To calculate the standard deviation for the data, we first need to organize the data into a frequency distribution and the corresponding values. Here is the data based on the table you provided:
-1 minute: 1 day 0 minutes: 2 days 1 minute: 4 days 2 minutes: 2 days 3 minutes: 1 day
Next, we can represent this data with a frequency (f) and the midpoint (x):
| Minutes (x) | Frequency (f) | |-------------|---------------| | -1 | 1 | | 0 | 2 | | 1 | 4 | | 2 | 2 | | 3 | 1 |
Now, let's calculate the mean (μ) of these minutes.
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Find the total number of days: \[ N = 1 + 2 + 4 + 2 + 1 = 10 \]
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Calculate the sum of (x * f): \[ (-1) * 1 + 0 * 2 + 1 * 4 + 2 * 2 + 3 * 1 = -1 + 0 + 4 + 4 + 3 = 10 \]
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Now, calculate the mean (μ): \[ \mu = \frac{\text{Sum of (x * f)}}{N} = \frac{10}{10} = 1 \]
Next, we will calculate the standard deviation (σ).
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The standard deviation formula: \[ \sigma = \sqrt{\frac{\sum f (x - \mu)^2}{N}} \]
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We will now compute \( (x - \mu)^2 \) and then find \( f (x - \mu)^2 \):
| Minutes (x) | Frequency (f) | (x - μ) | (x - μ)^2 | f(x - μ)^2 | |-------------|---------------|----------|-----------|------------| | -1 | 1 | -2 | 4 | 4 | | 0 | 2 | -1 | 1 | 2 | | 1 | 4 | 0 | 0 | 0 | | 2 | 2 | 1 | 1 | 2 | | 3 | 1 | 2 | 4 | 4 |
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Now sum the f(x - μ)^2: \[ 4 + 2 + 0 + 2 + 4 = 12 \]
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Now plug it into the standard deviation formula: \[ \sigma = \sqrt{\frac{12}{10}} = \sqrt{1.2} \approx 1.095 \]
Finally, rounding the answer to the nearest tenth, we get: \[ \text{Standard Deviation} \approx 1.1 \]
Therefore, the standard deviation for the number of minutes the bus arrived before or after 8:00 is 1.1 minutes.