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When the Stanford‐Binet IQ test came into use in 1932, it was adjusted so that scores for each age group of children followed roughly the Normal distribution with mean 100 and standard deviation 15 . The test is readjusted from time to time to keep the mean at 100 . If present‐day American children took the 1932 Stanford‐Binet test, their mean score would be about 120 . The reasons for the increase in IQ over time are not known but probably include better childhood nutrition and more experience in taking tests. IQ scores above 130 are often called "very superior." What percentage of children had very superior scores in 1932? (Enter your answer rounded to two decimal places.)

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Answer:

2.28% of children had very superior scores in 1932.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

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

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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:

[tex]\mu = 100, \sigma = 15[/tex]

IQ scores above 130 are often called "very superior." What percentage of children had very superior scores in 1932?

This is 1 subtracted by the pvalue of Z when X = 130. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{130 - 100}{15}[/tex]

[tex]Z = 2[/tex]

[tex]Z = 2[/tex] has a pvalue of 0.9772.

So 1-0.9772 = 0.0228 = 2.28% of children had very superior scores in 1932.

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