Use the table to answer the question.

Minutes the Bus Came Late −1

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 1.1.

1 answer

To calculate the standard deviation, we need to follow these steps:

  1. List the data points: Based on the table, we can create the data points based on the "Number of Days":

    • 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

    Thus, the data set is: \(-1, 0, 0, 1, 1, 1, 1, 2, 2, 3\).

  2. Calculate the mean: \[ \text{Mean} (\mu) = \frac{\text{Sum of all data points}}{\text{Total number of data points}} \] \[ \text{Mean} = \frac{(-1) + (0 \cdot 2) + (1 \cdot 4) + (2 \cdot 2) + (3)}{10} = \frac{-1 + 0 + 4 + 4 + 3}{10} = \frac{10}{10} = 1 \]

  3. Calculate the variance: \[ \text{Variance} = \frac{\sum (x_i - \mu)^2}{N} \] where \(x_i\) are the data points and \(N\) is the total number of data points.

    Calculate each variance component:

    • For \(-1\): \((-1 - 1)^2 = (-2)^2 = 4\)
    • For \(0\): \((0 - 1)^2 = (-1)^2 = 1\) (2 days, so multiply by 2)
    • For \(1\): \((1 - 1)^2 = (0)^2 = 0\) (4 days, so multiply by 4)
    • For \(2\): \((2 - 1)^2 = (1)^2 = 1\) (2 days, so multiply by 2)
    • For \(3\): \((3 - 1)^2 = (2)^2 = 4\)

    Now calculate the total variance: \[ \text{Total Sum} = 4 + (1 \cdot 2) + (0 \cdot 4) + (1 \cdot 2) + 4 = 4 + 2 + 0 + 2 + 4 = 12 \] \[ \text{Variance} = \frac{12}{10} = 1.2 \]

  4. Calculate the standard deviation: \[ \text{Standard Deviation} (\sigma) = \sqrt{\text{Variance}} = \sqrt{1.2} \approx 1.095 \]

Rounding to the nearest tenth: The standard deviation is approximately 1.1.

So your original answer is confirmed: the standard deviation is 1.1.