Question

Scores on a certain intelligence test for children between ages 13 and 15 years are approximately normally distributed with μ=111 and σ=21.

(a) What proportion of children aged 13 to 15 years old have scores on this test above 97 ? (NOTE: Please enter your answer in decimal form. For example, 45.23% should be entered as 0.4523.)

(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 ,

mean = = 111

standard deviation = = 21

(a)

P(x > 97) = 1 - P(x < 97)

= 1 - P((x - ) / < (97 - 111) / 21)

= 1 - P(z < -0.667)

= 1 - 0.2524   

= 0.7476

P(x > 97) = 0.7476

Proportion = 0.7476

(b)

P(Z < z) = 30% = 0.30

P(Z < -0.5244) = 0.30

z = -0.5244

Using z-score formula,

x = z * +

x = -0.5244 * 21 + 111 = 99.9876

Score = 99.9876

(c)

P(Z > z) = 15%

1 - P(Z < z) = 0.15

P(Z < z) = 1 - 0.15 = 0.85

P(Z < 1.036) = 0.85

z = 1.036

Using z-score formula,

x = z * +

x = 1.036 * 21 + 111 = 132.756

Score = 132.756