# 1. Who seems to have more variability in their shoe sizes, men or women? a) Men b) Women c) Neither group show variability d) Flag this Question 2. In

1. Who seems to have more variability in their shoe sizes, men or women?
a) Men
b) Women
c) Neither group show variability
d) Flag this Question
2. In general, why use the estimate of $n-1$ rather than n in the computation of the standard deviation and variance?
a) The estimate n-1 is better because it is used for calculating the population variance and standard deviation
b) The estimate n-1 is never used to calculate the sample variance and standard deviation
c) $n-1$ provides an unbiased estimate of the population and allows more variability when using a sample and gives a better mathematical estimate of the population
d) The estimate n-1 is better because it is use for calculation of both the population and sample variance as well as standard deviation.
$\begin{array}{|cc|}\hline \text{Shoe Size (in cm)}& \text{Gender (M of F)}\\ 25.7& M\\ 25.4& F\\ 23.8& F\\ 25.4& F\\ 26.7& M\\ 23.8& F\\ 25.4& F\\ 25.4& F\\ 25.7& M\\ 25.7& F\\ 23.5& F\\ 23.1& F\\ 26& M\\ 23.5& F\\ 26.7& F\\ 26& M\\ 23.1& F\\ 25.1& F\\ 27& M\\ 25.4& F\\ 23.5& F\\ 23.8& F\\ 27& M\\ 25.7& F\\ \hline\end{array}$
$\begin{array}{|cc|}\hline \text{Shoe Size (in cm)}& \text{Gender (M of F)}\\ 27.6& M\\ 26.9& F\\ 26& F\\ 28.4& M\\ 23.5& F\\ 27& F\\ 25.1& F\\ 28.4& M\\ 23.1& F\\ 23.8& F\\ 26& F\\ 25.4& M\\ 23.8& F\\ 24.8& M\\ 25.1& F\\ 24.8& F\\ 26& M\\ 25.4& F\\ 26& M\\ 27& M\\ 25.7& F\\ 27& M\\ 23.5& F\\ 29& F\\ \hline\end{array}$
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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
$\begin{array}{|cc|}\hline \text{Shoe size (in cm)}& \left(xi-\overline{x}{\right)}^{2}\\ 25.7& 0.9604\\ 26.7& 0.0004\\ 25.7& 0.9604\\ 26& 0.4624\\ 26& 0.4624\\ 27& 0.1024\\ 27& 0.1024\\ 27.6& 0.8464\\ 28.4& 4.6656\\ 28.4& 4.6656\\ 25.4& 1.6384\\ 24.8& 3.5344\\ 26& 0.4624\\ 26& 0.4624\\ 27& 0.1024\\ 27& 0.1024\\ 29& 5.3824\\ \hline\end{array}$
Total $=435.7$
$Mean\left(\overline{x}\right)=\frac{\text{Total}}{\text{Number of Observation}}$
$=\frac{453.7}{17}$
$=26.68$
$Variance\left({\sigma }^{2}\right)=\frac{1}{n}\sum \left({x}_{i}-\overline{x}{\right)}^{2}$
$=\frac{1}{17}\sum \left({x}_{i}-\overline{x}{\right)}^{2}$
$=\frac{26.7564}{17}$
$\cong 1.57=\sqrt{1.57}$
$CV=\frac{\sigma }{\overline{x}}×100$
$=\frac{\sqrt{1.57}}{26.68}×100=4.696$
Step 2
Now, we will compute for females
$\begin{array}{|cc|}\hline \text{Shoe size (in cm)}& \left(xi-\overline{x}{\right)}^{2}\\ 25.4& 0.7569\\ 23.8& 6.1009\\ 25.4& 0.7569\\ 23.8& 6.1009\\ 25.4& 0.7569\\ 25.4& 0.7569\\ 25.7& 0.3249\\ 23.5& 7.6729\\ 23.1& 10.0489\\ 23.5& 7.6729\\ 26.7& 0.1849\\ 23.1& 10.0489\\ 25.1& 1.3689\\ 25.4& 0.7569\\ 23.5& 7.6729\\ 23.8& 6.1009\\ 25.7& 0.3249\\ 26.9& 0.3969\\ 26& 0.0729\\ 23.5& 7.6729\\ 27& 0.5329\\ 25.1& 1.3689\\ 23.1& 10.0489\\ 23.8& 6.1009\\ 26& 0.0729\\ 23.8& 6.1009\\ 25.1& 1.3689\\ 24.8& 2.1609\\ 25.4& 0.7569\\ 25.7& 0.3249\\ 23.5& 7.6729\\ \hline\end{array}$
Total $=814.6$
$\overline{x}=\frac{814.6}{31}=26.27$
${\sigma }^{2}=\frac{1}{n}\sum \left({x}_{i}-\overline{x}{\right)}^{2}$
$=\frac{112.0599}{31}=3.614$
$\sigma =\sqrt{3.614}=1.90$
$CV=\frac{1.90}{26.27}×100$
$=7.23$
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): $\left(n-1\right)$ 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.