Question

a sample of blood - pressure measurements is taken for a group of adults, and those values (mm hg) are listed below. the values are matched so that 10 subjects each have a systolic and diastolic measurement. find the coefficient of variation for each of the two samples, then compare the variation. systolic 117 128 156 98 155 123 118 137 127 121 diastolic 79 77 75 53 91 89 58 66 70 81 the coefficient of variation for the systolic measurements is 13.7 %. (type an integer or decimal rounded to one decimal place as needed.) the coefficient of variation for the diastolic measurements is % (type an integer or decimal rounded to one decimal place as needed.)

Explanation:

Step1: Calculate the mean of diastolic measurements

Let the diastolic measurements be $x_1 = 79,x_2 = 77,x_3 = 75,x_4 = 53,x_5 = 91,x_6 = 89,x_7 = 58,x_8 = 66,x_9 = 70,x_{10}=81$.
The mean $\bar{x}=\frac{\sum_{i = 1}^{10}x_i}{n}$, where $n = 10$.
$\sum_{i=1}^{10}x_i=79 + 77+75 + 53+91+89+58+66+70+81=749$.
So, $\bar{x}=\frac{749}{10}=74.9$.

Step2: Calculate the standard - deviation of diastolic measurements

The formula for the sample standard - deviation $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}}$.
$(x_1-\bar{x})^2=(79 - 74.9)^2=4.1^2 = 16.81$
$(x_2-\bar{x})^2=(77 - 74.9)^2=2.1^2 = 4.41$
$(x_3-\bar{x})^2=(75 - 74.9)^2=0.1^2 = 0.01$
$(x_4-\bar{x})^2=(53 - 74.9)^2=(-21.9)^2 = 479.61$
$(x_5-\bar{x})^2=(91 - 74.9)^2=16.1^2 = 259.21$
$(x_6-\bar{x})^2=(89 - 74.9)^2=14.1^2 = 198.81$
$(x_7-\bar{x})^2=(58 - 74.9)^2=(-16.9)^2 = 285.61$
$(x_8-\bar{x})^2=(66 - 74.9)^2=(-8.9)^2 = 79.21$
$(x_9-\bar{x})^2=(70 - 74.9)^2=(-4.9)^2 = 24.01$
$(x_{10}-\bar{x})^2=(81 - 74.9)^2=6.1^2 = 37.21$
$\sum_{i = 1}^{10}(x_i-\bar{x})^2=16.81+4.41 + 0.01+479.61+259.21+198.81+285.61+79.21+24.01+37.21=1384.9$
$s=\sqrt{\frac{1384.9}{9}}\approx\sqrt{153.8778}\approx12.4$.

Step3: Calculate the coefficient of variation

The coefficient of variation $CV=\frac{s}{\bar{x}}\times100\%$.
$CV=\frac{12.4}{74.9}\times100\%\approx16.6\%$.

Answer:

$16.6$