Question

Scores on a certain intelligence test for children between ages 13 and 15 years are approximately normally distributed with μ=96μ=96 and σ=18σ=18.

(a) What proportion of children aged 13 to 15 years old have scores on this test above 86 ? (NOTE: Please enter your answer in decimal form. For example, 45.23% should be entered as 0.4523.)
Answer:

(b) Enter the score which marks the lowest 20 percent of the distribution.
Answer:

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

Answer 1

Solution :

Given ,

mean = = 96

standard deviation = = 18

P(x >86 ) = 1 - P(x<86 )

= 1 - P[(x -) / < (86 -96) / 18]

= 1 - P(z < -0.56)

Using z table

= 1 - 0.2877

= 0.7123

proportion=0.7123

(B)

Using standard normal table,

P(Z < z) =20 %

= P(Z < z) = 0.20  

= P(Z <- 0.84 ) = 0.20

z = - 0.84 Using standard normal z table,

Using z-score formula  

x= z * +

x= - 0.84*18+96

x= 80.88

x=81

(B)

Using standard normal table,

P(Z > z) = 5%

= 1 - P(Z < z) = 0.05

= P(Z < z ) = 1 - 0.05

= P(Z < z ) = 0.95

= P(Z < 1.64 ) = 0.95

z = 1.64 (using standard normal (Z) table )

Using z-score formula  

x = z * +

x= 1.64*18+96

x= 125.52

x=126