Let $X_{1}, X_{2}, \ldots X_{18}$ be eighteen observations such that $\sum_{i=1}^{18}\left(X_{i}-\alpha\right)=36$ and $\sum_{i=1}^{18}\left(X_{i}-\beta\right)^{2}=90$, where $\alpha$ and $\beta$ are distinct real numbers. If the standard deviation of these observations is 1, then the value of $|\alpha-\beta|$ is ____

JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Statistics

Let

X_{1}, X_{2}, \dots X_{18}

be eighteen observations such that

\sum_{i = 1}^{18} (X_{i} - α) = 36

and

\sum_{i = 1}^{18} (X_{i} - β)^{2} = 90

, where

α

and

β

are distinct real numbers. If the standard deviation of these observations is 1, then the value of

∣ α - β ∣

is ____

Answer: 4

From

\sum_{i = 1}^{18} (X_{i} - α) = 36

, we get

\sum X_{i} = 18 α + 36

, so the mean

\overset{ˉ}{X} = α + 2

. The variance is

σ^{2} = 1

, so

\sum (X_{i} - \overset{ˉ}{X})^{2} = 18

. Expanding

\sum (X_{i} - β)^{2} = \sum (X_{i} - \overset{ˉ}{X} + \overset{ˉ}{X} - β)^{2} = \sum (X_{i} - \overset{ˉ}{X})^{2} + 18 (\overset{ˉ}{X} - β)^{2} = 18 + 18 (\overset{ˉ}{X} - β)^{2} = 90

. Thus

(\overset{ˉ}{X} - β)^{2} = 4

, giving

∣ \overset{ˉ}{X} - β ∣ = 2

. Since

\overset{ˉ}{X} = α + 2

, we have

∣ α + 2 - β ∣ = 2

. This gives

∣ α - β ∣ = 4

∣ α - β ∣ = 0

. Since

α \neq = β

, we have

∣ α - β ∣ = 4

JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Integer answer