Question

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

a) What proportion of children aged 13 to 15 years old have scores on this test above 88 ? Answer in decimal form.

(b) Enter the score which marks the lowest 25 percent of the distribution.

c) Enter the score which marks the highest 10 percent of the distribution.

Answer 1

Solution :

Given ,

mean = = 106

standard deviation = = 15

P(x >88 ) = 1 - P(x<88 )

= 1 - P[(x -) / < (88-106) / 15]

= 1 - P(z <-1.2 )

Using z table

= 1 - 0.1151

= 0.8849

proportion= 0.8849

(B)

Using standard normal table,

P(Z < z) = 25%

= P(Z < z) = 0.25  

= P(Z <-0.67 ) = 0.25

z = -0.67 Using standard normal z table,

Using z-score formula  

x= z * +

x= -0.67 *15+106

x= 95.95

x=96

C)

Using standard normal table,

P(Z > z) = 10%

= 1 - P(Z < z) = 0.10

= P(Z < z ) = 1 - 0.10

= P(Z < z ) = 0.90

= P(Z < 1.28) = 0.90

z = 1.28 (using standard normal (Z) table )

Using z-score formula  

x = z * +

x= 1.28 *15+106

x= 125.2

x=125