Solution Found!
Listed below are the salaries, in $000, for a sample of 15
Chapter 4, Problem 12E(choose chapter or problem)
a. Determine the mean, median, and the standard deviation.
b. Determine the coefficient of skewness using Pearson’s method.
c. Determine the coefficient of skewness using the software method.
Listed below are the salaries, in $000, for a sample of 15 chief financial officers in the electronics industry.
\(\begin{array}{|l|l|l|l|l|l|} \hline \$ 516.0 & \$ 548.0 & \$ 566.0 & \$ 534.0 & \$ 586.0 & \$ 529.0 \\ \hline 546.0 & 523.0 & 538.0 & 523.0 & 551.0 & 552.0 \\ \hline 486.0 & 558.0 & 574.0 & & & \\ \hline \end{array}\)
Questions & Answers
QUESTION:
a. Determine the mean, median, and the standard deviation.
b. Determine the coefficient of skewness using Pearson’s method.
c. Determine the coefficient of skewness using the software method.
Listed below are the salaries, in $000, for a sample of 15 chief financial officers in the electronics industry.
\(\begin{array}{|l|l|l|l|l|l|} \hline \$ 516.0 & \$ 548.0 & \$ 566.0 & \$ 534.0 & \$ 586.0 & \$ 529.0 \\ \hline 546.0 & 523.0 & 538.0 & 523.0 & 551.0 & 552.0 \\ \hline 486.0 & 558.0 & 574.0 & & & \\ \hline \end{array}\)
ANSWER:
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