Let the mean and variance of 7 observations 2 | vedclass

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(B) Given mean $\overline{x} = 8$ for $7$ observations:$\frac{2+4+10+x+12+14+y}{7} = 8$ $\Rightarrow x+y+42 = 56$ $\Rightarrow x+y = 14$ ....$(1)$Give

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$\frac{4}{5}$

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(B) Given mean $\overline{x} = 8$ for $7$ observations:$\frac{2+4+10+x+12+14+y}{7} = 8$ $\Rightarrow x+y+42 = 56$ $\Rightarrow x+y = 14$ ....$(1)$Given variance $\sigma^2 = 16$:$\frac{2^2+4^2+10^2+x^2+12^2+14^2+y^2}{7} - 8^2 = 16$$\frac{4+16+100+x^2+144+196+y^2}{7} = 80$ $\Rightarrow x^2+y^2+460 = 560$ $\Rightarrow x^2+y^2 = 100$ ....$(2)$From $(1)$, $y = 14-x$. Substituting in $(2)$:$x^2 + (14-x)^2 = 100$ $\Rightarrow x^2 + 196 - 28x + x^2 = 100$ $\Rightarrow 2x^2 - 28x + 96 = 0$ $\Rightarrow x^2 - 14x + 48 = 0$$(x-8)(x-6) = 0$. Since $x > y$, we have $x=8$ and $y=6$.The set is $\{1, 2, 3, 8-4, 6, 5\} = \{1, 2, 3, 4, 6, 5\}$.Total ways to choose $2$ numbers without replacement is $6 	imes 5 = 30$.Let $S$ be the smaller number. We want $P(S $S \geq 4$ means both numbers chosen are from $\{4, 5, 6\}$.Number of ways to choose $2$ numbers from $\{4, 5, 6\}$ is $3 	imes 2 = 6$.$P(S \geq 4) = \frac{6}{30} = \frac{1}{5}$.Therefore, $P(S < 4) = 1 - \frac{1}{5} = \frac{4}{5}$.

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