The means of five observations is 4 and their | vedclass

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Question

(c) Let the two unknown items be $x$ and $y$, then

Mean $ = 4 \Rightarrow \frac{{1 + 2 + 6 + x + y}}{5} = 4$

==> $x + y = 11$ .....$(i)$

a

Vedclass · Accepted Answer

$4$ and $7$

Vedclass · Answer

(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 $x = 7,y = 4$.

Variate $( x )$	$x _{1}$	$x _{1}$	$x _{3} \ldots \ldots x _{15}$
Frequency $(f)$	$f _{1}$	$f _{1}$	$f _{3} \ldots f _{15}$

Classes	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$	$80-90$	$90-100$
${f_i}$	$3$	$7$	$12$	$15$	$8$	$3$	$2$

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

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