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=15σ=15.

(a) What proportion of children aged 13 to 15 years old have scores on this test above 83 ? (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 5 percent of the distribution.
Answer:

Answer 1

We know that the mean of this distribution is 101 and the standard deviation is 15.

a.

We must find the proportion of students above 83.

We can do this by finding the proportion below 83, and subtracting the value from 1.

Let us first find the z-score associated with this value.

=(83-101) / 15

= -18 / 15

= -1.2

The p value value with this is 0.11507.

Thus, the proportion of children above 83= 1-0.11507

= 0.88493

b.

We must the score which marks the lowest 30%. Thus, the p-value should be 0.30.

We can look this up from the z table to get the z value.

The value is around -0.525.

Thus,

(x-101) / 15 = -0.525

x-101= -7.866008

Thus, x= 93.13399

Therefore, the score that marks the lowest 30 percent of the distribution is 93.13399.

c.

We must the score which marks the highest 5%. Thus, the p-value should be 0.05, and convert the sign to + in the end.

We can look this up from the z table to get the z value.

The value is around 1.645.

Thus,

(x-101) / 15 = 1.645

x-101= 24.675

Thus, x= 125.675

Therefore, the score that marks the highest 5 percent of the distribution is 125.675.