Maharashtra State BoardHSC Commerce 11th
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Compute coefficient of variation for team A and team B. (Given:26=5.099,22=4.6904) No. of goals 0 1 2 3 4 No. of matches played by team A 18 7 5 16 14 No. of matches played by team B 14 16 5 - Mathematics and Statistics

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Sum

Compute coefficient of variation for team A and team B. `("Given": sqrt(26) = 5.099, sqrt(22) = 4.6904)`

No. of goals 0 1 2 3 4
No. of matches played by team A 18 7 5 16 14
No. of matches played by team B 14 16 5 18 17

Which team is more consistent?

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Solution

For team A

Let f1 denote no. of goals of team A.

No. of goals
(xi)
No. of matches
(f1i)
f1ixi f1ixi2
0 18 0 0
1 7 7 7
2 5 10 20
3 16 48 144
4 14 56 224
  N1 = 60 ∑f1ixi = 121 ∑f1ixi2 = 395

`bar("x"_1) = (sum"f"_"1i""x"_"i")/"N"_1`

= `(121)/(60)`
= 2.0167
Standard deviation,

`sigma_("x"_1)^2 = 1/"N"_1sum"f"_"1i""x"_"i"^2- (bar"x"_1)^2 `

= `395/60 - (2.0167)^2`

= 6.5833 − 4.0671
= 2.5162
∴ `sigma_"x1"=sqrt2.5162` = 1.5863 
Co-efficient of variance;

C.V (x1) = `sigma_"x1"/bar("x"_1) xx 100`

= `(1.5863)/(2.0167) xx 100`
= 78.66% 

For team B

Let f2 denote no. of goals of team B.

No. of goals
(xi)
No. of matches
(f2i)
f2ixi f2ixi2
0 14 0 0
1 16 16 16
2 5 10 20
3 18 544 162
4 17 68 272
  N2 = 70 ∑f2ixi = 148 ∑f2ixi2 = 470

`bar("x"_2) = (sum"f"_"2i""x"_"i")/"N"_2`

= `(148)/(70)`
= 2.1143
Standard deviation,

`sigma_("x"_2)^2=1/"N"_2sum"f"_"2i""x"_"i"^2- (bar"x"_2)^2 `

= `470/70- (2.1143)^2`

= (6.7143 − 4.46703)
= 2.244
∴ `sigma_"x2"= sqrt2.244` = 1.4980
Co-efficient of variance;

C.V. (x2) = `sigma_"x2"/bar("x"_2) xx 100`

= `1.4980/2.1143xx100`
= 70.85%

Since, C.V. of team A > C.V. of team B.
∴ team B is more consistent.

Concept: Coefficient of Variation
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