If the mean and variance of five observations | vedclass

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(C) Let the five observations be $x_1, x_2, x_3, x_4, x_5$. Given mean $\bar{X} = \frac{24}{5}$ and variance $\sigma^2 = \frac{194}{25}$. Sum of five

Vedclass · Accepted Answer

$\frac{5}{4}$

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(C) Let the five observations be $x_1, x_2, x_3, x_4, x_5$. Given mean $\bar{X} = \frac{24}{5}$ and variance $\sigma^2 = \frac{194}{25}$. Sum of five observations: $\sum_{i=1}^5 x_i = 5 	imes \frac{24}{5} = 24$. Mean of first four observations is $\frac{7}{2}$,so $\sum_{i=1}^4 x_i = 4 	imes \frac{7}{2} = 14$. Thus,$x_5 = 24 - 14 = 10$. Using the variance formula $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{X})^2$: $\frac{194}{25} = \frac{\sum_{i=1}^5 x_i^2}{5} - (\frac{24}{5})^2$. $\frac{194}{25} = \frac{\sum_{i=1}^5 x_i^2}{5} - \frac{576}{25}$ $\Rightarrow \frac{\sum_{i=1}^5 x_i^2}{5} = \frac{770}{25} = \frac{154}{5}$. $\sum_{i=1}^5 x_i^2 = 154$. Since $x_5 = 10$,$x_5^2 = 100$. $\sum_{i=1}^4 x_i^2 = 154 - 100 = 54$. Variance of first four observations: $	ext{Var} = \frac{\sum_{i=1}^4 x_i^2}{4} - (	ext{mean of first four})^2$. $	ext{Var} = \frac{54}{4} - (\frac{7}{2})^2 = \frac{27}{2} - \frac{49}{4} = \frac{54 - 49}{4} = \frac{5}{4}$.

$X$	$35$	$54$	$52$	$53$	$56$	$58$	$52$	$50$	$51$	$49$
$Y$	$108$	$107$	$105$	$105$	$106$	$107$	$104$	$103$	$104$	$101$

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:

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