To calculate the standard deviation for the number of minutes the bus arrived before or after 8:00, we'll follow these steps:
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List the data points, considering the frequency:
- From the table, we have:
- -1 minute: 1 day
- 0 minutes: 2 days
- 1 minute: 4 days
- 2 minutes: 2 days
- 3 minutes: 1 day
The data points, weighted by the number of days, are: \[ \text{Data points: } -1, 0, 0, 1, 1, 1, 1, 2, 2, 3 \]
- From the table, we have:
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Calculate the mean (\(\mu\)): \[ \mu = \frac{\text{Sum of all values}}{\text{Total number of days}} \] \[ \text{Sum of all values} = (-1) + 0 + 0 + 1 + 1 + 1 + 1 + 2 + 2 + 3 = 10 \] \[ \text{Total number of days} = 10 \] \[ \mu = \frac{10}{10} = 1 \]
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Calculate the squared differences from the mean: We find the squared difference for each data point:
- For -1: \((-1 - 1)^2 = (-2)^2 = 4\)
- For 0: \((0 - 1)^2 = (-1)^2 = 1\) (occurs twice, so count it twice)
- For 1: \((1 - 1)^2 = 0\) (occurs four times, so count it four times)
- For 2: \((2 - 1)^2 = (1)^2 = 1\) (occurs twice, so count it twice)
- For 3: \((3 - 1)^2 = (2)^2 = 4\)
Now we tally these:
- \[1 \times 4 = 4\]
- \[2 \times 1 = 2 \]
- \[4 \times 0 = 0 \]
- \[2 \times 1 = 2 \]
- \[1 \times 4 = 4 \]
Summing these squared differences: \[ 4 + 2 + 0 + 2 + 4 = 12 \]
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Calculate the variance (\( \sigma^2 \)): \[ \sigma^2 = \frac{\text{Sum of squared differences}}{\text{Total number of days}} = \frac{12}{10} = 1.2 \]
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Calculate the standard deviation (\( \sigma \)): \[ \sigma = \sqrt{\sigma^2} = \sqrt{1.2} \approx 1.095 \]
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Round to the nearest tenth: \[ \text{Standard deviation} \approx 1.1 \]
Therefore, the standard deviation is 1.1.
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