Answer:
The median of A is the same as the median of B.
The interquartile range of B is greater than the interquartile range of A.
Step-by-step explanation:
Given that:
A = number of runs allowed in first 9 games
A = {1, 4, 2, 2, 3, 1, 1, 2, 1}
Rearranging A : 1, 1, 1, 1, 2, 2, 2, 3, 4
Median A = 1/2(n + 1) th term
Median A = 1/2(10) = 5th term = 2
Q1 of A = 1/4(10) = 2.5th term = (1 + 1)/ 2 = 1
Q3 of A = 3/4(10) = 7.5th term = (2+3)/2 = 2.5
Interquartile range = Q3 - Q1 = 2.5 - 1 = 1.5
Number of runs allowed in 10th game = 9
B = {1, 4, 2, 2, 3, 1, 1, 2, 1, 9}
Rearranging B = 1, 1, 1, 1, 2, 2, 2, 3, 4, 9
Median A = 1/2(n + 1) th term
Median A = 1/2(11) = 5.5th term = (2+2)/2 = 2
Q1 of A = 1/4(11) = 2.75th tetm = (1 + 1)/ 2 = 1
Q3 of A = 3/4(11) = 8.25th term = (3+4)/2 = 3.5
Interquartile range = Q3 - Q1 = 3.5 - 1 = 2.5
Median A = 2 ; median B = 2
IQR B = 2.5 ; IQR A = 1.5 ; IQR B > IQR A
