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प्रश्न
Compute coefficient of variation for team A and team B.
No. of goals | 0 | 1 | 2 | 3 | 4 |
No. of matches played by team A | 19 | 6 | 5 | 16 | 14 |
No. of matches played by team B | 16 | 16 | 5 | 18 | 15 |
Which team is more consistent?
उत्तर
Let f1 denote no. of goals of team A and f2 denote no. of goals of team B.
No. of goals (xi) |
No. of matches (f1i) |
No. of matches (f2i) |
f1ixi | f2ixi | f1ixi2 | f2ixi2 |
0 | 19 | 16 | 0 | 0 | 0 | 0 |
1 | 6 | 16 | 6 | 16 | 6 | 16 |
2 | 5 | 5 | 10 | 10 | 20 | 20 |
3 | 16 | 18 | 48 | 54 | 144 | 162 |
4 | 14 | 15 | 56 | 60 | 224 | 240 |
N1 = 60 | N2 = 70 | f1ixi = 120 | f2ixi = 140 | f1ixi2 = 394 | f2ixi2 = 438 |
For Team A,
`bar("x"_1) = (sum"f"_"1i""x"_"i")/"N"_1 = (120)/(60) = 2`
Standard deviation,
`sigma_("x"_1)^2 = 1/"N"_1sum"f"_"1i""x"_"i"^2- (bar"x"_1)^2 `
`sigma_("x"_1)^2 = 394/60 - (2)^2`
`sigma_("x"_1)^2` = 6.57 − 4
`sigma_("x"_1)^2` = 2.57
∴ `sigma_"x1" = sqrt2.57 = 1.60`
Co-efficient of variance;
C.V (x1) = `sigma_"x1"/bar("x"_1) × 100 = (1.60)/(2) × 100 = 80%`
For Team B,
`bar("x"_2) = (sum"f"_"2i""x"_"i")/"N"_2 = (140)/(70) = 2`
Standard deviation,
`sigma_("x"_2)^2 = 1/"N"_2sum"f"_"2i""x"_"i"^2- (bar"x"_2)^2 `
`sigma_("x"_2)^2 = 438/70- (2)^2`
`sigma_("x"_2)^2` = 6.26 − 4
`sigma_("x"_2)^2` = 2.26
∴ `sigma_"x2"= sqrt(2.26)` = 1.50
Co-efficient of variance;
C.V. (x2) = `sigma_"x2"/bar("x"_2) × 100 = 1.50/2 × 100 = 75%`
Since C.V. of team A > C.V. of team B.
∴ Team B is more consistent.
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