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If mean and variance of 2, 3, 16, 20, 13, 7, x, y are 10 and 25 respectively, then xy =

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Solution

It is given that
$\frac{2 + 3 + 16 + 20 + 13 + 7 + x + y}{8} = 10$ and
$\frac{1}{8} (2^{2} + 3^{2} + 16^{2} + 20^{2} + 13^{2} + 7^{2} + x^{2} + y^{2}) - 10^{2} = 25$
$\Rightarrow x + y + 61 = 80$ and $\frac{x^{2} + y^{2} + 887}{8} = 125$
$\Rightarrow x + y = 19$ and $x^{2} + y^{2} = 113$
$\Rightarrow (x + y)^{2} = 361$ and $x^{2} + y^{2} = 113$
$\Rightarrow x^{2} + y^{2} + 2 x y = 361$ and $x^{2} + y^{2} = 113$
$\Rightarrow x^{2} + y^{2} + 2 x y = 361$ and $x^{2} + y^{2} = 113$
$\Rightarrow 113 + 2 x y = 361 \Rightarrow 2 x y = 248 \Rightarrow x y = 124$

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Introduction to statistics

Remember concepts with our Masterclasses.

If mean and variance of 2, 3, 16, 20, 13, 7, x, y are 10 and 25 respectively, then xy =

It is given that2+3+16+20+13+7+x+y8=10 and1822+32+162+202+132+72+x2+y2−102=25⇒ x+y+61=80 and x2+y2+8878=125⇒ x+y=19 and x2+y2=113⇒ (x+y)2=361 and x2+y2=113⇒ x2+y2+2xy=361 and x2+y2=113⇒ x2+y2+2xy=361 and x2+y2=113⇒ 113+2xy=361⇒2xy=248⇒xy=124

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