Question 289:

Asked on July 25, 2008

Tags: Z-Score , IQ Score

1

Answer:

No answer provided yet.

You'll need to find two areas under the normal curve given the means of 90 and 109 and standard deviation 15. First find the normal deviate or z-score for the larger mean: (109-100)/15 = a z-score of .6. Now, find the area under the normal curve using the z-score to percentile calculator, choose 1-sided area. This provides an area of about 72.5%.  Now we subtract the smaller area, which following the same steps above is (90-100)/15= -.6667 and has area of ~25.2%. Subtracting the smaller gets us 72.5-25.2 = 47.3

So the average region of IQ scores between 90 and 109 has a about a 47.3% probability of being selected.  You can also visualize this area by taking a look at the interactive graph of the standard normal curve and enter in the IQ mean of 100 and SD of 15.