Question

what is the standard deviation of the following sample: 5, 17, 12, 11, 10, 14, 8?
hint: use the following.
s = \sqrt{\frac{\sum (x - \bar{x})^2}{n - 1}}
3.62
15.33
3.92
13.14
3.76

Explanation:

Step1: Calculate the mean

The data set is \(5, 17, 12, 11, 10, 14, 8\). The mean \(\bar{x}=\frac{5 + 17+12+11+10+14+8}{7}=\frac{77}{7} = 11\).

Step2: Calculate \((x - \bar{x})^2\) for each data - point

For \(x = 5\): \((5 - 11)^2=(-6)^2 = 36\)
For \(x = 17\): \((17 - 11)^2=6^2 = 36\)
For \(x = 12\): \((12 - 11)^2=1^2 = 1\)
For \(x = 11\): \((11 - 11)^2=0^2 = 0\)
For \(x = 10\): \((10 - 11)^2=(-1)^2 = 1\)
For \(x = 14\): \((14 - 11)^2=3^2 = 9\)
For \(x = 8\): \((8 - 11)^2=(-3)^2 = 9\)

Step3: Calculate \(\sum(x - \bar{x})^2\)

\(\sum(x - \bar{x})^2=36 + 36+1+0+1+9+9=92\)

Step4: Calculate the standard deviation

We know \(n = 7\), and the formula for sample standard deviation \(s=\sqrt{\frac{\sum(x - \bar{x})^2}{n - 1}}\)
\(s=\sqrt{\frac{92}{7 - 1}}=\sqrt{\frac{92}{6}}\approx\sqrt{15.33}\approx3.92\)

Answer:

3.92