Question

Select ALL the correct answers.

Answer 1

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