The number of observations in the data is n = 10.
Determine the mean of the data.
[tex]\begin{gathered} \mu=\frac{514+508+502+497+495+507+458+477+464+515}{10} \\ =\frac{4937}{10} \\ =493.7 \end{gathered}[/tex]So mean of the data is 493.7.
Determine the sum of square difference between observation and mean.
[tex]\begin{gathered} \sum ^n_{i\mathop=1}(x_i-\mu)^2=(514-493.7)^2+(508-493.7)^2+(502-493.7)^2+(497-493.7)^2 \\ +(495-493.7)^2+(507-493.7)^2+(458-493.7)^2+(477-493.7)^2+(464-493.7)^2+(515-493.7)^2 \end{gathered}[/tex][tex]=412.09+204.49+68.89+10.89+1.69+176.89+1274.49+278.89+882.09+453.69[/tex][tex]=3764.1[/tex]The formula for the standard deviation is,
[tex]\sigma=\sqrt[]{\frac{\sum ^n_{i=1}(x_i-\mu)^2}{n-1}}[/tex]Substitute the value in the formula to determine the standard deviation of the data.
[tex]\begin{gathered} \sigma=\sqrt[]{\frac{3764.1}{10-1}} \\ =\sqrt[]{\frac{3764.1}{9}} \\ =20.4507 \\ \approx20.45 \end{gathered}[/tex]Answer:
Mean: 493.7
Standard deviation: 20.45
