 0.52 * 15+ 100

= 107.8

x = 107.8