Ramesh Suresh
No. of matches $10$ $20$
mean $53.3$ $45.3$
S.D. $41$ $17$
The table gives data about mean and SD. of runs in matches played by two players .
Find combined mean and variance.

47.96, 27.7

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49.47, 769.455

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49.00, 769.405

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4847, 768.455

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Solution

The correct option is A $47.96, 27.7$
$n_{1} = 10, {¯ x}_{1} = 53.3, σ_{1} = 41,$ $n_{2} = 20, {¯ x}_{2} = 45.3, σ_{2} = 17$
$¯ x = \frac{n_{1} {¯ x}_{1} + n_{2} {¯ x}_{2}}{n_{1} + n_{2}} = \frac{10 \times 53.3 + 20 \times 45.3}{30} = 47.96$
$d_{1} = {¯ x}_{1} - ¯ x = 53.3 - 47.96 = 5.37$
$d_{2} = {¯ x}_{2} - ¯ x = 45.3 - 47.96 = - 2.66$
$σ^{2} (n_{1} + n_{2}) = n_{1} (σ_{1}^{2} + d_{1}^{2}) + n_{2} (σ_{2}^{2} + d_{2}^{2})$
$σ^{2} (30) = 10 \times (1681 + 28.83) + 20 \times (289 + 7.076)$
$σ^{2} = 767.32$
$σ = 27.7$

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	Ramesh	Suresh
No. of matches	$10$	$20$
mean	$53.3$	$45.3$
S.D.	$41$	$17$

RameshSureshNo. of matches1020mean 53.345.3S.D.4117The table gives data about mean and SD. of runs in matches played by two players .Find combined mean and variance.

The correct option is A 47.96,27.7n1=10,¯x1=53.3,σ1=41, n2=20,¯x2=45.3,σ2=17¯x=n1¯x1+n2¯x2n1+n2=10×53.3+20×45.330=47.96d1=¯x1−¯x=53.3−47.96=5.37d2=¯x2−¯x=45.3−47.96=−2.66σ2(n1+n2)=n1(σ21+d21)+n2(σ22+d22)σ2(30)=10×(1681+28.83)+20×(289+7.076)σ2=767.32σ=27.7

Ramesh Suresh
No. of matches $10$ $20$
mean $53.3$ $45.3$
S.D. $41$ $17$
The table gives data about mean and SD. of runs in matches played by two players .
Find combined mean and variance.