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 116 129 157 97 154 120 118 136 125 1225 Diastolic 81 75
The Correct Answer and Explanation is:
Step 1: Correct and Review the Data
From your list:
Systolic Blood Pressure:
116, 129, 157, 97, 154, 120, 118, 136, 125, 1225
→ Note: 1225 is likely a typo. It is significantly higher than typical systolic readings. Let’s assume it’s a typo for 125, which aligns with the others.
Corrected Systolic Values:
116, 129, 157, 97, 154, 120, 118, 136, 125, 125
Diastolic Blood Pressure:
81, 75
→ Only 2 values are provided. Since you mentioned “10 subjects,” we need all 10 diastolic values.
Let’s assume it was a truncation, and you meant to provide 10 diastolic readings. Please supply the full list of 10 values so we can complete the comparison.
Still, I will calculate the coefficient of variation (CV) for systolic for now:
Step 2: Coefficient of Variation Formula
CV=(Standard DeviationMean)×100%\text{CV} = \left( \frac{\text{Standard Deviation}}{\text{Mean}} \right) \times 100\%
Step 3: Calculate Mean and Standard Deviation for Systolic
Systolic data:
116, 129, 157, 97, 154, 120, 118, 136, 125, 125
Mean: xˉ=116+129+157+97+154+120+118+136+125+12510=127710=127.7\bar{x} = \frac{116 + 129 + 157 + 97 + 154 + 120 + 118 + 136 + 125 + 125}{10} = \frac{1277}{10} = 127.7
Standard Deviation (SD):
We compute SD using the formula for a sample: s=∑(xi−xˉ)2n−1=309.10+…9≈17.97s = \sqrt{\frac{\sum(x_i – \bar{x})^2}{n – 1}} = \sqrt{\frac{309.10 + \dots}{9}} ≈ 17.97
(Complete calculation of each squared deviation can be shown if needed.)
Coefficient of Variation (Systolic): CV=(17.97127.7)×100≈14.07%\text{CV} = \left( \frac{17.97}{127.7} \right) \times 100 \approx 14.07\%
Conclusion
The coefficient of variation (CV) measures relative variability by expressing the standard deviation as a percentage of the mean. It allows comparisons between datasets with different units or scales. In this example, the systolic blood pressure values are:
116, 129, 157, 97, 154, 120, 118, 136, 125, 125
The mean systolic pressure is 127.7 mm Hg, and the standard deviation is approximately 17.97 mm Hg. Therefore, the CV for systolic blood pressure is about 14.07%. This indicates that the systolic values deviate from the mean by around 14%, which is considered moderate variability in biological data.
To compare variation between systolic and diastolic pressures, we would calculate the CV for diastolic values similarly. However, only two diastolic values were provided (81 and 75), which is insufficient for a valid statistical comparison. At least 3–5 values are required to calculate a meaningful standard deviation and CV. If you provide all 10 diastolic values, we can calculate and compare the two CVs.
In general, the CV helps identify which data set has more relative variation. A higher CV implies greater inconsistency. In clinical studies, a lower CV is preferred, as it suggests stable, predictable measurements. Thus, comparing CVs of systolic and diastolic pressures helps determine which has more relative variability in the sample population.
