IQ scores are standardized such that the population of scores has a mean of 100 and a variance of 225.

Question

IQ scores are standardized such that the population of scores has a mean of 100 and a variance of 225. Assuming IQ scores have a normal distribution, what proportion of scores will fall below 65?

What values occur in the top and bottom 9% of the distribution?

Answer 1

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Answer 2

Marnee,

 

In terms of the normal distribution, what we're looking for is that part of the normal curve represented by the area to the left of the score 65.

 

Unfortunately, we can't find that area using real scores; we need to convert real scores to normalized scores (z-scores).

 

To do that, use this formula:

 

z = (X - mean)/standard deviation (where X is the score in question)

 

For the score of 65:

 

z = (65 - 100)/15 (remember the standard deviation is the square root of the variance => √225 = 15)

 

z = -35/15

 

z = -2.33

 

Now, to find the value of the area (the proportion of scores) below the z-score of -2.33, you need to look this value up in a table of cumulative z-scores.  You can find such a table most anywhere online.

 

When you do, you'll find that the area of the curve less than -2.33 is .0099 or about .01 which is 1% or 1/100.

 

So, 1/100 of the scores will fall below 65.

 

Make sense?