Step 1

We have to show whose shoe size shows maximum variability ic (man of female).

We will compute CV (coefficient of variation) for this purpase and accordingly give the result.

Firstly we will compute for males

\(\begin{array}{|c|c|}\hline \text{Shoe size (in cm)} & (xi-\bar{x})^{2} \\ \hline 25.7 & 0.9604 \\ \hline 26.7 & 0.0004 \\ \hline 25.7 & 0.9604 \\ \hline 26 & 0.4624 \\ \hline 26 & 0.4624 \\ \hline 27 & 0.1024 \\ \hline 27 & 0.1024 \\ \hline 27.6 & 0.8464 \\ \hline 28.4 & 4.6656 \\ \hline 28.4 & 4.6656 \\ \hline 25.4 & 1.6384 \\ \hline 24.8 & 3.5344 \\ \hline 26 & 0.4624 \\ \hline 26 & 0.4624 \\ \hline 27 & 0.1024 \\ \hline 27 & 0.1024 \\ \hline 29 & 5.3824 \\ \hline \end{array}\)

Total \(=435.7\)

\(Mean(\bar{x})=\frac{\text{Total}}{\text{Number of Observation}}\)

\(=\frac{453.7}{17}\)

\(=26.68\)

\(Variance(\sigma^{2})=\frac{1}{n}\sum(x_{i}-\bar{x})^{2}\)

\(=\frac{1}{17}\sum(x_{i}-\bar{x})^{2}\)

\(=\frac{26.7564}{17}\)

\(\cong1.57=\sqrt{1.57}\)

\(CV=\frac{\sigma}{\bar{x}}\times100\)

\(=\frac{\sqrt{1.57}}{26.68}\times100=4.696\)

Step 2

Now, we will compute for females

\(\begin{array}{|c|c|}\hline \text{Shoe size (in cm)} & (xi-\bar{x})^{2} \\ \hline 25.4 & 0.7569 \\ \hline 23.8 & 6.1009 \\ \hline 25.4 & 0.7569 \\ \hline 23.8 & 6.1009 \\ \hline 25.4 & 0.7569 \\ \hline 25.4 & 0.7569 \\ \hline 25.7 & 0.3249 \\ \hline 23.5 & 7.6729 \\ \hline 23.1 & 10.0489 \\ \hline 23.5 & 7.6729 \\ \hline 26.7 & 0.1849 \\ \hline 23.1 & 10.0489 \\ \hline 25.1 & 1.3689 \\ \hline 25.4 & 0.7569 \\ \hline 23.5 & 7.6729 \\ \hline 23.8 & 6.1009 \\ \hline 25.7 & 0.3249 \\ \hline 26.9 & 0.3969 \\ \hline 26 & 0.0729 \\ \hline 23.5 & 7.6729 \\ \hline 27 & 0.5329 \\ \hline 25.1 & 1.3689 \\ \hline 23.1 & 10.0489 \\ \hline 23.8 & 6.1009 \\ \hline 26 & 0.0729 \\ \hline 23.8 & 6.1009 \\ \hline 25.1 & 1.3689 \\ \hline 24.8 & 2.1609 \\ \hline 25.4 & 0.7569 \\ \hline 25.7 & 0.3249 \\ \hline 23.5 & 7.6729 \\ \hline \end{array}\)

Total \(=814.6\)

\(\bar{x}=\frac{814.6}{31}=26.27\)

\(\sigma^{2}=\frac{1}{n}\sum(x_{i}-\bar{x})^{2}\)

\(=\frac{112.0599}{31}=3.614\)

\(\sigma=\sqrt{3.614}=1.90\)

\(CV=\frac{1.90}{26.27}\times100\)

\(=7.23\)

Step 3

Answer (1): If we look at CV of the both males and females.

The more variability is in female shoe size as compared to male shoe size.

So option (B) is correct.

Answer (2): \((n-1)\) provides an umbiased estimate of population and allouss more variability when using a sample and gives a better mathematical estimate of the population.

Option (C) is correct.