The variability of IQ within a population, as with many other measurable characteristics subject to manifold influences (e.g. the height and weight of individuals), appears to follow what is referred to as the normal (gaussian) distribution, otherwise known colloquially as the the bell curve.
The area under this curve from one standard deviation below the mean to one standard deviation above the mean is very close to two thirds. More accurately it is about 0.6827.
So the answer to your question is 7000 x 0.6827 = 4779.
John Smith ---- Hi John Smith, Your answer works for the range of 85 to 115, but the question asked about the range of 85 to 120. I would tackle it this way: First, I would assess the Z-scores for both 85 and 120 with a mean of 100 and standard deviation of 15. The Z-score is the area under the normal curve including and to the left of the statistic in question. Or perhaps more clearly defined, it is the percent chance that your test statistic, or any statistic less than it, has of occurring in the given distribution. The Z-score for 85 in this distribution is 0.158655, and the Z-score for 120 is 0.908788. However, since the score for 120 accounts for everything less than and including 120, it already accounts for the chance of 85 occurring. So, since we know the chance of anything less than or including 85 occurring, we can subtract this from the Z-score for 120 to get the chance of finding a statistic within the range of 85-120. This percentage is 0.750133. Now, our sample size is 7000, and we know that 75.0133% of that sample falls within the range in question, so simply multiply the two to get the answer: 7000 x 0.750133 = approximately 5251 people.
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