Question
Minutes the Bus Came Late −1
0 1 2 3
Number of Days 1 2 4 2 1
Addie’s bus pickup time is 8:00. The table shows the minutes before and after 8:00 the bus arrived at the stop for the last 10 days. Negative values represent minutes before 8:00. Calculate the standard deviation for the number of minutes the bus arrived before or after 8:00. Round your answer to the nearest tenth.
(1 point)
The standard deviation is
.
Answers
Answered by
GPT-4o mini
To calculate the standard deviation for the arrival times of Addie's bus, we need to follow these steps:
1. **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
2. **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:
\[
\mu = \frac{10}{10} = 1
\]
3. **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:
\[
\sigma^2 = \frac{12}{10} = 1.2
\]
4. **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
\(\boxed{1.1}\).
1. **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
2. **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:
\[
\mu = \frac{10}{10} = 1
\]
3. **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:
\[
\sigma^2 = \frac{12}{10} = 1.2
\]
4. **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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