Question

Which of the following lists of data has the smallest standard deviation?

Answer 1

To determine which data set has the smallest standard deviation you have to determine the standard deviation of each sample and then compare them. The standard deviation is the square root of the variance, which means that you have to calculate the variance of each sample and then their standard deviation.

The formula for the sample variance is:

`S^2=(1)/(n-1)\lbrack\Sigma x^2_i-((\Sigma x_i)^2)/(n)\rbrack`

Where

∑xi represents the sum of all observations of the sample

∑xi² represents the sum of the squares of all observations of the sample

n represents the sample size

Sample 1

Variance

n= 10

`\begin{gathered} \Sigma x_i=30+21+19+17+16+32+26+25+19+16 \\ \Sigma x_i=221 \end{gathered}`
`\begin{gathered} \Sigma x^2_i=30^2+21^2+19^2+17^2+16^2+32^2+26^2+25^2+19^2+16^2 \\ \Sigma x^2_i=5189 \end{gathered}`
`\begin{gathered} S^2_1=(1)/(10-1)\lbrack5189-((221)^2)/(10)\rbrack \\ S^2_1=(1)/(9)\lbrack5189-4884.1\rbrack \\ S^2_1=(1)/(9)\cdot304.9 \\ S^2_1=33.88 \end{gathered}`

Standard deviation

`\begin{gathered} S_1=\sqrt[]{S^2_1} \\ S_1=\sqrt[]{33.88} \\ S_1=5.82 \end{gathered}`

Sample 2

Variance

n=10

`\begin{gathered} \Sigma x_i=5+11+15+7+5+9+8+16+14+11 \\ \Sigma x_i=101 \end{gathered}`
`\begin{gathered} \Sigma x^2_i=5^2+11^2+15^2+7^2+5^2+9^2+8^2+16^2+14^2+11^2 \\ \Sigma x^2_i=1163 \end{gathered}`
`\begin{gathered} S^2_2=(1)/(10-1)\lbrack1163-((101)^2)/(10)\rbrack \\ S^2_2=(1)/(9)\lbrack1163-1020.1\rbrack \\ S^2_2=(1)/(9)\cdot142.9 \\ S^2_2=15.88 \end{gathered}`

Standard deviation

`\begin{gathered} S_2=\sqrt[]{S^2_2} \\ S_2=\sqrt[]{15.88} \\ S_2=3.98 \end{gathered}`

Sample 3

Variance

n=10

`\begin{gathered} \Sigma x_i=25+24+28+18+32+34+34+22+28+19 \\ \Sigma x_i=264 \end{gathered}`
`\begin{gathered} \Sigma x^2_i=25^2+24^2+28^2+18^2+32^2+34^2+34^2+22^2+28^2+19^2 \\ \Sigma x^2_i=7274 \end{gathered}`
`\begin{gathered} S^2_3=(1)/(10-1)\lbrack7274-((264)^2)/(10)\rbrack \\ S^2_3=(1)/(9)\lbrack7274-6969.6\rbrack \\ S^2_3=(1)/(9)\cdot304.4 \\ S^2_3=33.82 \end{gathered}`

Standard deviation

`\begin{gathered} S_3=\sqrt[]{S^2_3} \\ S_3=\sqrt[]{33.82} \\ S_3=5.82 \end{gathered}`

Sample 4

Variance

n=10

`\begin{gathered} \Sigma x_i=17+19+17+18+17+16+16+16+17+20 \\ \Sigma x_i=173 \end{gathered}`
`\begin{gathered} \Sigma x^2_i=17^2+19^2+17^2+18^2+17^2+16^2+16^2+16^2+17^2+20^2 \\ \Sigma x^2_i=3009 \end{gathered}`
`\begin{gathered} S^2_4=(1)/(10-1)\lbrack3009-((173)^2)/(10)\rbrack \\ S^2_4=(1)/(9)\lbrack3009-2992.9\rbrack \\ S^2_4=(1)/(9)\cdot16.1 \\ S^2_4=1.79 \end{gathered}`

Standard deviation

`\begin{gathered} S_4=\sqrt[]{S^2_4} \\ S_4=\sqrt[]{1.79} \\ S_4=1.34 \end{gathered}`

Sample 5

Variance

n=10

`\begin{gathered} \Sigma x_i=9+16+14+22+20+9+19+16+21+8 \\ \Sigma x_i=154 \end{gathered}`
`\begin{gathered} \Sigma x^2_i=9^2+16^2+14^2+22^2+20^2+9^2+19^2+16^2+21^2+8^2 \\ \Sigma x^2_i=2620 \end{gathered}`
`\begin{gathered} S^2_5=(1)/(10-1)\lbrack2620-((154)^2)/(10)\rbrack \\ S^2_5=(1)/(9)\lbrack2620-2371.6\rbrack \\ S^2_5=(1)/(9)\cdot248.4 \\ S^2_5=27.60 \end{gathered}`

Standard deviation

`\begin{gathered} S_5=\sqrt[]{S^2_5} \\ S_5=\sqrt[]{27.60} \\ S_5=5.25 \end{gathered}`

So, the standard deviation of the five samples are:

`\begin{gathered} S_1=5.82 \\ S_2=3.98 \\ S_3=5.82 \\ S_4=1.34 \\ S_5=5.25 \end{gathered}`

Sample 4 has the smallest standard deviation.