To calculate the standard deviation for the arrival times of Addie's bus, we need to follow these steps:
-
List the data: From the table, we have:
- For -1 minutes: 1 day
- For 0 minutes: 2 days
- For 1 minute: 4 days
- For 2 minutes: 2 days
- For 3 minutes: 1 day
This corresponds to the following values and frequencies:
- \(-1\): 1
- \(0\): 2
- \(1\): 4
- \(2\): 2
- \(3\): 1
-
Calculate the mean: The mean (\(\mu\)) is calculated as follows:
\[ \mu = \frac{\sum (x \cdot f)}{N} \]
Where \(x\) is the value, \(f\) is the frequency, and \(N\) is the total number of days.
First, we calculate \(\sum (x \cdot f)\):
\[ = (-1 \cdot 1) + (0 \cdot 2) + (1 \cdot 4) + (2 \cdot 2) + (3 \cdot 1) = -1 + 0 + 4 + 4 + 3 = 10 \]
The total number of days:
\[ N = 1 + 2 + 4 + 2 + 1 = 10 \]
Now, calculate the mean:
\[ \mu = \frac{10}{10} = 1 \]
- Calculate the variance (\(\sigma^2\)): The variance is calculated using the formula:
\[ \sigma^2 = \frac{\sum f \cdot (x - \mu)^2}{N} \]
Calculate \( (x - \mu)^2 \) for each value:
- For \(x = -1\): \((-1 - 1)^2 = 4 \) → Frequency = 1 → Contribution: \(1 \cdot 4 = 4\)
- For \(x = 0\): \((0 - 1)^2 = 1 \) → Frequency = 2 → Contribution: \(2 \cdot 1 = 2\)
- For \(x = 1\): \((1 - 1)^2 = 0 \) → Frequency = 4 → Contribution: \(4 \cdot 0 = 0\)
- For \(x = 2\): \((2 - 1)^2 = 1 \) → Frequency = 2 → Contribution: \(2 \cdot 1 = 2\)
- For \(x = 3\): \((3 - 1)^2 = 4 \) → Frequency = 1 → Contribution: \(1 \cdot 4 = 4\)
Now sum these contributions:
\[ \sum f \cdot (x - \mu)^2 = 4 + 2 + 0 + 2 + 4 = 12 \]
Now calculate variance:
\[ \sigma^2 = \frac{12}{10} = 1.2 \]
- Calculate the standard deviation (\(\sigma\)):
\[ \sigma = \sqrt{1.2} \approx 1.095 \]
Rounding to the nearest tenth:
\[ \sigma \approx 1.1 \]
Thus, the standard deviation is
\(\boxed{1.1}\).
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