http://www.ck12.org Chapter 6. Normal Distribution Curves
You can see that the mean of the test scores was approximately 80.9, while the standard deviation was approximately
12.1. This means that Justin’s test score was less than 1 standard deviation above the mean. To calculate the variance
of the test scores, square the exact value of the standard deviation, which is 12.09488592. This can be done as
follows:
Therefore, the variance of the test scores was approximately 146.3.
Practice
- If the standard deviation of a population is 6, the population variance is:
a. 2.44
b. 3
c. 6
d. 36 - If the variance of a sample is 25, the sample standard deviation is:
a. 5
b. 25
c. 125
d. 625 - Suppose data are normally distributed, with a mean of 50 and a standard deviation of 10. What would be the
variance?
a. 10
b. 40
c. 50
d. 100 - Suppose data in a population are normally distributed, with a variance of 64. What would be the population
standard deviation?
a. 4
b. 8
c. 16
d. 64 - What is the population variance of the following data values? 40 38 42 47 35
a. 4.03
b. 4.51
c. 16.24
d. 20.34 - What is the sample variance of the following data values? 114 120 138 125 119
a. 8.18