Minutes the bus came late negative one 0 1 2 3 number of days 1 2 4 2 1 Andrews 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 this standard deviation for the number of minutes the bus arrived before or after 8:00 round your answer to the nearest tenth

1 answer

To calculate the standard deviation, we first need to determine the mean (average) of the data set, and then compute the variance. Finally, we take the square root of the variance to get the standard deviation.

Here's the data we have:

  • Minutes the bus came late (X): \[-1, 0, 1, 2, 3\]
  • Number of days (frequency, f): \[1, 2, 4, 2, 1\]

First, let’s list the minutes along with their frequencies:

  • -1 minutes: 1 day
  • 0 minutes: 2 days
  • 1 minute: 4 days
  • 2 minutes: 2 days
  • 3 minutes: 1 day

To calculate the mean:

  1. Calculate the total number of days:
    \(1 + 2 + 4 + 2 + 1 = 10\)

  2. Calculate the weighted sum of minutes:
    \[(-1 \times 1) + (0 \times 2) + (1 \times 4) + (2 \times 2) + (3 \times 1)\]
    \[= -1 + 0 + 4 + 4 + 3 = 10\]

  3. Calculate the mean (μ):
    \[\mu = \frac{\text{Total Sum}}{\text{Total Number of Days}} = \frac{10}{10} = 1\]

Next, we compute the variance:

  1. Calculate the squared differences from the mean:
    \[ (-1 - 1)^2, (0 - 1)^2, (1 - 1)^2, (2 - 1)^2, (3 - 1)^2 \]
    \[ = 4, 1, 0, 1, 4 \]

  2. Multiply each squared difference by its frequency:
    \[ 4 \times 1, 1 \times 2, 0 \times 4, 1 \times 2, 4 \times 1 \]
    \[ = 4, 2, 0, 2, 4 \]

  3. Sum these values:
    \[ 4 + 2 + 0 + 2 + 4 = 12 \]

  4. Calculate the variance (σ²):
    \[\sigma^2 = \frac{\text{Total of weighted squared differences}}{\text{Total number of days}} = \frac{12}{10} = 1.2\]

  5. Calculate the standard deviation (σ):
    \[\sigma = \sqrt{1.2} \approx 1.095\]

Now rounding to the nearest tenth, we get:
\(\sigma \approx 1.1\)

Therefore, the standard deviation for the number of minutes the bus arrived before or after 8:00 is 1.1 minutes.