smith mediu as alturas de alguns alunos em sua aula de matemática. alturas dos alunos (pol.) 56 61 55 63 66…

Explanation:

Step1: Arrange data in ascending order

52, 55, 56, 56, 61, 63, 63, 64, 64, 66

Step2: Find the minimum value

The minimum value is 52.

Step3: Find the first - quartile ($Q_1$)

Since $n = 10$, the position of $Q_1$ is $\frac{10 + 1}{4}=2.75$. So $Q_1=56$.

Step4: Find the median ($Q_2$)

The position of the median is $\frac{10}{2}=5$ and $\frac{10}{2}+1 = 6$. So $Q_2=\frac{61 + 63}{2}=62$.

Step5: Find the third - quartile ($Q_3$)

The position of $Q_3$ is $\frac{3(10 + 1)}{4}=8.25$. So $Q_3=64$.

Step6: Find the maximum value

The maximum value is 66.

Step7: Analyze box - and - whisker plots

The box - and - whisker plot should have a whisker starting at 52, the left - hand side of the box at 56, the line inside the box at 62, the right - hand side of the box at 64, and a whisker ending at 66.

Answer:

The box - and - whisker plot with a minimum of 52, $Q_1 = 56$, median of 62, $Q_3=64$, and maximum of 66. (Since no options are given in text, we describe the correct box - and - whisker plot characteristics).