The mean and variance of n observations are 8 and 16, respectively.

Question

The mean and variance of n observations are 8 and 16, respectively. If the sum of the first (n - 1) observations is 48 and the sum of squares of the first (n - 1) observations is 496, then the value of n is :

Accepted Answer

${X}_{1}+{X}_{2}+.......+{X}_{n-1}+{X}_{n}=8n$

$48+{x}_{n}=8n$

$\Rightarrow {x}_{n}=8n-48\, \, \ldots (1)$

$\Rightarrow 16=\frac{496+{x}_{2}^{n}}{n}-(8{)}^{2}$

$\Rightarrow 80n=496+{x}_{2}^{n}$

$\Rightarrow {x}_{2}^{n}=80n-496\, \, \ldots (2)$

$\Rightarrow (8n-48{)}^{2}=80n-496$

$\Rightarrow 64(n-6{)}^{2}=8(10n-62)$

$\Rightarrow 8(n-6{)}^{2}=10n-62$

$\Rightarrow 4(n-6{)}^{2}=5n-31$

$\Rightarrow 4({n}^{2}-36-12n)=5n-31$

$\Rightarrow 4{n}^{2}+144-48n=5n-31$

$\Rightarrow 4{n}^{2}-53n+175=0$

$\Rightarrow 4{n}^{2}-28n-25n+175=0$

$\Rightarrow 4n(n-7)-25(n-7)=0$

$n=7$

Solutions · 1