michael recorded the number of points his team scored for the first seven basketball games.\n64, 58, 60, 52…

michael recorded the number of points his team scored for the first seven basketball games.\n64, 58, 60, 52, 56, 62, 54\nwhich box plot correctly represents the data?
Answer
Answer:
To determine the correct box - plot, we first need to find the five - number summary (minimum, first quartile $Q_1$, median (second quartile $Q_2$), third quartile $Q_3$, maximum) of the data set ${52,54,56,58,60,62,64}$.
- Sort the data: The data set ${64,58,60,52,56,62,54}$ in ascending order is ${52,54,56,58,60,62,64}$.
- Find the minimum and maximum:
- The minimum value of the data set is $52$.
- The maximum value of the data set is $64$.
- Find the median ($Q_2$):
- Since there are $n = 7$ data points, the median is the $\left(\frac{n + 1}{2}\right)$-th value.
- $\frac{7+1}{2}=4$-th value. So the median $Q_2=58$.
- Find the first quartile ($Q_1$):
- The lower half of the data set is ${52,54,56}$. Since there are $n_1 = 3$ data points in the lower half, the first quartile is the $\left(\frac{3 + 1}{2}\right)$-th value, which is the 2 - nd value. So $Q_1 = 54$.
- Find the third quartile ($Q_3$):
- The upper half of the data set is ${60,62,64}$. Since there are $n_2=3$ data points in the upper half, the third quartile is the $\left(\frac{3 + 1}{2}\right)$-th value, which is the 2 - nd value of the upper - half. So $Q_3 = 62$.
In a box - plot, the left - end of the box is at $Q_1$, the right - end of the box is at $Q_3$, the line inside the box is at the median, and the whiskers extend to the minimum and maximum values. So the box - plot should have the left - end of the box at $54$, the line inside the box at $58$, the right - end of the box at $62$, and the whiskers extending to $52$ and $64$.
Without seeing the actual visual options clearly (assuming the correct one has these characteristics), we have found the key values for the box - plot construction.
Explanation:
Step1: Sort the data
${52,54,56,58,60,62,64}$
Step2: Find minimum and maximum
Min = $52$, Max = $64$
Step3: Calculate median
$n = 7$, $\frac{n + 1}{2}=4$, Median $=58$
Step4: Calculate first quartile
Lower - half ${52,54,56}$, $n_1 = 3$, $\frac{n_1+1}{2}=2$, $Q_1 = 54$
Step5: Calculate third quartile
Upper - half ${60,62,64}$, $n_2 = 3$, $\frac{n_2+1}{2}=2$, $Q_3 = 62$