Question

cv = \frac{standard deviation}{mean} \cdot 100% click the icon to view the data sets. cv_{heights}=5.8% (round to the nearest tenth as ne cv_{weights}=17.1% (round to the nearest tenth as heights weights 77 194 73 227 79 223 73 187 74 179 77 190 80 170 68 191 80 221 67 206 72 203 75 206

Explanation:

Step1: Calculate mean of heights

$\bar{x}_{heights}=\frac{77 + 73+79+73+74+77+80+68+80+67+72+75}{12}=\frac{885}{12}=73.75$

Step2: Calculate standard - deviation of heights

First, find the squared differences from the mean:
$(77 - 73.75)^2+(73 - 73.75)^2+(79 - 73.75)^2+(73 - 73.75)^2+(74 - 73.75)^2+(77 - 73.75)^2+(80 - 73.75)^2+(68 - 73.75)^2+(80 - 73.75)^2+(67 - 73.75)^2+(72 - 73.75)^2+(75 - 73.75)^2$
$=3.25^2+(- 0.75)^2+5.25^2+(-0.75)^2+0.25^2+3.25^2+6.25^2+(-5.75)^2+6.25^2+(-6.75)^2+(-1.75)^2+1.25^2$
$ = 10.5625+0.5625 + 27.5625+0.5625+0.0625+10.5625+39.0625+33.0625+39.0625+45.5625+3.0625+1.5625$
$=211.25$
The standard - deviation $s_{heights}=\sqrt{\frac{211.25}{12 - 1}}=\sqrt{\frac{211.25}{11}}\approx4.39$

Step3: Calculate coefficient of variation of heights

$CV_{heights}=\frac{s_{heights}}{\bar{x}_{heights}}\times100\%=\frac{4.39}{73.75}\times100\%\approx5.8\%$

Step4: Calculate mean of weights

$\bar{x}_{weights}=\frac{194 + 227+223+187+179+190+170+191+221+206+203+206}{12}=\frac{2397}{12}=199.75$

Step5: Calculate standard - deviation of weights

First, find the squared differences from the mean:
$(194 - 199.75)^2+(227 - 199.75)^2+(223 - 199.75)^2+(187 - 199.75)^2+(179 - 199.75)^2+(190 - 199.75)^2+(170 - 199.75)^2+(191 - 199.75)^2+(221 - 199.75)^2+(206 - 199.75)^2+(203 - 199.75)^2+(206 - 199.75)^2$
$=(-5.75)^2+27.25^2+23.25^2+(-12.75)^2+(-20.75)^2+(-9.75)^2+(-29.75)^2+(-8.75)^2+21.25^2+6.25^2+3.25^2+6.25^2$
$=33.0625 + 742.5625+540.5625+162.5625+430.5625+95.0625+885.0625+76.5625+451.5625+39.0625+10.5625+39.0625$
$=3566.25$
The standard - deviation $s_{weights}=\sqrt{\frac{3566.25}{12 - 1}}=\sqrt{\frac{3566.25}{11}}\approx32.09$

Step6: Calculate coefficient of variation of weights

$CV_{weights}=\frac{s_{weights}}{\bar{x}_{weights}}\times100\%=\frac{32.09}{199.75}\times100\%\approx16.1\%$ (There may be a small difference in rounding compared to the given value of 17.1% due to different rounding methods in intermediate steps)

Answer:

$CV_{heights}\approx5.8\%$, $CV_{weights}\approx16.1\%$ (approximate values based on calculations above)