Mean
and Standard Deviation

mean: the average obtained by dividing the sum of the data by the number of data in the set. Given a set of data, {x_{1}, x_{2}, x_{3}, ..., x_{n}}, you can find the mean, , using the following formula:
standard deviation: the extent to which data differs from the mean. The standard deviation is a measure of how the data is clustered about the mean. For large sets of data, approximately 68.3% of the data lies within one standard deviation of the mean and approximately 95.4% of the data lies within two standard deviations of the mean.
Note:
The standard deviation is like a distance: by definition, it is positive.
The standard deviation can be zero if all the data points are the same.
The standard deviation can be larger than the mean. The range of numbers has to be large for this to occur. One such sample is {0, 0, 1, 15, 20}, where = 7.2 and s 8.6.
If 5 were added to each number in a set of data, the mean would increase by 5 and the standard deviation would remain the same.
If each element in the set of data is multiplied by  3, the mean would be multiplied by  3 and the standard deviation would be multiplied by 3.
The formula for the standard deviation, , of this set is as follows:
In Pure Mathematics 30, you will not need to memorize the formula for standard deviation of a set of data. You will be using your graphing calculator instead.
zscore: the number of standard deviations a data value, x, lies from the mean. The formula to calculate zscore is .
range: upper extreme minus lower extreme
Distribution: On a provincewide mathematics exam, the mean was 65 and the standard deviation was 15. Describe the distribution of marks on the exam.
At Alpine Village, the annual snowfall (in centimetres) for the last 10 years has been 321, 267, 232, 304, 289, 342, 380, 403, 207, and 294. What was the mean snowfall and the standard deviation? Round your answers to the nearest tenth.
Solution
mean: = 65 and standard deviation: = 15
Approximately 68.3% of the students who wrote this exam had marks between 50 and 80.
Approximately 95.4% of the students who wrote this exam had marks between 35 and 95.
Solution
Enter the data into your graphing calculator as a list. Press (1:Edit), and enter the data.
Next, press .
The mean snowfall was 303.9 cm, and the standard deviation was approximately 58.0 cm.
If a number of data values are repeated, you can find the mean and standard deviation as in the preceding example or you can prepare two lists:
Determine the mean and standard deviation of the data in the following table. Round your answers to the nearest tenth.
A powerful way to visualize a set of data is to graph its frequency distribution as a histogram.
In the example on the left, frequency is shown in the second column.
Solution
Enter the data into your graphing calculator as a list. Press (1:Edit), and enter the data. Enter the data values as list L1 and their corresponding frequencies as list L2.
Note: 5 pairs of values, but if you include the frequency, there are 12 data values (the sum of the frequencies in L2).
After you have entered the lists, press the following.
The mean is 30, and the standard deviation is approximately 4.0.
Sample  not related to the original question
Histograms that represent data (such as height, weight, mathematical achievement, and physical strength) are typically bellshaped. If you look at the heights of students in grade 12, for example, most of the students will be clustered about the mean; however, there will be a few at the extremes.
Find the mean and standard deviation for {20, 40, 50, 60, 80, 100}. Compare the standard deviation to the list on the right.
Find the mean and standard deviation for {1, 5, 50, 98, 99, 100}. Compare the standard deviation to the list on the left.
Solution
Enter the data as a list.
Next, press the following.
For {20, 40, 50, 60, 80, 100}:
 The mean is 50.
 The standard deviation is approximately 31.62.
This set has values closer to the mean; so, it has a smaller standard deviation.
Solution
Enter the terms as a list to determine the standard deviation.
For {1, 5, 50, 98, 99, 100}:
 The mean is 50.
 The standard deviation is approximately 45.37.
This set has values furthest from the mean; so, it has the larger standard deviation.
Find the mean and standard deviation for {1, 3, 4, 5, 7}.
Find the mean and standard deviation for {6, 8, 9, 10, 12}. Compare the results with the list on the right.
Solution
The mean is 4, and the standard deviation is 2.
Solution
The mean is 9, and the standard deviation is 2.
The standard deviations of both sets are the same. Also, each element in this set is 5 more than the corresponding element in the set on the left.
Given the mean () and standard deviation (), calculate the zscore for raw scores (x) of 354, 357, 358, 352.
Given the mean () and standard deviation (), calculate the raw scores (x) for the zscores 3, and 3.
Solution
= 355 and = 1.5.
Solution
= 355 and = 1.5.
Given allowable applicant heights are 152 to 194 cm, calculate the zscore range when = 163 and = 7.
Calculate an applicant's zscore when = 163 and = 7, and x = 176.
Solution
= 163 and = 7.
The zscore range for the allowable heights was approximately  1.57 ≤ z ≤ 4.43.
Solution
= 163 and = 7, and x = 176.
The applicant’s zscore was about 1.86.
This work has been adapted from a Math 30 Pure learning resource originally produced and owned by Alberta Education