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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
$40$
$45$
$49$
$48$
Solution
Let $7$ observation be ${x_1},{x_2},{x_3},{x_4},{x_5},{x_6},{x_7}$
$\bar x = 8 \Rightarrow \sum\limits_{i = 1}^7 {{x_i}} = 56\,\,\,\,\,\,…….\left( 1 \right)$
Also ${\sigma ^2} = 16$
$ \Rightarrow 16 = \frac{1}{7}\left( {\sum\limits_{i = 1}^7 {x_i^2} } \right) – {\left( {\bar x} \right)^2}$
$ \Rightarrow 16 = \frac{1}{7}\left( {\sum\limits_{i = 1}^7 {x_i^2} } \right) – 64$
$ \Rightarrow \left( {\sum\limits_{i = 1}^7 {x_i^2} } \right) = 560\,\,\,\,\,\,\,\,\,…….\left( 2 \right)$
Now, ${x_1} = 2,{x_2} = 4,{x_3} = 10,{x_4} = 12,{x_5} = 14$
$ \Rightarrow {x_6} + {x_7} = 14$ (from $(1)$) and $x_6^2 + x_7^2 = 100$ (from$(2)$)
$\therefore x_6^2 + x_7^2 = {\left( {{x_6} + {x_7}} \right)^2} – 2{x_6}{x_7} \Rightarrow {x_6}{x_7} = 48$
Similar Questions
Let the mean and variance of the frequency distribution
| $\mathrm{x}$ | $\mathrm{x}_{1}=2$ | $\mathrm{x}_{2}=6$ | $\mathrm{x}_{3}=8$ | $\mathrm{x}_{4}=9$ |
| $\mathrm{f}$ | $4$ | $4$ | $\alpha$ | $\beta$ |
be $6$ and $6.8$ respectively. If $x_{3}$ is changed from $8$ to $7 ,$ then the mean for the new data will be:
Let $\mathrm{X}$ be a random variable with distribution.
| $\mathrm{x}$ | $-2$ | $-1$ | $3$ | $4$ | $6$ |
| $\mathrm{P}(\mathrm{X}=\mathrm{x})$ | $\frac{1}{5}$ | $\mathrm{a}$ | $\frac{1}{3}$ | $\frac{1}{5}$ | $\mathrm{~b}$ |
If the mean of $X$ is $2.3$ and variance of $X$ is $\sigma^{2}$, then $100 \sigma^{2}$ is equal to :
