Practise sample mean, sample variance, sample standard deviation and coefficient of variation. Enter your answers, check them instantly, and use the detailed solution only after attempting the problem.
The arithmetic mean tells us the average level of the data. It gives an idea of how big or small the values are.
The standard deviation tells us how far the data values usually move away from the mean. A larger standard deviation means the data are more spread out.
The variance is the square of the standard deviation, so it magnifies the spread of the data.
The coefficient of variation compares the spread with the mean. It is very useful for checking consistency. The data set with the smaller CV is more consistent.
\[ \bar{x}=\frac{\sum x}{n} \]
\[ s^2=\frac{\sum x^2-\frac{(\sum x)^2}{n}}{n-1} \]
\[ s=\sqrt{s^2} \]
\[ CV=\frac{s}{\bar{x}}\times 100\% \]
The following table shows the number of hours 45 hospital patients slept following the administration of a certain anesthetic.
| 7 | 10 | 12 | 4 | 8 | 7 | 3 | 8 | 5 |
| 12 | 11 | 3 | 8 | 1 | 1 | 13 | 10 | 4 |
| 4 | 5 | 5 | 8 | 7 | 7 | 3 | 2 | 3 |
| 8 | 13 | 1 | 7 | 17 | 3 | 4 | 5 | 5 |
| 3 | 1 | 17 | 10 | 4 | 7 | 7 | 11 | 8 |
Find the sample mean, sample variance, sample standard deviation and coefficient of variation.
Even a wrong attempt will help your brain remember the method. Enter at least one answer first, or open the solution only if you are really stuck.
For the given data:
Sample mean:
\[ \bar{x}=\frac{\sum x}{n} \]
\[ \bar{x}=\frac{302}{45} \]
\[ \bar{x}=6.71 \]
Sample variance:
\[ s^2=\frac{\sum x^2-\frac{(\sum x)^2}{n}}{n-1} \]
\[ s^2=\frac{2740-\frac{302^2}{45}}{44} \]
\[ s^2=16.21 \]
Sample standard deviation:
\[ s=\sqrt{s^2} \]
\[ s=\sqrt{16.21} \]
\[ s=4.03 \]
Coefficient of variation:
\[ CV=\frac{s}{\bar{x}}\times 100 \]
\[ CV=\frac{4.03}{6.71}\times 100 \]
\[ CV=59.99\% \]
Following data reveals 50 samples of speed, in km/hr, of vehicles travelling in an intersection of a busy road. Estimate the average speed of vehicles. Also test consistency by applying a suitable formula.
| 54 | 65 | 58 | 59 | 53 | 67 | 68 | 59 | 64 | 68 |
| 69 | 72 | 75 | 48 | 44 | 42 | 53 | 52 | 51 | 49 |
| 48 | 46 | 54 | 52 | 50 | 51 | 64 | 45 | 42 | 49 |
| 54 | 71 | 72 | 58 | 59 | 52 | 55 | 55 | 56 | 54 |
| 54 | 68 | 66 | 65 | 67 | 67 | 65 | 65 | 64 | 64 |
Even a rough attempt is useful. Try entering at least one value first, then compare your work with the detailed solution.
For the given speed data:
Average speed:
\[ \bar{x}=\frac{\sum x}{n} \]
\[ \bar{x}=\frac{2902}{50} \]
\[ \bar{x}=58.04 \]
Sample variance:
\[ s^2=\frac{\sum x^2-\frac{(\sum x)^2}{n}}{n-1} \]
\[ s^2=\frac{172158-\frac{2902^2}{50}}{49} \]
\[ s^2=76.04 \]
Sample standard deviation:
\[ s=\sqrt{s^2} \]
\[ s=\sqrt{76.04} \]
\[ s=8.72 \]
Coefficient of variation:
\[ CV=\frac{s}{\bar{x}}\times 100 \]
\[ CV=\frac{8.72}{58.04}\times 100 \]
\[ CV=15.02\% \]
Interpretation: Since the CV is about \(15.02\%\), the speed data may be treated as fairly consistent. In general, a smaller CV indicates greater consistency.