A 10-year-old child scores 140 on an IQ test with a mean of 100 and a standard deviation of 15. Based on the normal distribution, you can draw the following conclusion about this child:

A. The child has scored 40% higher than an average 10-year-old.
B. Ten percent of 10-year-olds score higher than this child.
C. This child is as bright as an average 8-year-old.
D. This child scores in the gifted and talented range.

The 10-year-old's score of 140 on the IQ test is a lot greater than the mean (100). By calculating the Z-score, we find out that the child's score is 2.67 standard deviations above the mean, indicating a score in the top 1%, and therefore in the gifted and talented range.

Explanation:

In this scenario, the 10-year-old child has scored 140 on an IQ test, which is considerably higher than the average score of 100. The standard deviation of 15 measures the variability of scores, but to determine where this particular score stands, we need to use the concept of Z-scores.

The Z-score represents how many standard deviations the particular score or value is from the mean. In this case, 140-100 = 40, so the child's score is 40 points higher than the mean. We then divide these 40 points by the standard deviation of 15, resulting in a Z-score of approximately 2.67. This means that the child's score is 2.67 standard deviations above the mean score. In the context of IQ scores and normal distribution, scoring 2.67 standard deviations above the mean places this child in the top 1% of the population, therefore, the statement 'This child scores in the gifted and talented range' would be correct.

Learn more about Z-score here:

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