Question

question #6
find the standard deviation of a group of men whose height in centimeters is given below.
mens heights in centimeters
178 175 176 173 171 178
178 173 184 169 179 180
173 177 179

3.9
4.36
4.57
4.08

Explanation:

Step1: Calculate the mean

Let the data - set be \(x_1,x_2,\cdots,x_n\). Here \(n = 18\), and \(\sum_{i = 1}^{18}x_i=178 + 175+176+173+171+178+178+173+184+169+179+180+173+177+179\)
\(\sum_{i = 1}^{18}x_i = 3150\)
The mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}=\frac{3150}{18}=175\)

Step2: Calculate the squared - differences

For each data - point \(x_i\), calculate \((x_i-\bar{x})^2\). For example, when \(x_1 = 178\), \((x_1-\bar{x})^2=(178 - 175)^2=9\).

Step3: Calculate the variance

The variance \(s^2=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}\)
\(\sum_{i = 1}^{18}(x_i - 175)^2=(178 - 175)^2+(175 - 175)^2+(176 - 175)^2+(173 - 175)^2+(171 - 175)^2+(178 - 175)^2+(178 - 175)^2+(173 - 175)^2+(184 - 175)^2+(169 - 175)^2+(179 - 175)^2+(180 - 175)^2+(173 - 175)^2+(177 - 175)^2+(179 - 175)^2\)
\(\sum_{i = 1}^{18}(x_i - 175)^2=9 + 0+1 + 4+16+9+9+4+81+36+16+25+4+4+16\)
\(\sum_{i = 1}^{18}(x_i - 175)^2=274\)
The variance \(s^2=\frac{274}{18 - 1}=\frac{274}{17}\approx16.1176\)

Step4: Calculate the standard deviation

The standard deviation \(s=\sqrt{s^2}=\sqrt{\frac{274}{17}}\approx4.0147\approx4.08\)

Answer:

4.08