# The mean and variance of seven observations are $8 \; and \; 16$, respectively. If $5$ of the observations are $2,4,10,12,14,$ then the product of the remaining two observations is :   Option 1) $49$ Option 2) $48$ Option 3) $45$   Option 4) $40$

Mean $\bar{a}=\frac{\sum x_ i}{n}$

Variance $=\sigma ^{2}=\frac{\sum x_i^{2}}{n}-(\bar{x})^{2}$

Given $\bar{x}=8$

$\bar{x}=\frac{2+4+10+12+14+a+b}{7}=8$

$=\frac{16+26+a+b}{7}=8$

$=42+a+b=56$

$a+b=14\; \; \; (1)$

$\sigma ^{2}=\frac{4+16+100+144+196+a^{2}+b^{2}}{7}-64=16$

$=460-448+a^{2}+b^{2}=112$

$=12+a^{2}+b^{2}=112$

$a^{2}+b^{2}=100$

$(a+b)^{2}=a^{2}+b^{2}+2ab$

$\\a+b=14 \; \; \; \; \; from (1)\\\; \; \; \\a^{2}+b^{2}=100\; \; \; \; \; \; \; \; from (11)$

$\\196=100+2ab\\\; \; \; \\96=2ab\\\; \; \; \\ab=48$

Option 1)

$49$

Option 2)

$48$

Option 3)

$45$

Option 4)

$40$

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