the coefficient of variation cv describes the standard deviation as a percent of the mean. because it has no…

the coefficient of variation cv describes the standard deviation as a percent of the mean. because it has no units, you can use the coefficient of variation to compare data with different units. find the coefficient of variation for each sample data set. what can you conclude? cv = standard deviation / mean · 100% click the icon to view the data sets. cv_heights = % (round to the nearest tenth as needed.) data table heights 79 76 66 65 77 75 65 76 66 73 66 79 weights 200 189 214 206 221 222 218 218 180 181 187 185

the coefficient of variation cv describes the standard deviation as a percent of the mean. because it has no units, you can use the coefficient of variation to compare data with different units. find the coefficient of variation for each sample data set. what can you conclude? cv = standard deviation / mean · 100% click the icon to view the data sets. cv_heights = % (round to the nearest tenth as needed.) data table heights 79 76 66 65 77 75 65 76 66 73 66 79 weights 200 189 214 206 221 222 218 218 180 181 187 185

Answer

Explanation:

Step1: Calculate mean of heights

Let (x_{1},x_{2},\cdots,x_{n}) be the height - data values. (n = 12), and (\sum_{i = 1}^{12}x_{i}=66 + 79+66 + 73+76+65+75+77+65+66+76+79 = 865). The mean (\bar{x}{h}=\frac{\sum{i = 1}^{n}x_{i}}{n}=\frac{865}{12}\approx72.083).

Step2: Calculate standard - deviation of heights

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}{h})^{2}=(66 - 72.083)^{2}), ((x{2}-\bar{x}{h})^{2}=(79 - 72.083)^{2},\cdots). (\sum{i = 1}^{12}(x_{i}-\bar{x}{h})^{2}=(66 - 72.083)^{2}+(79 - 72.083)^{2}+\cdots+(79 - 72.083)^{2}=274.917). (s{h}=\sqrt{\frac{274.917}{11}}\approx4.99).

Step3: Calculate coefficient of variation for heights

The coefficient of variation (CV_{h}=\frac{s_{h}}{\bar{x}{h}}\times100%). (CV{h}=\frac{4.99}{72.083}\times100%\approx6.9%).

Step4: Calculate mean of weights

Let (y_{1},y_{2},\cdots,y_{n}) be the weight - data values. (n = 12), and (\sum_{i = 1}^{12}y_{i}=185 + 187+181+180+218+218+222+221+206+214+189+200 = 2421). The mean (\bar{x}{w}=\frac{\sum{i = 1}^{n}y_{i}}{n}=\frac{2421}{12}=201.75).

Step5: Calculate standard - deviation of weights

(\sum_{i = 1}^{12}(y_{i}-\bar{x}{w})^{2}=(185 - 201.75)^{2}+(187 - 201.75)^{2}+\cdots+(200 - 201.75)^{2}=1989.25). (s{w}=\sqrt{\frac{1989.25}{11}}\approx13.4).

Step6: Calculate coefficient of variation for weights

The coefficient of variation (CV_{w}=\frac{s_{w}}{\bar{x}{w}}\times100%). (CV{w}=\frac{13.4}{201.75}\times100%\approx6.6%).

Answer:

The coefficient of variation for heights is approximately (6.9%) and for weights is approximately (6.6%). We can conclude that the relative variability in heights is slightly higher than the relative variability in weights.