Question

two soccer teams join a league in which every team plays in 12 games per season. the number of points that the two teams score each game is recorded in the dot - plots shown. complete the statement to correctly compare the dot plots. based on the data, blank is more likely to score the same number of points during a game as the mean of its data because the teams data blank.

Explanation:

Step1: Calculate the mean for Team A

To find the mean of Team A's scores, we first note the number of data - points at each value. Let \(x_i\) be the score value and \(f_i\) be the frequency. The scores and their frequencies for Team A: \(x_1 = 0,f_1=2\); \(x_2 = 1,f_2 = 3\); \(x_3=2,f_3 = 2\); \(x_4 = 4,f_4=1\); \(x_5 = 5,f_5 = 1\); \(x_6=6,f_6 = 2\); \(x_7 = 8,f_7=1\). The sum of the products \(\sum_{i = 1}^{7}x_if_i=0\times2 + 1\times3+2\times2 + 4\times1+5\times1+6\times2+8\times1=3 + 4+4 + 5+12+8=36\). The total number of data - points \(n=\sum_{i = 1}^{7}f_i=2 + 3+2+1+1+2+1 = 12\). The mean \(\bar{x}_A=\frac{36}{12}=3\).

Step2: Calculate the mean for Team B

For Team B, the scores and their frequencies: \(x_1 = 0,f_1=1\); \(x_2 = 2,f_2 = 2\); \(x_3=3,f_3 = 3\); \(x_4 = 4,f_4=6\); \(x_5 = 6,f_5 = 2\). The sum of the products \(\sum_{i = 1}^{5}x_if_i=0\times1+2\times2 + 3\times3+4\times6+6\times2=0 + 4+9+24+12=49\). The total number of data - points \(n=\sum_{i = 1}^{5}f_i=1+2 + 3+6+2=14\). The mean \(\bar{x}_B=\frac{49}{14}=3.5\).

Step3: Analyze the spread of data

The spread of data can be analyzed by looking at how clustered the data is around the mean. Team B's data is more clustered around its mean (more dots are concentrated around 3 and 4, close to the mean of 3.5) compared to Team A's data which is more spread out.

Answer:

Team B; is more clustered around the mean.