Question

(1 point) Scores on a certain intelligence test for children between ages 13 and 15 years are approximately normally distributed with 27 u= 106 and o (a) What proportion of children aged 13 to 15 years old have scores on this test above 92? (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 30 percent of the distribution Answer: (c) Enter the score which marks the highest 10 percent of the distribution Answer:

Answer 1

Solution :

Given that ,

mean = $\mu$ = 106

standard deviation = $\sigma$ = 27

a)

P(x > 92) = 1 - P(x < 92)

= 1 - P((x - $\mu$ ) / $\sigma$ < (92 - 106) / 27)

= 1 - P(z < -0.52)

= 1 - 0.3015 Using standard normal table.

= 0.6985

Probability = 0.6985

b)

The z - distribution of the 30 % is,

P( Z < z ) = 30%

P( Z < z ) = 0.30

P( Z < -0.52) = 0.30

z = -0.52

Using z - score formula,

X = z * $\sigma$ + $\mu$

= -0.52 * 27 + 106

= 91.96

The lowest 30 % value is 91.96

c)

The z - distribution of the 10 % is,

P( Z > z ) = 10 %

1 - P( Z < z ) = 0.10

P( Z <  ) = 1 - 0.10

P( Z < z ) = 0.90

P( Z < 1.28) = 0.90

z = 1.28

Using z - score formula,

X = z * $\sigma$ + $\mu$

= 1.28 * 27 + 106  

= 141.56

The highest 10% value is 141.56