Final Answer
The standard deviation of the stopping times for the 4 r/s vehicles is approximately 1.87 seconds. Therefore, the correct option is (a) 1.87.
Step-by-step explanation
The standard deviation (σ) is a measure of the amount of variation or dispersion in a set of values. It can be calculated using the formula:
![\[ \sigma = \sqrt{\frac{\sum_(i=1)^(N)(X_i - \bar{X})^2}{N}} \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/9v0v1040dhwnnp7ujrbe7v0peen1ids08d.png)
Where:
- N is the number of data points,
-
is each individual data point,
-
is the mean of the data set.
In this case, the stopping times are 3, 4, 5, and 8 seconds. The mean (\
is calculated as:
![\[ \bar{X} = (3 + 4 + 5 + 8)/(4) = (20)/(4) = 5 \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/d2m6ynlug2e15ns8pzj6xnyjg3jsmkwcs2.png)
Now, substitute the values into the standard deviation formula:
![\[ \sigma = \sqrt{((3-5)^2 + (4-5)^2 + (5-5)^2 + (8-5)^2)/(4)} \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/1jzt9gb9at0235tg8r12obm3sqkezus8jg.png)
![\[ \sigma = \sqrt{(4 + 1 + 0 + 9)/(4)} \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/fjs2awjnjnwsp9o59p2cwv3n2kemn74aa4.png)
![\[ \sigma = \sqrt{(14)/(4)} \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/g2pvh9pf1x4dt9f0ikvgcoc4wm1t35qrkc.png)
![\[ \sigma = √(3.5) \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/rs9i5un064yqn7pb1me1jfsp4a5n7ug0q4.png)
![\[ \sigma \approx 1.87 \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/nixwmbjzlw42baqwu1clwmh3086oldqzlg.png)
Therefore, the standard deviation of the stopping times is approximately 1.87 seconds. This value represents the average amount of deviation or spread from the mean stopping time, providing insight into the variability of the braking performance of the 4 r/s vehicles. The correct option is a.