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

Given: X = 11, 12, 13,, 41 and Y = 61, 62, 63,, 91. Find: | x + y - ^2|. From the solution, first compute the means: x = 11 + 412 = 26, y = 61 + 912 = 76 Next, the combined variance is taken as: ^2 = (41 - 11)^2 + (91 - 61)^212 + 31 - 1 = 705 Then, | x + y - ^2| = |26 + 76 - 705| = |-603| = 603 Therefore, the required value is 603.

Let X = 11, 12, 13,, 41 and Y = 61, 62, 63,, 91 be two sets of ob… — Solution | JEE Mathematics

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

Answer

Correct answer:603

Step-by-step solution

Standard Method

Given: $X = \{11, 12, 13, \dots, 41\}$ and $Y = \{61, 62, 63, \dots, 91\}$ .

Find: $|\bar{x} + \bar{y} - \sigma^2|$ .

From the solution, first compute the means:

\bar{x} = \frac{11 + 41}{2} = 26, \quad \bar{y} = \frac{61 + 91}{2} = 76

Next, the combined variance is taken as:

\sigma^2 = \frac{(41 - 11)^2 + (91 - 61)^2}{12 + 31 - 1} = 705

Then,

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

Therefore, the required value is $603$ .

Common mistakes

Using the mean formula incorrectly for an arithmetic sequence. The mean of consecutive integers from the first term to the last term is the average of the two extremes. Use $\frac{a+l}{2}$ for both sets here.
Computing the variance separately for $X$ and $Y$ but not for all observations in $X \cup Y$ . The question asks for the variance of the combined data set, so the calculation must use all observations together.
Dropping the modulus sign at the final step. After obtaining $26 + 76 - 705 = -603$ , the absolute value must be taken, giving $603$ instead of $-603$ .

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