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

Question

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]

Smart Achievers · Accepted Answer

(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 $= \frac{194}{25}$

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

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

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

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

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

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

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 $\frac{4}{5}$

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Solution

Q. If the mean and variance of five observations are 245 and 19425 respectively and the mean of the first four observations is 72, then the variance of the first four observations is equal to: [2024]

1 7712

2 1054

3 54