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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Statistics
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The mean of the numbers $ a, b, 8, 5, 10$ is $6$ and the variance is $6.80$. Then which one of the following gives possible values of $a$ and $b$ ?

$\begin {array} {1 1} (A)\;a=0\: b=7 & \quad (B)\;a=5\: b=2 \\ (C)\;a=1\: b=6 & \quad (D)\;a=3\: b=4 \end {array}$

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1 Answer

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According to the given condition
$ 6.80 = \large\frac{(6-a)^2+(6-b)^2+(6-8)^2+(6-5)^2+(6-10)^2}{5}$
$ \Rightarrow (6-a)^2+(6-b)^2+4+1+16=34$
$ (6-a)^2+(6-b)^2=34-21$
$ (6-a)^2+(6-b)^2=13$
$(6-a)^2+(6-b)^2=9+4$
$(6-a)^2+(6-b)^2=3^2+2^2$
$(6-a)^2 = 3^2 \: (6-b)^2 = 2^2$
$ 6-a = 3\: \: \:\: \: \: \: \:\; \: \: 6-b=2$
$ -a=-3\: \: \: \: \: \: \: \:\: \: \: \: -b=-4$
$ \: \: \: a = 3\: \: \: \: \: \: \: \:\: \: \: \: \: \: \: \: \: \: \: \: \: b=4$
Ans : (D)
answered Jan 29, 2014 by thanvigandhi_1
 

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