Listed below are the salaries, in $000, for a sample of 15

Step 1 of 3

a) We have to find mean, median and standard deviation for the given data

Arrange the given values in the table form

\(\begin{array}{|c|c|} \hline X & (X-\bar{X})^{2} \\ \hline 516 & 676 \\ \hline 548 & 36 \\ \hline 566 & 576 \\ \hline 534 & 64 \\ \hline 586 & 1936 \\ \hline 529 & 169 \\ \hline 546 & 16 \\ \hline 523 & 361 \\ \hline 538 & 16 \\ \hline 523 & 361 \\ \hline 551 & 81 \\ \hline 552 & 100 \\ \hline 486 & 3136 \\ \hline 558 & 256 \\ \hline 574 & 1024 \\ \hline \text { Total } & 8808 \\ \hline \end{array}\)

Here n=no.of value=15

Mean

\(\begin{aligned} \bar{X} & =\Sigma X / n \\ & =8130 / 15 \\ & =542 \end{aligned}\)

To find the median value arrange the values into ascending order

Median

\(\begin{aligned} (M) & =\left(\frac{n+1}{2}\right)^{\text {th }} \text { Value } \\ & =\left(\frac{15+1}{2}\right)^{\text {th }} \text { Value } \\ & =8^{\text {th }} \text { Value } \\ & =546 \end{aligned}\)

Standard deviation

\(\begin{aligned} (s) & =\sqrt{\frac{\sum(x-\bar{X})^{2}}{n-1}} \\ & =\sqrt{\frac{8808}{15-1}} \\ & =25.08 \end{aligned}\)

Hence Mean=542

Median=546

standard deviation =25.08