Question

Scores on a certain intelligence test for children between ages 13 and 15 years are approximately normally distributed with ?=101 and ?=24.

(a) What proportion of children aged 13 to 15 years old have scores on this test above 91? (Reminder: proportions are between 0 and 1 - don't put in percentages!)

(b) Enter the score which marks the lowest 30 percent of the distribution.

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

Answer 1

Solution :

Given that ,

a) P(x > 91) = 1 - p( x< 91)

=1- p P[(x - ) / < (91 - 101) / 24 ]

=1- P(z < -0.42)

= 1 - 0.3372

= 0.6628

b) Using standard normal table,

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 * 24 + 101

x = 88.52

c) Using standard normal table,

P(Z > z) = 15%

= 1 - P(Z < z) = 0.15  

= P(Z < z) = 1 - 0.15

= P(Z < z ) = 0.85

= P(Z < 1.04 ) = 0.85  

z = 1.04

Using z-score formula,

x = z * +

x = 1.04 * 24 + 101

x = 125.96