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A person must score in the upper 2% of the population on an IQ test to qualify for membership in Mensa, the international high-IQ society. If IQ scores are normally distributed with a mean of 100 and a standard deviation of 15, what score must a person have to qualify for Mensa?

## Answer

First we standardize the score with

z=(x-u)/o

where z is the distance in standard deviations from the mean, x is the score to be standardized, u is the mean of the population, and o is the standard deviation of the population [or sample].

Now, we can suspect from a percentile rank chart that the upper 2% is 2 standard deviations from the mean. In fact, however there is approximately only 100-2.27=97.73 within two standard deviations above the mean. Chances are, however, that your instructor only wants 2 standard deviations above the mean:

z=(x-100)/15

15z+100=x

This gives a score, x, as a function of its distance from the mean [z]. So if we want 2 standard deviations above the mean, 15(2)+100=130. You will need an IQ of 130 to get into MENSA accordingly. Strictly speaking, the score would be a little higher than 130 (in order for it to be ahead of 98% of the population) due to the reasoning above.

A person must score in the upper 2% of the population on an IQ test to qualify for membership in Mensa, the international high-IQ society (US Airways Attache, September 2000). There are 110,000 Mensa members in 100 countries throughout the world (Mensa International website, January 8, 2013). If IQ scores are normally distributed with a mean of 100 and a standard deviation of 15, what score must a person have to qualify for Mensa (to the nearest whole number)? | Find z for p (z ≤ ?) = .98; z=2.05 x = 2.05*15+100=130.75 Answer: 131 |

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