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

Using z-scores and the properties of a normal distribution, it was calculated that approximately 71.90% of children score above 92. The score marking the lowest 25 percent is approximately 90.20, and the score that marks the highest 5 percent of the distribution is around 145.88.

Step-by-step explanation:

To solve the problems about normal distribution and interpreting IQ scores, we use the properties of the normal curve and z-scores. Z-scores help us understand how far away a particular score is from the mean, in terms of standard deviations.

Part (a): Proportion of Children With Scores Above 92

We first calculate the z-score for 92 using the formula: z = (X - μ) / σ, where X is the score, μ is the mean, and σ is the standard deviation. With μ = 106 and σ = 24, the z-score for 92 is (92 - 106) / 24 = -0.5833. Using a standard normal distribution table, we find that the proportion of children scoring above 92 corresponds to the area to the right of the z-score, which is approximately 0.7190. Therefore, the proportion of children aged 13 to 15 with scores above 92 is 0.7190.

Part (b): Lowest 25 Percent of the Distribution

The score marking the lowest 25 percent of the distribution corresponds to the 25th percentile or a z-score of about -0.675. We convert this z-score back to the original scale using the formula: X = μ + zσ, which yields X = 106 + (-0.675)(24) = 90.20. Thus, the score marking the lowest 25 percent is approximately 90.20.

Part (c): Highest 5 Percent of the Distribution

To find the score that marks the highest 5 percent, we locate the z-score that corresponds to the 95th percentile, which is about 1.645. Applying the conversion formula, we get X = 106 + (1.645)(24) = 145.88. Therefore, the score marking the highest 5 percent is approximately 145.88.

Answer 2

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 = (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 = (X - \mu)/(\sigma)`

`Z = (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 = (X - \mu)/(\sigma)`

`-0.675 = (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 = (X - \mu)/(\sigma)`

`1.645 = (X - 106)/(24)`

`X - 106 = 1.645*24`

`X = 145.48`

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