Answer :
Final answer:
Calculating the z-score for the top 14% of a normal distribution results in an IQ score of 116.2, which cuts off the top 14% from the other scores. Hence, the correct answer is a. 116.2.
Explanation:
To answer the question of finding the IQ score that cuts off the top 14% from the others in a normal distribution with a mean of 100 and a standard deviation of 15, we need to consider the properties of the normal distribution and the concept of Z-scores.
We know that in a normal distribution, certain percentages of scores fall within a certain number of standard deviations from the mean. For instance, roughly 68% of scores fall within one standard deviation of the mean, and about 95% fall within two standard deviations.
We can find the Z-score corresponding to the top 14% by referencing Z-score tables or using calculator functions. It is 1.08. We multiply that Z-score by the standard deviation (15) and then add the mean score (100).
Calculations are as follows: (1.08 * 15) + 100 = 116.2
Therefore, the correct answer is a. 116.2.
Learn more about Z-scores and normal distribution here:
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