Browsing as Guest. Progress, bookmarks and attempts are disabled.
Log in to track your work.
June 2006 p6 q6
2440
32 teams enter for a knockout competition, in which each match results in one team winning and the other team losing. After each match the winning team goes on to the next round, and the losing team takes no further part in the competition. Thus 16 teams play in the second round, 8 teams play in the third round, and so on, until 2 teams play in the final round.
How many teams play in only 1 match?
How many teams play in exactly 2 matches?
Draw up a frequency table for the numbers of matches which the teams play.
Calculate the mean and variance of the numbers of matches which the teams play.
Solution
(i) In the first round, 32 teams play, resulting in 16 matches. Therefore, 16 teams lose and play only 1 match.
(ii) In the second round, 16 teams play, resulting in 8 matches. Therefore, 8 teams lose and play exactly 2 matches.
(iii) The frequency table is constructed as follows:
Matches
1
2
3
4
5
Frequency
16
8
4
2
2
(iv) To calculate the mean and variance: Total matches played = 16(1) + 8(2) + 4(3) + 2(4) + 2(5) = 62 Mean = \(\frac{62}{32} = 1.9375 \approx 1.94\) For variance, calculate \(\sum f x^2 = 16(1^2) + 8(2^2) + 4(3^2) + 2(4^2) + 2(5^2) = 166\) Variance = \(\frac{166}{32} - \left(\frac{62}{32}\right)^2 = 1.43\)