Answer:
For 2.5 standard deviations above the average, only 0.621% should qualify for membership at Mensa.
Explanation:
To answer this question, we need to make use of some concepts:
- The meaning of the <em>standard normal distribution</em>.
- The purpose of the <em>cumulative standard normal distribution</em>.
- The <em>z-scores</em>.
<h3>Standard normal distribution</h3>
We can obtain the probabilities for any normal distribution using the values of the <em>standard normal distribution</em>, which are given by the <em>cumulative distribution function</em>, and we can also consult all these values, by general, from the <em>cumulative standard normal table</em> using <em>z-scores</em>.
<h3>Z-scores</h3>
A z-score is a "transformation" of a "raw" value using the next formula:
(1)
Where
<em>x</em> is a normally distributed value to be transformed in a z-score.
.
.
Well, a <em>z-score</em>, as we can see from (1) represents <em>how far are the values from the population mean</em> in terms of <em>standard deviations</em>. When a z-score is <em>positive</em>, it means that the value of the raw score is <em>above</em> the mean, whereas a <em>negative</em> value indicates that the raw score is <em>below</em> the mean.
<h3>2.5 Standard Deviations and Solution</h3>
Therefore, <em>2.5 standard deviations above the mean</em> is a z-score = 2.5.
Consulting the cumulative standard normal table, a z-score = 2.5 has a cumulative probability P(z<2.5) = 0.99379 (from negative infinity to this value).
Then, to determine the probability for P(z>2.5), that is, the probability for those values greater than 2.5 standard deviations (Mensa membership), we need to subtract from 1 the probability P(z<2.5) (because the sum of all probabilities equals 1, and we are looking for the remaining probability), or mathematically:

Or P(z>2.5) = 0.621%.
In words, only 0.621% of people have IQs 2.5 standard deviations <em>above</em> the mean and should qualify for membership at Mensa.
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