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 (%) (type an integer or decimal rounded to one decimal place as needed.)

Explanation:

Step1: Calculate the mean of systolic measurements

Let \(x_1 = 117,x_2=128,\cdots,x_{10}=121\). The mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}\), where \(n = 10\).
\(\sum_{i=1}^{10}x_i=117 + 128+156+98+155+123+118+137+127+121=1280\)
\(\bar{x}=\frac{1280}{10}=128\)

Step2: Calculate the standard - deviation of systolic measurements

The formula for the sample standard - deviation \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}}\)
\((x_1-\bar{x})^2=(117 - 128)^2=(-11)^2 = 121\)
\((x_2-\bar{x})^2=(128 - 128)^2=0\)
\((x_3-\bar{x})^2=(156 - 128)^2=28^2 = 784\)
\((x_4-\bar{x})^2=(98 - 128)^2=(-30)^2 = 900\)
\((x_5-\bar{x})^2=(155 - 128)^2=27^2 = 729\)
\((x_6-\bar{x})^2=(123 - 128)^2=(-5)^2 = 25\)
\((x_7-\bar{x})^2=(118 - 128)^2=(-10)^2 = 100\)
\((x_8-\bar{x})^2=(137 - 128)^2=9^2 = 81\)
\((x_9-\bar{x})^2=(127 - 128)^2=(-1)^2 = 1\)
\((x_{10}-\bar{x})^2=(121 - 128)^2=(-7)^2 = 49\)
\(\sum_{i = 1}^{10}(x_i-\bar{x})^2=121+0 + 784+900+729+25+100+81+1+49=2790\)
\(s=\sqrt{\frac{2790}{9}}\approx\sqrt{310}\approx17.6\)

Step3: Calculate the coefficient of variation for systolic measurements

The coefficient of variation \(CV=\frac{s}{\bar{x}}\times100\%\)
\(CV=\frac{17.6}{128}\times100\%\approx13.7\%\)

Answer:

13.7