Question

The intelligence quotients of 400 children have a mean value of 100 and a standard deviation of 14. Assuming that I.Q.’s are normally distributed, determine the number of children likely to have I.Q.’s of between

(a) 80 and 90,

(b) 90 and 110 and

(c) 110 and 130.

Answer 1

Given,

= 100, = 14

We convert to standard normal as

P(X < x) = P( Z < x - / )

a)

P(80 < X < 90) = P( X < 90) - P( X < 80)

= P( Z < 90 - 100 / 14) - P( Z < 80 - 100 / 14)

= P( Z < -0.7143) - P(Z < -1.4286)

= ( 1 - P( Z < 0.7143) ) - P( 1 - P( Z < 1.4286) )

= (1 - 0.7625) - ( 1 - 0.9234)

= 0.1609

Out of 400, number of children between 80 and 90 = 0.1609 * 400

= 64.36

= 64 cheldren

b)

P( 90 < X < 110) = P( X < 110) - P( X < 90)

= P( Z < 110 - 100 / 14) - P( Z < 90 - 100 / 14)

= P( Z < 0.7143) - P( Z < -0.7143)

= 0.7625 - ( 1 - 0.7625)

= 0.5250

Number of students out of 400 are between 90 and 110 = 0.5250 * 400

= 210 children.

c)

P(110 < X < 130) = P( X < 130) - P( X < 110)

= P( Z < 130 - 100 / 14) - P( Z < 110 - 100 / 14)

= P( Z < 2.1429) - P( Z < 0.7143)

= 0.9839 - 0.7625

= 0.2214

Number of students out of 400 are between 110 and 130 = 0.2214 * 400

= 88.56

= 89 children