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13.Statistics
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The means of five observations is $4$ and their variance is $5.2$. If three of these observations are $1, 2$ and $6$, then the other two are
A
$2$ and $9$
B
$3$ and $8$
C
$4$ and $7$
D
$5$ and $6$
Solution
(c) Let the two unknown items be $x$ and $y$, then
Mean $ = 4 \Rightarrow \frac{{1 + 2 + 6 + x + y}}{5} = 4$
==> $x + y = 11$ …..$(i)$
and variance = $5. 2$
==> $\frac{{{1^2} + {2^2} + {6^2} + {x^2} + {y^2}}}{5} – {({\rm{mean}})^2} = 5.2$
$41 + {x^2} + {y^2} = 5[5.2 + {(4)^2}]$
$41 + {x^2} + {y^2} = 106$
${x^2} + {y^2} = 65$…..$(ii)$
Solving $(i)$ and $(ii)$ for $x$ and $y$, we get
$x = 4,y = 7$ or $x = 7,y = 4$.
Standard 11
Mathematics
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