Sample Variance & Standard Deviation | Introduction to Statistics and Data Analysis | Probability & Statistics for Engineers & Scientists | Walpole | Problem 1.7


Consider the drying time data for Exercise 1.1 on page 13. Compute the sample variance and sample standard deviation.


Solution:

We know, based on our answer in Exercise 1.1, that the sample mean is \displaystyle \overline{x}=3.787.

To compute for the sample variance, we shall use the formula

\displaystyle s^2=\sum _{i=1}^n\frac{\left(x_i-\overline{x}\right)^2}{n-1}\:

The formula states that we need to get the sum of \displaystyle \left(x_i-\overline{x}\right)^2, so we can use a table to solve \displaystyle \left(x_i-\overline{x}\right)^2 for every sample.

\displaystyle x\displaystyle x_i-\overline{x}\displaystyle \left(x_i-\overline{x}\right)^2
\displaystyle 3.4\displaystyle 3.4-3.787=-0.387\displaystyle \left(-0.387\right)^2=0.150
\displaystyle 2.5\displaystyle 2.5-3.787=-1.287\displaystyle \left(-1.287\right)^2=1.656
\displaystyle 4.8\displaystyle 4.8-3.787=1.013\displaystyle \left(1.013\right)^2=1.026
\displaystyle 2.9\displaystyle 2.9-3.787=-0.887\displaystyle \left(-0.887\right)^2=0.787
\displaystyle 3.6\displaystyle 3.6-3.787=-0.187\displaystyle \left(-0.187\right)^2=0.035
\displaystyle 2.8\displaystyle 2.8-3.787=-0.987\displaystyle \left(-0.987\right)^2=0.974
\displaystyle 3.3\displaystyle 3.3-3.787=-0.487\displaystyle \left(-0.487\right)^2=0.237
\displaystyle 5.6\displaystyle 5.6-3.787=1.813\displaystyle \left(1.813\right)^2=3.287
\displaystyle 3.7\displaystyle 3.7-3.787=-0.087\displaystyle \left(-0.087\right)^2=0.008
\displaystyle 2.8\displaystyle 2.8-3.787=-0.987\displaystyle \left(-0.987\right)^2=0.974
\displaystyle 4.4\displaystyle 4.4-3.787=0.613\displaystyle \left(0.613\right)^2=0.376
\displaystyle 4.0\displaystyle 4.0-3.787=0.213\displaystyle \left(0.213\right)^2=0.045
\displaystyle 5.2\displaystyle 5.2-3.787=1.413\displaystyle \left(1.413\right)^2=1.997
\displaystyle 3.0\displaystyle 3-3.787=-0.787\displaystyle \left(-0.787\right)^2=0.619
\displaystyle 4.8\displaystyle 4.8-3.787=1.013\displaystyle \left(1.013\right)^2=1.026
\displaystyle SUM \displaystyle 13.197

The table above shows that 

\displaystyle \sum _{i=1}^{15}=13.197

Therefore, the variance is

\displaystyle s^2=\sum _{i=1}^n\frac{\left(x_i-\overline{x}\right)^2}{n-1}=\frac{13.197}{15-1}=0.9426\:

The standard deviation is just the square root of the variance. That is

\displaystyle s=\sqrt{s^2}=\sqrt{0.9426}=0.971