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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?

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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?

योग
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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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Coefficient of Variation
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 2: Measures of Dispersion - Exercise 2.3 [पृष्ठ ३४]

APPEARS IN

बालभारती Mathematics and Statistics (Commerce) Part 2 [English] Standard 11 Maharashtra State Board
अध्याय 2 Measures of Dispersion
Exercise 2.3 | Q 9 | पृष्ठ ३४

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