To calculate the standard deviation of the data set \({20, 35, 41, 48, 79, 83, 83, 97, 120}\), follow these steps:
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Calculate the mean (average) of the data set: \[ \text{Mean} = \frac{\sum x_i}{n} \] Where \(n\) is the number of values. Here, \(n = 9\):
\[ \text{Mean} = \frac{20 + 35 + 41 + 48 + 79 + 83 + 83 + 97 + 120}{9} = \frac{506}{9} \approx 56.2 \]
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Calculate each data point's deviation from the mean, square it, and find the average of these squared deviations: \[ \text{Variance} = \frac{\sum (x_i - \text{Mean})^2}{n} \]
Calculate deviations:
- For \(20\): \((20 - 56.2)^2 \approx 1296.64\)
- For \(35\): \((35 - 56.2)^2 \approx 449.64\)
- For \(41\): \((41 - 56.2)^2 \approx 231.36\)
- For \(48\): \((48 - 56.2)^2 \approx 67.24\)
- For \(79\): \((79 - 56.2)^2 \approx 517.96\)
- For \(83\): \((83 - 56.2)^2 \approx 711.84\)
- For \(83\): \((83 - 56.2)^2 \approx 711.84\)
- For \(97\): \((97 - 56.2)^2 \approx 1685.44\)
- For \(120\): \((120 - 56.2)^2 \approx 4052.64\)
Now sum these squared deviations: \[ \sum (x_i - \text{Mean})^2 \approx 1296.64 + 449.64 + 231.36 + 67.24 + 517.96 + 711.84 + 711.84 + 1685.44 + 4052.64 \approx 10096.6 \]
Next, divide by the number of values: \[ \text{Variance} \approx \frac{10096.6}{9} \approx 1122.96 \]
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Calculate the standard deviation by taking the square root of the variance: \[ \text{Standard Deviation} = \sqrt{1122.96} \approx 33.5 \]
Rounding to the nearest tenth, the standard deviation is approximately 33.5.
None of the options exactly match this calculated value, but if the choices given are:
- 29.8
- 79.0
- 31.0
- 32.9
The closest rounded standard deviation to 33.5 is 32.9.
Thus, the answer is: 32.9