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Answer :
Final answer:
The shortest interval that covers 50% of scores on the Wechsler Adult Intelligence Scale, given that these scores are normally distributed with a mean of 100 and a standard deviation of 15, is approximately from 90 to 110.
Explanation:
The question is asking for the shortest interval that covers 50% of scores on the Wechsler Adult Intelligence Scale (WAIS), given that these scores are normally distributed with a mean (μ) of 100 and a standard deviation (σ) of 15.
In a normal distribution, about 68% of scores lie within one standard deviation of the mean. Approximately 95% of scores are within two standard deviations, and nearly all (99.7%) are within three standard deviations. However, we are looking for the interval that covers 50% of the scores. This range falls within approximately 0.675 standard deviations of the mean, as this is the z-score corresponding to a cumulative probability of 0.75 (which is the mean of 0.50 and 1, since we are looking for the middle 50% of scores).
By using the formula X = μ + zσ, where X represents the value on the WAIS, we get the values X = 100 + 0.675(15) and X = 100 - 0.675(15), which simplifies to around 110.13 and 89.87 respectively. Therefore, the shortest interval covering 50% of WAIS scores is approximately from 90 to 110, which is option b) 90 to 110.
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