# The mean and variance of 5 observations of an experiment are 4 and 5.2 respectively. If from these observations three are 1, 2 and 6, then remaining will be

## Solution :

As given $$\bar{x}$$ = 4, n = 5 and $${\sigma}^2$$ = 5.2. If the remaining observations are $$x_1$$, $$x_2$$ then

$${\sigma}^2$$ = $$\sum{(x_i – \bar{x})}^2\over n$$ = 5.2

$$\implies$$ $${(x_1-4)}^2 + {(x_2-4)}^2 + {(1-4)}^2 + {2-4)}^2 + {(6-4)}^2\over 5$$ = 5.2

$$\implies$$ $${(x_1-4)}^2 + {(x_2-4)}^2$$ = 9  …..(1)

Also $$\bar{x}$$ = 4 $$\implies$$ $$x_1 + x_2 + 1 + 2 + 6\over 5$$ = 4 $$\implies$$ $$x_1 + x_2$$ = 11  ….(2)

from eq.(1), (2)  $$x_1$$, $$x_2$$ = 4, 7

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