Let \(\mathrm{X}=\{11,12,13, \ldots, 40,41\}\) and \(\mathrm{Y}=\{61,62, 63, \ldots, 90,91\}\) be the two sets of observations. If \(\bar{x}\) and \(\bar{y}\) are their respective means and \(\sigma^{2}\) is the variance of all the observations in \(\mathrm{X} \cup \mathrm{Y}\), then \(|\overline{\mathrm{x}}+\overline{\mathrm{y}}-\sigma^{2}|\) is equal to ______.

JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Statistics

Let

X = {11, 12, 13, \dots, 40, 41}

and

Y = {61, 62, 63, \dots, 90, 91}

be the two sets of observations. If

\overset{x}{ˉ}

and

\overset{y}{ˉ}

are their respective means and

σ^{2}

is the variance of all the observations in

X \cup Y

, then

∣ \overline{x} + \overline{y} - σ^{2} ∣

is equal to ______.

Answer: 603

Each set has

31

terms, so

\overset{x}{ˉ} = \frac{11 + 41}{2} = 26

and

\overset{y}{ˉ} = \frac{61 + 91}{2} = 76

, and the mean of all

62

values is

μ = \frac{26 + 76}{2} = 51

. Measured from

51

, the entries of

X

are

- 40, - 39, \dots, - 10

and those of

Y

are

10, 11, \dots, 40

, so each set contributes

\sum_{k = 10}^{40} k^{2}

to the total squared deviation. Now

\sum_{k = 10}^{40} k^{2} = \frac{40 \cdot 41 \cdot 81}{6} - \frac{9 \cdot 10 \cdot 19}{6} = 22140 - 285 = 21855

, hence

σ^{2} = \frac{2 \times 21855}{62} = 705

. Therefore

∣ \overset{x}{ˉ} + \overset{y}{ˉ} - σ^{2} ∣ = ∣26 + 76 - 705∣ = 603

JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Integer answer