Find the minimum IQ needed to be a Mensa member.
Mensa is an organization whose members possess IQs that are in the top 2% of the population. It is known that IQs are normally distributed with a mean of 100 and a standard deviation of 16.
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To answer this question, you have to have a table with z scores in front of you.
The z-score is based on a different normalized normal distribution. The mean in this case will be 0, and the standard deviation is 1. We can actually find the z-score for our case and use it to determine the desired value with the following formula (`mu` is our mean IQ and `sigma` is our IQ standard deviation, `L` is the lower limit for IQ's in the problem):
`L = mu + z*sigma`
Now, it's going to take a little bit of critical thinking to determine which z-score for which you're looking. We are trying to find an upper bound, so that 2% of the area under the normal distribution curve is above that IQ, and 98% of the area is below that IQ. Based on how many z-tables are set up, we will be looking for the z-score at which 98% of the area and behind that value.
If you look at the table in the link below, you'll see that our z-value must lie between 2.05 and 2.06. Based on the score being slightly closer to 2.05 (`2/5` of the distance between the two numbers, based on a linear approximation), we can approximate the correct z-score as 2.054.
Now, let's go back to our formula for the lower limit of IQ to get into Mensa. Let's plug our values into the equation to find L. Remember, `mu` will be the mean (100), `sigma` will be the standard deviation (16), and z will be our z-score (2.054).
`L = mu + z*sigma = 100 + 2.054*16 = 132.864`
So you must have an IQ of roughly 133 to get into Mensa! I hope that helps!
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