Questions: Blood pressure: The following table presents systolic blood pressure readings (in millimeters of mercury) for older adults (age 60+). Younger: 110, 101, 92, 121, 101, 121, 121, 120, 112, 109 Older: 120, 131, 157, 124, 155, 121, 122, 100, 140, 119 Part 1 of 3 (a) Find the sample standard deviation of the blood pressures for the younger adults. Round the answer to one The sample standard deviation of the blood pressures for the younger adults is .

To find the sample standard deviation of the blood pressures for the younger adults, we need to follow these steps:

1. Calculate the mean (average) of the data set.
2. Subtract the mean from each data point and square the result.
3. Sum all the squared results.
4. Divide this sum by the number of data points minus one (this is the variance).
5. Take the square root of the variance to get the standard deviation.

Let's go through these steps with the given data for younger adults: 110, 101, 92, 121, 101, 121, 121, 120, 112, 109.

\[ \text{Mean} = \frac{110 + 101 + 92 + 121 + 101 + 121 + 121 + 120 + 112 + 109}{10} = \frac{1108}{10} = 110.8 \]

\[ (110 - 110.8)^2 = (-0.8)^2 = 0.64 \] \[ (101 - 110.8)^2 = (-9.8)^2 = 96.04 \] \[ (92 - 110.8)^2 = (-18.8)^2 = 353.44 \] \[ (121 - 110.8)^2 = (10.2)^2 = 104.04 \] \[ (101 - 110.8)^2 = (-9.8)^2 = 96.04 \] \[ (121 - 110.8)^2 = (10.2)^2 = 104.04 \] \[ (121 - 110.8)^2 = (10.2)^2 = 104.04 \] \[ (120 - 110.8)^2 = (9.2)^2 = 84.64 \] \[ (112 - 110.8)^2 = (1.2)^2 = 1.44 \] \[ (109 - 110.8)^2 = (-1.8)^2 = 3.24 \]

\[ 0.64 + 96.04 + 353.44 + 104.04 + 96.04 + 104.04 + 104.04 + 84.64 + 1.44 + 3.24 = 947.6 \]

\[ \text{Variance} = \frac{947.6}{10 - 1} = \frac{947.6}{9} \approx 105.29 \]

\[ \text{Standard Deviation} = \sqrt{105.29} \approx 10.26 \]

So, the sample standard deviation of the blood pressures for the younger adults is approximately 10.3 (rounded to one decimal place).