Practise sample mean, sample standard deviation and coefficient of variation. This page trains you to compare variability between data sets and decide which data set is more uniform.
A statistic is a measurable characteristic calculated from a sample. Since most engineering data are samples, we normally use sample mean and sample standard deviation.
The coefficient of variation is useful when we compare two or more data sets whose means are not exactly the same. It measures variation as a percentage of the mean.
For comparison, the data set with the smaller CV is treated as more consistent or more uniform.
\[ \bar{x}=\frac{\sum x}{n} \]
\[ s^2=\frac{\sum x^2-\frac{(\sum x)^2}{n}}{n-1} \]
\[ s=\sqrt{s^2} \]
\[ CV=\frac{s}{\bar{x}}\times 100\% \]
A semiconductor manufacturer produces devices used as central processing units in personal computers. The speed of the device, in Megahertz, is important because it determines the price that the manufacturer can charge for the devices. The following table contains measurements on 48 devices.
| 717 | 727 | 653 | 637 | 660 | 693 | 679 | 682 | 724 | 642 | 704 | 695 |
| 704 | 652 | 664 | 702 | 661 | 720 | 695 | 670 | 656 | 718 | 660 | 648 |
| 683 | 723 | 710 | 680 | 684 | 705 | 681 | 748 | 697 | 703 | 660 | 722 |
| 662 | 709 | 683 | 705 | 678 | 674 | 656 | 667 | 683 | 691 | 750 | 685 |
Find the sample mean, sample standard deviation and coefficient of variation.
Enter at least one answer first. Even a wrong attempt helps you remember the calculation method.
For the given data:
Sample mean:
\[ \bar{x}=\frac{\sum x}{n} \]
\[ \bar{x}=\frac{33002}{48} \]
\[ \bar{x}=687.54 \]
Sample variance:
\[ s^2=\frac{\sum x^2-\frac{(\sum x)^2}{n}}{n-1} \]
\[ s^2=\frac{22724906-\frac{33002^2}{48}}{47} \]
\[ s^2=737.36 \]
Sample standard deviation:
\[ s=\sqrt{s^2} \]
\[ s=\sqrt{737.36} \]
\[ s=27.15 \]
Coefficient of variation:
\[ CV=\frac{s}{\bar{x}}\times 100 \]
\[ CV=\frac{27.15}{687.54}\times 100 \]
\[ CV=3.95\% \]
As part of a study monitoring acid rain, measurements of sulfate deposits, in kg/hectare, are recorded for different locations on the Eastern Terai of Nepal. The results are listed for 15 recent and consecutive years.
| Year | Location 1 \(x\) | Location 2 \(y\) | Location 3 \(z\) |
|---|---|---|---|
| 1 | 11.94 | 13.09 | 7.96 |
| 2 | 11.28 | 10.88 | 12.84 |
| 3 | 10.38 | 12.19 | 7.38 |
| 4 | 8.00 | 10.75 | 7.26 |
| 5 | 12.12 | 17.21 | 10.12 |
| 6 | 10.27 | 10.26 | 8.89 |
| 7 | 14.80 | 15.49 | 11.60 |
| 8 | 13.52 | 11.61 | 9.02 |
| 9 | 10.55 | 10.53 | 7.78 |
| 10 | 9.81 | 12.50 | 8.70 |
| 11 | 11.27 | 9.94 | 10.50 |
| 12 | 12.12 | 11.21 | 9.95 |
| 13 | 11.68 | 9.71 | 15.59 |
| 14 | 11.77 | 9.37 | 10.54 |
| 15 | 17.29 | 13.87 | 13.64 |
Find the sample mean, sample standard deviation and coefficient of variation for each location. Then give your conclusion about variability and uniformity.
Try at least one location first. The important idea is to compare CV values, not only standard deviations.
For each location, use:
\[ \bar{x}=\frac{\sum x}{n} \]
\[ s^2=\frac{\sum x^2-\frac{(\sum x)^2}{n}}{n-1} \]
\[ s=\sqrt{s^2} \]
\[ CV=\frac{s}{\bar{x}}\times 100 \]
Location 1:
\[ \bar{x}=11.79 \]
\[ s=2.19 \]
\[ CV=18.59\% \]
Location 2:
\[ \bar{y}=11.91 \]
\[ s_y=2.23 \]
\[ CV_y=18.69\% \]
Location 3:
\[ \bar{z}=10.12 \]
\[ s_z=2.43 \]
\[ CV_z=24.04\% \]
Comparison of coefficient of variation:
\[ CV_x=18.59\% \]
\[ CV_y=18.69\% \]
\[ CV_z=24.04\% \]
Since Location 1 has the smallest CV, Location 1 is the most uniform. Location 3 has the largest CV, so it shows the greatest variability.