Question

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

Answer 1

To determine which data set has the largest standard deviation you have to calculate and compare the standard deviation of each sample. The standard deviation is the square root of the variance, so first, you have to calculate the variance of each sample using the following formula:

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

Where

∑xi is the sum of the observations

∑xi² is the sum of the squares of the observations

n is the sample size

Sample 1

Variance

n=10

`\begin{gathered} \Sigma x_i=22+25+25+22+25+23+24+23+26+24 \\ \Sigma x_i=239 \end{gathered}`
`\begin{gathered} \Sigma x^2_i=22^2+25^2+25^2+22^2+25^2+23^2+24^2+23^2+26^2+24^2 \\ \Sigma x^2_i=5729 \end{gathered}`
`\begin{gathered} S^2_1=(1)/(10-1)\lbrack5729-((239)^2)/(10)\rbrack \\ S^2_1=(1)/(9)\lbrack5729-5712.1\rbrack \\ S^2_1=(1)/(9)16.9 \\ S^2_1=1.88 \end{gathered}`

Standard deviation

`\begin{gathered} S_1=\sqrt[]{S^2_1} \\ S_1=\sqrt[]{1.88} \\ S_1=1.37 \end{gathered}`

Sample 2

Variance

n=10

`\begin{gathered} \Sigma x_i=13+11+11+11+10+14+14+10+11+11 \\ \Sigma x_i=116 \end{gathered}`
`\begin{gathered} \Sigma x^2_i=13^2+11^2+11^2+11^2+10^2+14^2+14^2+10^2+11^2+11^2 \\ \Sigma x^2_i=1366 \end{gathered}`
`\begin{gathered} S^2_2=(1)/(10-1)\lbrack1366-((116)^2)/(10)\rbrack \\ S^2_2=(1)/(9)\lbrack1366-1345.6\rbrack \\ S^2_2=(1)/(9)\cdot20.4 \\ S^2_2=2.27 \end{gathered}`

Standard deviation

`\begin{gathered} S_2=\sqrt[]{S^2_2} \\ S_2=\sqrt[]{2.27} \\ S_2=1.51 \end{gathered}`

Sample 3

Variance

n=10

`\begin{gathered} \Sigma x_i=16+14+15+15+16+16+15+16+18+14 \\ \Sigma x_i=155 \end{gathered}`
`\begin{gathered} \Sigma x^2_i=16^2+14^2+15^2+15^2+16^2+16^2+15^2+16^2+18^2+14^2 \\ \Sigma x^2_1=2415 \end{gathered}`
`\begin{gathered} S^2_3=(1)/(10-1)\lbrack2415-((155)^2)/(10)\rbrack \\ S^2_3=(1)/(9)\lbrack2415-2402.5\rbrack \\ S^2_3=(1)/(9)\cdot12.5 \\ S^2_3=1.39 \end{gathered}`

Standard deviation

`\begin{gathered} S_3=\sqrt[]{S^2_3} \\ S_3=\sqrt[]{1.39} \\ S_3=1.18 \end{gathered}`

Sample 4

Variance

n=10

`\begin{gathered} \Sigma x_i=25+23+25+22+21+25+22+21+25+23 \\ \Sigma x_i=232 \end{gathered}`
`\begin{gathered} \Sigma x^2_i=25^2+23^2+25^2+22^2+21^2+25^2+22^2+21^2+25^2+23^2 \\ \Sigma x^2_i=5408 \end{gathered}`
`\begin{gathered} S^2_4=(1)/(10-1)\lbrack5408-((232)^2)/(10)\rbrack \\ S^2_4=(1)/(9)\lbrack5408-5382.4\rbrack \\ S^2_4=(1)/(9)\cdot25.6 \\ S^2_4=2.84 \end{gathered}`

Standard deviation

`\begin{gathered} S_4=\sqrt[]{S^2_4} \\ S_4=\sqrt[]{2.84} \\ S_4=1.69 \end{gathered}`

Sample 5

Variance

n=10

`\begin{gathered} \Sigma x_i=7+14+19+18+4+16+5+11+11+12 \\ \Sigma x_i=117 \end{gathered}`
`\begin{gathered} \Sigma x^2_i=7^2+14^2+19^2+18^2+4^2+16^2+5^2+11^2+11^2+12^2 \\ \Sigma x^2_i=1613 \end{gathered}`
`\begin{gathered} S^2_5=(1)/(10-1)\lbrack1613-((117)^2)/(10)\rbrack \\ S^2_5=(1)/(9)\lbrack1613-1368.9\rbrack \\ S^2_5=(1)/(9)\cdot244.1 \\ S^2_5=27.12 \end{gathered}`

Standard deviation

`\begin{gathered} S_5=\sqrt[]{S^2_5} \\ S_5=\sqrt[]{27.12} \\ S_5=5.21 \end{gathered}`

The standard deviations of the five samples are:

`\begin{gathered} S_1=1.37 \\ S_2=1.51 \\ S_3=1.18 \\ S_4=1.69 \\ S_5=5.21 \end{gathered}`

The sample with the largest standard deviation is the fifth one