Two middle-order batsmen are compared based on their performance in their previous cricket match. Batsman A got 9 runs more than Batsman B and Batsman A’s runs are 56% of the sum of both their runs.
Answers
Answer:Batsman A
x d=x−
x
ˉ
d
2
18 −16 256
33 −1 1
34
0 0
38 4 16
47 13 169
170 0 442
Now
x
ˉ
=
5
170
=34
σ=
n
∑d
2
=
5
442
=
88.4
≃9.4
Coefficient of variation, C.V=
x
ˉ
σ
×100
=
34
9.4
×100
=
34
940
=27.65
∴ The coefficient of variation for the runs scored by batsman A is 27.65 ............(1)
Batsman B
x d=x−
x
ˉ
d
2
27 −8 64
35 0 0
35
$$0$4 0
37 2 4
41 6 36
175 0 104
x
ˉ
=
5
175
=35
σ=
n
∑d
2
=
5
104
=
20.8
≃4.6
Coefficient of variation
=
x
ˉ
σ
×100
=
35
4.6
×100
=
35
460
=
7
92
=13.14
∴ The coefficient of varaition for the runs scored by batsman B is =13.14 ...........(2)
From (1) and (2), the coefficient of variation for B is less than the coefficient of variation for A.
∴ Batsman B is more consistent than the batsman A in scoring the runs.
Explanation:
Answer:
players: A B
runs: x+9 x
percent: 56% 44%
(x + 9) = [(x + 9) + x]* 56%
x + 9 = [2x + 9] * 0.56
x + 9 = 1.12x + 5.04
9 - 5.04 = 1.12x - 1x
3.96 = 0.12x
x = 33
replace x for player A 33 + 9 = 42
player A 42 runs, B 33runs
Explanation: