the coefficient of variation cv describes the standard variation for each sample data set. what can you…
the coefficient of variation cv describes the standard variation for each sample data set. what can you co\ncv = \\frac{standard deviation}{mean} \\cdot 100% \nclick the icon to view the data sets.\ncv_{heights} = % (round to the nearest tenth as nee\ndata table\n| heights | weights |\n| ---- | ---- |\n| 77 | 194 |\n| 73 | 227 |\n| 79 | 223 |\n| 73 | 187 |\n| 74 | 179 |\n| 77 | 190 |\n| 80 | 170 |\n| 68 | 191 |\n| 80 | 221 |\n| 67 | 206 |\n| 72 | 203 |\n| 75 | 206 |
Answer
Explanation:
Step1: Calculate the mean of heights
The mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$, where $x_{i}$ are the height - values and $n = 15$. $\sum_{i=1}^{15}x_{i}=77 + 73+79+73+74+77+80+68+80+67+72+75 = 1093$ $\bar{x}=\frac{1093}{15}\approx72.87$
Step2: Calculate the standard - deviation of heights
The formula for the sample standard deviation $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}$. First, calculate $(x_{i}-\bar{x})^{2}$ for each $x_{i}$: $(77 - 72.87)^{2}=(4.13)^{2}=17.0569$ $(73 - 72.87)^{2}=(0.13)^{2}=0.0169$ $\cdots$ $\sum_{i = 1}^{15}(x_{i}-\bar{x})^{2}=293.7333$ $s=\sqrt{\frac{293.7333}{14}}\approx4.58$
Step3: Calculate the coefficient of variation
Using the formula $CV=\frac{s}{\bar{x}}\times100%$, substitute $s\approx4.58$ and $\bar{x}\approx72.87$. $CV=\frac{4.58}{72.87}\times100%\approx6.3%$
Answer:
$6.3$