Question

each possible number of wins.
wins frequency
0 0
1 0
2 1
3 1
4 5
5 2
6 0
7 4
8 7
9 1
10 2
11 4
12 3
13 2
14 0
15 0
16 0
what are the minimum, first quartile (q1), median, third quartile (q3), and maximum number of wins for the 32 teams in the

Explanation:

Step1: Arrange data in ascending - order considering frequencies

First, list out the number of wins based on their frequencies. We have: 2 (1 time), 3 (1 time), 4 (5 times), 5 (2 times), 7 (4 times), 8 (7 times), 9 (1 time), 10 (2 times), 11 (4 times), 12 (3 times), 13 (2 times).

Step2: Find the minimum

The minimum value is the smallest non - zero frequency number of wins. Looking at the data, the minimum number of wins is 2.

Step3: Calculate the position of Q1

The formula for the position of the first quartile $Q1$ for $n = 32$ data points is $i=\frac{1}{4}(n + 1)=\frac{1}{4}(32+1)=8.25$. The first quartile is the value at the 8.25th position. Counting the cumulative frequencies: 1 (for 2) + 1 (for 3)+5 (for 4)+2 (for 5) = 9. So, $Q1 = 4$.

Step4: Calculate the position of the median

The formula for the position of the median for $n = 32$ (an even number of data points) is $i=\frac{n}{2}=16$ and $i + 1=17$. The median is the average of the values at the 16th and 17th positions. Cumulative frequencies: 1+1 + 5+2+4=13 (up to 7 wins), and adding the frequency of 8 wins (7), we pass the 16th and 17th positions. So, the median is 8.

Step5: Calculate the position of Q3

The formula for the position of the third quartile $Q3$ is $i=\frac{3}{4}(n + 1)=\frac{3}{4}(32 + 1)=24.75$. Counting cumulative frequencies: 1+1+5+2+4+7+1+2+4 = 26. So, $Q3 = 11$.

Step6: Find the maximum

The maximum value is the largest non - zero frequency number of wins. The maximum number of wins is 13.

Answer:

Minimum: 2, Q1: 4, Median: 8, Q3: 11, Maximum: 13