Let X = { 11 , 12 , 13 , … , 40 , 41 } and Y = { 61 , 62 , 63 , … , 90 , 91 } be the two sets of observations. If x ¯ and y ¯ are their respective means and 2 is the variance of all the observations in X Y , then | x ¯ + y ¯ - 2 | is equal to

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Solution

Q.
Let $X = {11, 12, 13, \dots, 40, 41}$ and $Y = {61, 62, 63, \dots, 90, 91}$ be the two sets of observations. If $\bar{x}$ and $\bar{y}$ are their respective means and $σ^{2}$ is the variance of all the observations in $X \cup Y$ , then $| \bar{x} + \bar{y} - σ^{2} |$ is equal to _______ . [2023]

Ans.
(603)

$Mean \bar{x} = \frac{\sum_{i = 11}^{41} i}{31} = \frac{11 + 41}{2} = 26$

$\bar{y} = \frac{\sum_{j = 61}^{91} j}{31} = \frac{61 + 91}{2} = 76$

$Combined mean μ = \frac{31 \times 26 + 31 \times 76}{31 + 31} = \frac{26 + 76}{2} = 51$

$σ^{2} = \frac{1}{62} (\sum_{i = 1}^{31} {(x_{i} - μ)}^{2} + \sum_{i = 1}^{31} {(y_{i} - μ)}^{2}) = 705$

$∴$ $| \bar{x} + \bar{y} - σ^{2} |$ $=$ $| 26 + 76 - 705 |$ $= 603$

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