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The mean and variance of $7$ observations are $8$ and $16$ respectively. If two observations are $6$ and $8 ,$ then the variance of the remaining $5$ observations is:
$\frac{92}{5}$
$\frac{134}{5}$
$\frac{536}{25}$
$\frac{112}{5}$
Solution
Let $8,16, \mathrm{x}_{1}, \mathrm{x}_{2}, \mathrm{x}_{3}, \mathrm{x}_{4}, \mathrm{x}_{5}$ be the observations.
Now $\frac{x_{1}+x_{2}+\ldots+x_{5}+14}{7}=8….(i)$
$\Rightarrow \sum_{i=1}^{5} x_{i}=42$
Also $\frac{x_{1}^{2}+x_{2}^{2}+\ldots x_{5}^{2}+8^{2}+6^{2}}{7}-64=16$
$\Rightarrow \sum_{i=1}^{5} x_{i}^{2}=560-100=460….(ii)$
So variance of $x_{1}, x_{2}, \ldots, x_{5}$
$=\frac{460}{5}-\left(\frac{42}{5}\right)^{2}=\frac{2300-1764}{25}=\frac{536}{25}$
Similar Questions
Let $\mu$ be the mean and $\sigma$ be the standard deviation of the distribution
| $X_i$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ |
| $f_i$ | $k+2$ | $2k$ | $K^{2}-1$ | $K^{2}-1$ | $K^{2}-1$ | $k-3$ |
where $\sum f_i=62$. if $[x]$ denotes the greatest integer $\leq x$, then $\left[\mu^2+\sigma^2\right]$ is equal $………$.
