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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
If the variance of the following frequency distribution is $50$ then $x$ is equal to:
| Class | $10-20$ | $20-30$ | $30-40$ |
| Frequency | $2$ | $x$ | $2$ |
