If the mean and variance of five observations are and respectively and the mean of the first four observations is , then the variance of the first four observations is equal to

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Solution

Q.
If the mean and variance of five observations are $\frac{24}{5}$ and $\frac{194}{25}$ respectively and the mean of the first four observations is $\frac{7}{2}$ , then the variance of the first four observations is equal to: [2024]

1 $\frac{77}{12}$

2 $\frac{105}{4}$

3 $\frac{5}{4}$

4 $\frac{4}{5}$

Ans.
(3)

Let $a, b, c, d$ and $e$ be the five observations.

$∴$ $Mean of 5 observations, \bar{x} = \frac{24}{5}$

$\Rightarrow a + b + c + d + e = 24$ ...(i)

$Mean of four observations, {\bar{x}}_{1} = \frac{7}{2}$

$\Rightarrow a + b + c + d = 14$ ...(ii)

From (i) and (ii), we have $e = 10$

Now, variance of 5 observations $= \frac{194}{25}$

$\Rightarrow \frac{\sum x_{i}^{2}}{5} - {(\bar{x})}^{2} = \frac{194}{25}$

$\Rightarrow a^{2} + b^{2} + c^{2} + d^{2} + e^{2} = 5 (\frac{194}{25} + \frac{576}{25}) = 154$

$\Rightarrow a^{2} + b^{2} + c^{2} + d^{2} = 154 - 100 (∵ e = 10)$

$\Rightarrow a^{2} + b^{2} + c^{2} + d^{2} = 54$

$∴ Variance of 4 observations = \frac{a^{2} + b^{2} + c^{2} + d^{2}}{4} - {({\bar{x}}_{1})}^{2}$

$= \frac{54}{4} - \frac{49}{4} = \frac{5}{4}$

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