Saturday, March 7, 2009

Practice problems for standard deviation

The data for IQ test scores is normally distributed. The average IQ is 100 and the standard deviation is 15.

1. What are the z-scores for the following IQ scores?

a. 90
b. 112
c. 120

2. Using the answers from part 1 and the Positive and Negative z-score tables.

a. What percentage of the population has an IQ less than 90?

b. What percentage of the population has an IQ higher than 120?

c. What percentage of the population has an IQ between 112 and 120?

3. Percentiles

a. Find the z-score that corresponds to the 63rd percentile. Find the IQ that corresponds to the 63rd percentile, rounded to the nearest whole number.

b. Find the z-score that corresponds the 13th percentile. Find the IQ that corresponds to the 13th percentile, rounded to the nearest whole number.

Answers in the comments.

1 comment:

Prof. Hubbard said...

The data for IQ test scores is normally distributed. The average IQ is 100 and the standard deviation is 15.

1. What are the z-scores for the following IQ scores?

a. 90
(90-100)/15 = -.666... = -.67

b. 112
(112-100)/15 = .8

c. 120
(120-100)/15 = 1.333... = 1.33


2. Using the answers from part 1 and the Positive and Negative z-score tables.

a. What percentage of the population has an IQ less than 90?

row -0.6, column .07 on the Negative z-scores table: .2514 This mean 25.14% of the population has IQs at 90 or under.

b. What percentage of the population has an IQ higher than 120?
row 1.3, column .03 on the Positive z-scores table: .9082
This means that (100 - 90.82)% or 9.18% of the population has an IQ of 120 or more.


c. What percentage of the population has an IQ between 112 and 120?

row 0.8, column .00 on the Positive z-scores table: .7881
.9082 - .7881 = .1201 12.01% of the population has IQs between 112 and 120.


3. Percentiles

a. Find the z-score that corresponds to the 63rd percentile. Find the IQ that corresponds to the 63rd percentile, rounded to the nearest whole number.

The z-score 0.33 corresponds to .6293, the z-score 0.34 corresponds to .6331. Halfway in-between these two at z = 0.335, the percentile should be (.6293+.6331)/2 = .6312

.6293 is .0007 below the 63rd percentile
.6312 is .0012 above the 63rd percentile
.6331 is .0031 above the 63rd percentile

Since .6293 is the closest, we use 0.33 as the 63rd percentile z-score.

100 + 0.33*15 = 100 + 4.95 = 104.95, which rounds to 105, so the 63rd percentile IQ score is 105.



b. Find the z-score that corresponds the 13th percentile. Find the IQ that corresponds to the 13th percentile, rounded to the nearest whole number.


The z-score -1.12 corresponds to .1314, the z-score -1.13 corresponds to .1292. Halfway in-between these two at z = -1.125, the percentile should be (.1314+.1292)/2 = .1303

.1292 is .0008 below the 13th percentile
.1303 is .0003 above the 13th percentile
.1314 is .0014 above the 13th percentile

Since .1303 is the closest, we use -1.125 as the 13th percentile z-score.

100 + (-1.125*15) = 100 - 16.875 = 83.125, which rounds to 83, so the 13th percentile IQ score is 83.