Question

1.Scores on a certain intelligence test for children between ages 13 and 15 years are approximately normally distributed with μ=106 and σ=18

(a) What proportion of children aged 13 to 15 years old have scores on this test above 96? 0.710743

(b) Enter the score which marks the lowest 20 percent of the distribution. 90 85

help with part c
(c) Enter the score which marks the highest 15 percent of the distribution.

Answer 1

Solution :

Given that,

mean = = 106

standard deviation = = 18

a ) P (x > 96 )

= 1 - P (x < 96 )

= 1 - P ( x -  / ) < ( 96 - 106 / 18)

= 1 - P ( z <- 10 / 18 )

= 1 - P ( z < -0.55 )

Using z table

= 1 - 0.2912

= 0.9558

Proporation = 0.7088

b )P(Z < z) = 20%

P(Z < z) = 0.20

P(Z < -0.84) = 0.20

z = -0.84

Using z-score formula,

x = z * +

x = -0.84 * 18 + 106

x = 90.88

c ) P( Z > z) = 15%

P(Z > z) = 0.15

1 - P( Z < z) = 0.15

P(Z < z) = 1 - 0.85

P(Z < z) = 0.85

z = 1.04

Using z-score formula,

x = z * +

x = 1.04 * 18 + 106

x = 124.72