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 23.6 %.
(round to one decimal place as needed.)
the coefficient of variation for females is □%.
(round to one decimal place as needed.)

Explanation:

Step1: Calculate the mean of female data

The formula for the mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\).
For female data \(x=\{2.6,3.7,2.2,4.1,3.6,3.9,2.2,3.8,3.7,2.3\}\), \(n = 10\).
\(\sum_{i=1}^{10}x_{i}=2.6 + 3.7+2.2 + 4.1+3.6+3.9+2.2+3.8+3.7+2.3=32.1\)
\(\bar{x}=\frac{32.1}{10}=3.21\)

Step2: Calculate the standard deviation of female data

The formula for the sample standard deviation \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}\)
\((x_{1}-\bar{x})^{2}=(2.6 - 3.21)^{2}=(- 0.61)^{2}=0.3721\)
\((x_{2}-\bar{x})^{2}=(3.7 - 3.21)^{2}=(0.49)^{2}=0.2401\)
\((x_{3}-\bar{x})^{2}=(2.2 - 3.21)^{2}=(-1.01)^{2}=1.0201\)
\((x_{4}-\bar{x})^{2}=(4.1 - 3.21)^{2}=(0.89)^{2}=0.7921\)
\((x_{5}-\bar{x})^{2}=(3.6 - 3.21)^{2}=(0.39)^{2}=0.1521\)
\((x_{6}-\bar{x})^{2}=(3.9 - 3.21)^{2}=(0.69)^{2}=0.4761\)
\((x_{7}-\bar{x})^{2}=(2.2 - 3.21)^{2}=(-1.01)^{2}=1.0201\)
\((x_{8}-\bar{x})^{2}=(3.8 - 3.21)^{2}=(0.59)^{2}=0.3481\)
\((x_{9}-\bar{x})^{2}=(3.7 - 3.21)^{2}=(0.49)^{2}=0.2401\)
\((x_{10}-\bar{x})^{2}=(2.3 - 3.21)^{2}=(-0.91)^{2}=0.8281\)
\(\sum_{i = 1}^{10}(x_{i}-\bar{x})^{2}=0.3721+0.2401 + 1.0201+0.7921+0.1521+0.4761+1.0201+0.3481+0.2401+0.8281 = 5.489\)
\(s=\sqrt{\frac{5.489}{10 - 1}}=\sqrt{\frac{5.489}{9}}\approx\sqrt{0.6099}\approx0.781\)

Step3: Calculate the coefficient of variation for females

The formula for the coefficient of variation \(CV=\frac{s}{\bar{x}}\times100\%\)
\(CV=\frac{0.781}{3.21}\times100\%\approx24.3\%\)

Answer:

\(24.3\)