Question

Scores on a certain intelligence test for children between ages 13 and 15 years are approximately normally distributed with μ=101 and σ=24. (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 1

Answer:

a) 0.7734

Step-by-step explanation:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

$\mu = 101, \sigma = 24$

(a) what proportion of children aged 13 to 15 years old have scores on this test above 83 ?

This is 1 subtracted by the pvalue of Z when X = 83. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{83 - 101}{24}$

$Z = -0.75$

$Z = -0.75$ has a pvalue of 0.2266

1 - 0.2266 = 0.7734

The answer is 0.7734