Standard deviation (\sigma) is a measure of spread, which helps give us a meaningful estimate of the variability in a data set. It is a weighted average of the distance of each data point from the mean. A small standard deviation indicates that most scores are close to the mean, while a large standard deviation indicates that the scores are more spread out away from the mean value.

Note that the mean will determine where the scores are clustered, while the standard deviation tells us how tightly they are clustered. The two sets of data above have a similar mean but different standard deviations.

When comparing data sets, to find which data set had higher scores, we would choose the set with the higher mean. To find the set with the more consistent scores, we would choose the set with the lower standard deviation.

Examples

Example 1

Two machines A and B are producing chocolate bars with the following mean and standard deviation for the weight of the bars.

Machine	Mean (g)	Standard deviation (g)
\text{A}	52	1.5
\text{B}	56	0.65

What does a comparison of the mean of the two machines tell us?

Worked Solution

Create a strategy

Compare the means from the table.

Apply the idea

We can see that Machine B has a greater mean weight than Machine A. This means that Machine B generally produces heavier chocolate bars.

What does a comparison of the standard deviation of the two machines tell us?

Worked Solution

Create a strategy

Compare the standard deviations from the table.

Apply the idea

Machine B has a smaller standard deviation than Machine A. This means that Machine B produces chocolate bars with a more consistent weight.

Example 2

Han, a cricketer, has made scores of 52,\,20,\,68,\,70, and 150 in all his innings this season. In his next innings, he scores no runs.

What is the change in his season batting average before and after the sixth inning?

Worked Solution

Create a strategy

Subtract the old batting mean from the new batting mean.

To find the mean, we use the formula: \text{Mean}=\dfrac{\text{Sum of the scores}}{\text{Number of scores}}

Apply the idea

The old list is: 52,\,20,\,68,\,70,\,150.

The new list is: 52,\,20,\,68,\,70,\,150,\,0.

\displaystyle \text{Old mean}	\displaystyle =	\displaystyle \dfrac{52+20+68+70+150}{5}	Find the mean for 5 innings
	\displaystyle =	\displaystyle 72	Evaluate
\displaystyle \text{New mean}	\displaystyle =	\displaystyle \dfrac{52+20+68+70+150+0}{6}	Find the mean for 6 innings
	\displaystyle =	\displaystyle 60	Evaluate

\displaystyle \text{Change in mean}	\displaystyle =	\displaystyle 60 - 72	Subtract the means
	\displaystyle =	\displaystyle -12	Evaluate

This means that the batting average dropped by 12.

What is the change in his standard deviation before and after his sixth innings? Give your answer correct to two decimal places.

Worked Solution

Create a strategy

Subtract the old standard deviation from the new standard deviation.

Apply the idea

We use the same lists from part (a) to find each standard deviation.

\displaystyle \text{Change in } \sigma	\displaystyle =	\displaystyle \sigma_{\text{new}}-\sigma_{\text{old}}	Subtract the standard deviations
	\displaystyle =	\displaystyle 47.48-42.91	Calculate the standard deviations
	\displaystyle =	\displaystyle 4.57	Evaluate the subtraction

Idea summary

Outcomes

VCMSP372 (10a)

Calculate and interpret the mean and standard deviation of data and use these to compare data sets. Investigate the effect of individual data values including outliers, on the standard deviation

11.05 Comparisons using standard deviation

Ideas

Comparisons using standard deviation

Examples

Example 1

Example 2

Outcomes

VCMSP372 (10a)

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