Question

Scores on a certain intelligence test for children between ages 13 and 15 years are approximately normally distributed with μ=101μ=101 and σ=18σ=18.

(a) What proportion of children aged 13 to 15 years old have scores on this test above 87 ? (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 20 percent of the distribution.

(c) Enter the score which marks the highest 5 percent of the distribution.

Answer 1

Answer:

a)

Given,

Mean = μ = 101

Standard deviation = σ = 18

To determine the proportion of children aged 13 to 15 years old have score on this test above 87

P(X > 87) = 1 - P(X <= 87)

= 1 - P((X - )/ < (87 - )/)

substitute the known values

= 1 - P(Z <= (87 - 101)/18)

= 1 - P(Z <= -14/18)

= 1 - P(Z <= - 0.7778)

= 1 - 0.2184 [ since from z table]

P(X > 87) = 0.7816

b)

To determine the score which marks the lowest 20% of the distribution

i.e.,

P(X < x) = 0.20

P((X - )/ < (x- )/) = 0.20

P(Z < z) = 0.20

z = - 0.84162

(X - ) / = - 0.84162

X - =  - 0.84162

X = - 0.84162

= 101 - 0.84162(18)

= 101 - 15.14916

X = 85.85084

c)

To determine the score which marks the highest 5 percent of the distribution

P(X > x) = 0.05

P((X - )/ > (x- )/) = 0.05

1 - P(Z <= z) = 0.05

P(Z <= z) = 1 - 0.05

P(Z <= z) = 0.95

Here z = 1.645

(X - ) / = 1.645

X - = 1.645

X = + 1.645

= 101 + 1.645(18)

= 101 + 29.61

X = 130.61