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Nimfa-mama [501]
3 years ago
8

Mensa is a society for "geniuses." one way to qualify for membership is having an iq at least 2.5 standard deviations above aver

age. iqs are approximately normally distributed, according to the wais scale. what percent of people should qualify?
Business
2 answers:
7nadin3 [17]3 years ago
7 0
According to the scale, an average person would posses the mean of 100 IQ <span>and standard deviation of  15. If to be a member of mensa one should have </span><span>an iq at least 2.5 standard deviations above average, the minimum iq should be: 
</span>
2.5 = (x-100)/15

x = 137.5 >>>>> Less than 1% population belong to this IQ group or higher.


EleoNora [17]3 years ago
5 0

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:

  1. The meaning of the <em>standard normal distribution</em>.
  2. The purpose of the <em>cumulative standard normal distribution</em>.
  3. 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:

\\ z = \frac{x - \mu}{\sigma} (1)

Where

<em>x</em> is a normally distributed value to be transformed in a z-score.

\\ \mu\;is\;the\;population\;mean.

\\ \sigma\;is\;the\;population\;standard\;deviation.

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:

\\ P(z>2.5) = 1 - P(z

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.

<h3 />

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