To calculate the standard deviation for the bus arrival times, we'll follow these steps:
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List the Data Points: Based on the table provided:
- \(-1\) minute: 1 day
- \(0\) minutes: 2 days
- \(1\) minute: 4 days
- \(2\) minutes: 2 days
- \(3\) minutes: 1 day
The data points can be expressed in a list:
- \([-1, 0, 0, 1, 1, 1, 1, 2, 2, 3]\)
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Calculate the Mean: \[ \text{Mean} = \frac{\sum{\text{data points}}}{\text{number of data points}} = \frac{-1 + 0 + 0 + 1 + 1 + 1 + 1 + 2 + 2 + 3}{10} = \frac{10}{10} = 1 \]
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Calculate Variance: \[ \text{Variance} = \frac{\sum{(x_i - \text{Mean})^2}}{N} \] Where \(x_i\) are the data points and \(N\) is the number of data points (10).
Now we calculate \((x_i - \text{Mean})^2\):
- For \(-1\): \((-1 - 1)^2 = (-2)^2 = 4\)
- For \(0\): \((0 - 1)^2 = (-1)^2 = 1\)
- For another \(0\): \((0 - 1)^2 = 1\)
- For \(1\): \((1 - 1)^2 = (0)^2 = 0\)
- For another \(1\): \(0\)
- For another \(1\): \(0\)
- For another \(1\): \(0\)
- For \(2\): \((2 - 1)^2 = (1)^2 = 1\)
- For another \(2\): \(1\)
- For \(3\): \((3 - 1)^2 = (2)^2 = 4\)
Now we sum these squared differences: \[ 4 + 1 + 1 + 0 + 0 + 0 + 0 + 1 + 1 + 4 = 12 \]
Divide by the number of data points (10): \[ \text{Variance} = \frac{12}{10} = 1.2 \]
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Calculate Standard Deviation: \[ \text{Standard Deviation} = \sqrt{\text{Variance}} = \sqrt{1.2} \approx 1.095 \]
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Round to the Nearest Tenth: \[ \text{Standard Deviation} \approx 1.1 \]
Thus, the standard deviation for the number of minutes the bus arrived before or after 8:00 is approximately 1.1.