Question

Scores on a certain intelligence test for children between ages 13 and 15 years are approximately normally distributed with MEAN = 106 and standard deviation = 15.

(a) What proportion of children aged 13 to 15 years old have scores on this test above 96 ? (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 25 percent of the distribution.
Answer:
(c) Enter the score which marks the highest 10 percent of the distribution.
Answer:

Answer 1

The proportion of children aged 13 to 15 with IQ scores above 96 is approximately 0.7486. The score marking the lowest 25% is around 96, and the score marking the highest 10% is approximately 125.

Step-by-step explanation:

To solve these problems involving the normal distribution of IQ scores, we'll use the properties of the normal curve where the mean score is 106 and the standard deviation is 15.

Part (a)

To find the proportion of children with scores above 96, we calculate the z-score and refer to standard normal distribution tables or use a statistical software:

Z = (X - mean) / SD = (96 - 106) / 15 ≈ -0.67

Looking up the z-score of -0.67 on the standard normal distribution table or software gives the cumulative probability below 96 as 0.2514. Therefore, the proportion above 96 is 1 - 0.2514 = 0.7486.

Part (b)

To find the score that marks the lowest 25 percent, we look for a z-score that correspond to a cumulative probability of 0.25. This z-score is approximately -0.675. Using the z-score formula:

X = mean + (z * SD) = 106 + (-0.675 * 15) ≈ 96.

Part (c)

To find the score for the highest 10 percent, we need the z-score corresponding to the cumulative probability of 0.90. This z-score is approximately 1.28. Using the z-score formula:

X = mean + (z * SD) = 106 + (1.28 * 15) ≈ 125.