The mean and variance of a data of 10 observa | vedclass

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(D) Let the $10$ observations be $x_1, x_2, \ldots, x_9, \alpha$. Given mean $\bar{x} = 10$, so $\frac{\sum_{i=1}^9 x_i + \alpha}{10} = 10 \Rightarrow

Vedclass · Accepted Answer

$20$

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(D) Let the $10$ observations be $x_1, x_2, \ldots, x_9, \alpha$. Given mean $\bar{x} = 10$, so $\frac{\sum_{i=1}^9 x_i + \alpha}{10} = 10 \Rightarrow \sum_{i=1}^9 x_i + \alpha = 100$. Given variance $\sigma^2 = 2$, so $\frac{\sum_{i=1}^9 x_i^2 + \alpha^2}{10} - (10)^2 = 2 \Rightarrow \sum_{i=1}^9 x_i^2 + \alpha^2 = 1020$. When $\alpha$ is replaced by $\beta$, the new mean is $10.1$, so $\frac{\sum_{i=1}^9 x_i + \beta}{10} = 10.1 \Rightarrow \sum_{i=1}^9 x_i + \beta = 101$. Subtracting the first mean equation from this, $\beta - \alpha = 101 - 100 = 1 \Rightarrow \beta = \alpha + 1$. The new variance is $1.99$, so $\frac{\sum_{i=1}^9 x_i^2 + \beta^2}{10} - (10.1)^2 = 1.99$. $\sum_{i=1}^9 x_i^2 + \beta^2 = 10(1.99 + 102.01) = 10(104) = 1040$. Subtracting the first variance equation from this, $\beta^2 - \alpha^2 = 1040 - 1020 = 20$. Since $\beta^2 - \alpha^2 = (\beta - \alpha)(\beta + \alpha) = 20$ and $\beta - \alpha = 1$, we have $1 	imes (\beta + \alpha) = 20$. Therefore, $\alpha + \beta = 20$.

\text{Parameter}	\text{Firm } $A$ \text{ and Firm } $B$
\text{No. of wage earners}	$A: 586, B: 648$
\text{Mean of monthly wages}	$Rs. 5253$
\text{Variance of wages}	$A: 100, B: 121$

List-$I$	List-$II$
$(P)$ The mean of the above data is	$(1) 2.5$
$(Q)$ The median of the above data is	$(2) 5$
$(R)$ The mean deviation about the mean of the above data is	$(3) 6$
$(S)$ The mean deviation about the median of the above data is	$(4) 2.7$
	$(5) 2.4$

The mean and variance of a data of $10$ observations are $10$ and $2$, respectively. If an observation $\alpha$ in this data is replaced by $\beta$, then the mean and variance become $10.1$ and $1.99$, respectively. Then $\alpha + \beta$ equals.

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