Question

(1 point) Scores on a certain intelligence test for children between ages 13 and 15 years are approximately normally distributed with µ=106μ=106 and σ=24σ=24. (a) What proportion of children aged 13 to 15 years old have scores on this test above 92 ? Answer: Round to four decimal places. (b) What score which marks the lowest 25 percent of the distribution? Answer: Round to two decimal places. (c) Enter the score that marks the highest 5 percent of the distribution. Answer: Round to two decimal places.

Answer 1

Answer:

a) 0.719 = 71.90% of children aged 13 to 15 years old have scores on this test above 92

b) A score of 89.8 marks the lowest 25 percent of the distribution

c) A score of 145.48 marks the highest 5 percent of the distribution

Step-by-step explanation:

Problems of normally distributed samples are 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 problem, we have that:

$\mu = 106, \sigma = 24$

(a) What proportion of children aged 13 to 15 years old have scores on this test above 92 ?

This is 1 subtracted by the pvalue of Z when X = 92. So

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

$Z = \frac{92 - 106}{24}$

$Z = -0.58$

$Z = -0.58$ has a pvalue of 0.2810

1 - 0.2810 = 0.719

0.719 = 71.90% of children aged 13 to 15 years old have scores on this test above 92

(b) What score which marks the lowest 25 percent of the distribution?

The 25th percentile, which is X when Z has a pvalue of 0.25. So it is X when Z = -0.675.

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

$-0.675 = \frac{X - 106}{24}$

$X - 106 = -0.675*24$

$X = 89.8$

A score of 89.8 marks the lowest 25 percent of the distribution

(c) Enter the score that marks the highest 5 percent of the distribution

The 100-5 = 95th percentile, which is X when Z has a pvalue of 0.95. So it is X when Z = 1.645

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

$1.645 = \frac{X - 106}{24}$

$X - 106 = 1.645*24$

$X = 145.48$

A score of 145.48 marks the highest 5 percent of the distribution