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Question
blood pressure: the following table presents systolic blood - pressure readings (in millimeters of mercury) for samples of 10 young adults (age 18 - 29) and 10 older adults (age 60+).
younger 120 91 92 124 121 121 122 100 110 117
older 142 135 119 124 157 121 122 100 130 158
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your answer
part 1 of 3
(a) find the sample standard deviation of the blood pressures for the younger adults. round the answer to one decimal place as needed.
the sample standard deviation of the blood pressures for the younger adults is 12.8.
part: 1 / 3
part 2 of 3
(b) find the sample standard deviation of the blood pressures for the older adults. round the answer to one decimal place as needed.
the sample standard deviation of the blood pressures for the older adults is
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Step1: Recall sample - standard - deviation formula
The formula for the sample standard deviation $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}$, where $n$ is the sample size, $x_{i}$ are the individual data - points, and $\bar{x}$ is the sample mean.
Step2: Calculate the mean for older adults
The data for older adults is $142,135,119,124,157,121,122,100,130,158$.
$n = 10$.
$\bar{x}=\frac{142 + 135+119+124+157+121+122+100+130+158}{10}=\frac{1308}{10}=130.8$.
Step3: Calculate $(x_{i}-\bar{x})^{2}$ for each data - point
$(142 - 130.8)^{2}=(11.2)^{2}=125.44$;
$(135 - 130.8)^{2}=(4.2)^{2}=17.64$;
$(119 - 130.8)^{2}=(-11.8)^{2}=139.24$;
$(124 - 130.8)^{2}=(-6.8)^{2}=46.24$;
$(157 - 130.8)^{2}=(26.2)^{2}=686.44$;
$(121 - 130.8)^{2}=(-9.8)^{2}=96.04$;
$(122 - 130.8)^{2}=(-8.8)^{2}=77.44$;
$(100 - 130.8)^{2}=(-30.8)^{2}=948.64$;
$(130 - 130.8)^{2}=(-0.8)^{2}=0.64$;
$(158 - 130.8)^{2}=(27.2)^{2}=739.84$.
Step4: Calculate $\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}$
$\sum_{i = 1}^{10}(x_{i}-\bar{x})^{2}=125.44 + 17.64+139.24+46.24+686.44+96.04+77.44+948.64+0.64+739.84=2877.6$.
Step5: Calculate the sample standard deviation
$s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}=\sqrt{\frac{2877.6}{9}}\approx\sqrt{319.733}\approx17.9$.
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$17.9$