Question

Use the table to answer the 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.


The standard deviation is

Answer 1

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.

1. Find the total number of days:
\[
N = 1 + 2 + 4 + 2 + 1 = 10
\]

2. Calculate the sum of (x * f):
\[
(-1) * 1 + 0 * 2 + 1 * 4 + 2 * 2 + 3 * 1 = -1 + 0 + 4 + 4 + 3 = 10
\]

3. 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).

4. The standard deviation formula:
\[
\sigma = \sqrt{\frac{\sum f (x - \mu)^2}{N}}
\]

5. 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 |

6. Now sum the f(x - μ)^2:
\[
4 + 2 + 0 + 2 + 4 = 12
\]

7. 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**.