Answer 6
- Assume the mean on an IQ test is 100 with a standard deviation of 16. Using a table of areas under the normal curve, estimate each of the following:
ANSWERS
To answer these, we need the formula for a standard (z) score. Recall that this is:
a. Find the percentage of individuals having an IQ of 130 or higher.
First compute the z-score
z = (130-100)/16 = 30/16 = 1.875 (round to 1.88).
Then look up the proporton in tail (Column C) for a z-score of 1.88. The proportion is .0301. Hence, about 3% have IQs of 130 or higher.
b. Find the proportion of individuals with an IQ of 90 or lower.
c. Find the percentage of individuals with IQs between 85 and 115.z = (90 - 100)/16 = -10/16 = -.625
According to the Unit Normal Curve Table (Appendix B), the proportion in the tail to the left (hence, below, Column C) a z-score of -.62 is .2676. Hence, a little less than 27% of the individuals have IQs lower than 90.
The z-score for an IQ of 115 is .94 (rounded from .9375); the z-score for an IQ of 85 is -.94.
The area to the left (below) a z-score of .94 is .8264 (Column B).The area to the left (below) a z-score of -94 is .1736 (Column C).
Hence the proportional area between -.94 and .94 is the difference between the two proportions (.8264 - .1736) or .6528.
d. Find the proportion of individuals with IQs between the first and third quartiles.
e. Find the percentage of individuals with IQs above 130.The first and third quartiles correspond the the 25th and 75th percentiles, respectively. Furthermore, percentiles are directly related to proportions (or areas) of a normal distribution: .25 (or 25%) of the distribution lies below the 25th percentile; .75 (or 75%) lies below the 75th percentile. Hence .5 lie between the 25th and 75th percentile. )
An IQ score of 130 corresponds to a z-score of 1.875 (or 1.87).
Using Column C in the Unit Normal Curve Table we see that the proportion above a z of 1.87 is .0397.
Hence, approximately 4% of individuals have IQ scores above 130.
f. Find the IQ of individuals at the 85th percentile.
An individual at the 85th percentile has a z-score of 1.04 (Note: since the 85th percentile is above the median we use Column B and scan down until we find .8508. The corresponding to z-score is 1.04).
Now, solve the z-score equation for X:
Then compute, X = 1.04(16) + 100 = 116.64, or, after rounding to the nearest whole score, an IQ of 117.
g. Find the IQ of individuals at the 35th percentile.
35% of the distribution lies below the 35th percentile. We can use the Table of Unit Normal Curve Table to determine that this corresponds to a z-score of -.38 (using Column C).
Next, using the formula above to solve for X, we obtain,
X = -.38(16) + 100 = 83.36, or an IQ of 94.