a sixth - grade class had a bowling party. below are the scores of the 11 students who bowled. 46, 49, 55…

Explanation:

Step1: Find the minimum value

The minimum value in the data - set 46, 49, 55, 56, 56, 58, 60, 68, 70, 78, 82 is 46.

Step2: Find the first quartile ($Q_1$)

The data - set has $n = 11$ values. The position of $Q_1$ is $\frac{n + 1}{4}=\frac{11+1}{4}=3$. So, $Q_1$ is the 3rd value in the ordered data - set, which is 55.

Step3: Find the median ($Q_2$)

The position of the median is $\frac{n + 1}{2}=\frac{11 + 1}{2}=6$. So, the median $Q_2$ is 58.

Step4: Find the third quartile ($Q_3$)

The position of $Q_3$ is $\frac{3(n + 1)}{4}=\frac{3\times(11 + 1)}{4}=9$. So, $Q_3$ is the 9th value in the ordered data - set, which is 70.

Step5: Find the maximum value

The maximum value in the data - set is 82.

The box - plot should have a minimum of 46, $Q_1 = 55$, median = 58, $Q_3 = 70$, and maximum of 82.

Answer:

C