Question

sample grade point averages for ten male students and ten female students are listed. find the coefficient of variation for each of the two data sets. then compare the results.
males 2.4 3.8 3.8 3.8 2.8 2.7 3.8 3.3 3.8 1.7
females 2.6 3.7 2.2 4.1 3.6 3.9 2.2 3.8 3.7 2.3
the coefficient of variation for males is %
(round to one decimal place as needed.)

Explanation:

Step1: Calculate the mean for males

The formula for the mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\).
For males, \(x=\{2.4,3.8,3.8,3.8,2.8,2.7,3.8,3.3,3.8,1.7\}\), \(n = 10\)
\(\sum_{i=1}^{10}x_{i}=2.4 + 3.8+3.8+3.8+2.8+2.7+3.8+3.3+3.8+1.7=31.9\)
\(\bar{x}_{males}=\frac{31.9}{10}=3.19\)

Step2: Calculate the standard deviation for males

The formula for the sample standard deviation \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}\)
\((x_{1}-\bar{x})^{2}=(2.4 - 3.19)^{2}=(- 0.79)^{2}=0.6241\)
\((x_{2}-\bar{x})^{2}=(3.8 - 3.19)^{2}=(0.61)^{2}=0.3721\)
\((x_{3}-\bar{x})^{2}=(3.8 - 3.19)^{2}=0.3721\)
\((x_{4}-\bar{x})^{2}=(3.8 - 3.19)^{2}=0.3721\)
\((x_{5}-\bar{x})^{2}=(2.8 - 3.19)^{2}=(-0.39)^{2}=0.1521\)
\((x_{6}-\bar{x})^{2}=(2.7 - 3.19)^{2}=(-0.49)^{2}=0.2401\)
\((x_{7}-\bar{x})^{2}=(3.8 - 3.19)^{2}=0.3721\)
\((x_{8}-\bar{x})^{2}=(3.3 - 3.19)^{2}=(0.11)^{2}=0.0121\)
\((x_{9}-\bar{x})^{2}=(3.8 - 3.19)^{2}=0.3721\)
\((x_{10}-\bar{x})^{2}=(1.7 - 3.19)^{2}=(-1.49)^{2}=2.2201\)
\(\sum_{i = 1}^{10}(x_{i}-\bar{x})^{2}=0.6241+0.3721+0.3721+0.3721+0.1521+0.2401+0.3721+0.0121+0.3721+2.2201 = 5.109\)
\(s_{males}=\sqrt{\frac{5.109}{10 - 1}}=\sqrt{\frac{5.109}{9}}\approx\sqrt{0.5677}\approx0.753\)

Step3: Calculate the coefficient of variation for males

The formula for the coefficient of variation \(CV=\frac{s}{\bar{x}}\times100\%\)
\(CV_{males}=\frac{0.753}{3.19}\times100\%\approx23.6\%\)

Answer:

\(23.6\)