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Statistics question

2021 · 26 Feb · Shift 2 · Q46
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Statistics question

2021 · 26 Feb · Shift 2 · Q46

JEE MainMathematicsStatisticsNumerical+4 / −1
Let X1, X2, ......., X18 be eighteen observations such that ∑i=118(Xi−α)=36\sum\limits_{i = 1}^{18} {({X_i} - } \alpha ) = 36i=1∑18​(Xi​−α)=36 and ∑i=118(Xi−β)2=90\sum\limits_{i = 1}^{18} {({X_i} - } \beta {)^2} = 90i=1∑18​(Xi​−β)2=90, where α\alphaα and β\betaβ are distinct real numbers. If the standard deviation of these observations is 1, then the value of |α−β\alpha-\betaα−β | is ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 4

  1. Use the given sum to find the mean

We are given ∑i=118(Xi−α)=36.\sum_{i=1}^{18}(X_i-\alpha)=36.∑i=118​(Xi​−α)=36.

Expanding, ∑i=118Xi−18α=36.\sum_{i=1}^{18}X_i-18\alpha=36.∑i=118​Xi​−18α=36.

If the mean is Xˉ\bar XXˉ, then ∑i=118Xi=18Xˉ.\sum_{i=1}^{18}X_i=18\bar X.∑i=118​Xi​=18Xˉ.

So, 18Xˉ−18α=3618\bar X-18\alpha=3618Xˉ−18α=36 18(Xˉ−α)=3618(\bar X-\alpha)=3618(Xˉ−α)=36 Xˉ−α=2.\bar X-\alpha=2.Xˉ−α=2.

Hence, Xˉ=α+2.\bar X=\alpha+2.Xˉ=α+2.

  1. Use the standard deviation information

The standard deviation is 111, so the variance is 111. For n=18n=18n=18 observations, 118∑i=118(Xi−Xˉ)2=1.\frac{1}{18}\sum_{i=1}^{18}(X_i-\bar X)^2=1.181​∑i=118​(Xi​−Xˉ)2=1.

Therefore, ∑i=118(Xi−Xˉ)2=18.\sum_{i=1}^{18}(X_i-\bar X)^2=18.∑i=118​(Xi​−Xˉ)2=18.

  1. Use the identity for shift of origin

We are given ∑i=118(Xi−β)2=90.\sum_{i=1}^{18}(X_i-\beta)^2=90.∑i=118​(Xi​−β)2=90.

Using the identity ∑i=1n(Xi−a)2=∑i=1n(Xi−Xˉ)2+n(Xˉ−a)2,\sum_{i=1}^{n}(X_i-a)^2=\sum_{i=1}^{n}(X_i-\bar X)^2+n(\bar X-a)^2,∑i=1n​(Xi​−a)2=∑i=1n​(Xi​−Xˉ)2+n(Xˉ−a)2,

with a=βa=\betaa=β and n=18n=18n=18, 90=18+18(Xˉ−β)2.90=18+18(\bar X-\beta)^2.90=18+18(Xˉ−β)2.

So, 72=18(Xˉ−β)272=18(\bar X-\beta)^272=18(Xˉ−β)2 (Xˉ−β)2=4.(\bar X-\beta)^2=4.(Xˉ−β)2=4.

Thus, ∣Xˉ−β∣=2.|\bar X-\beta|=2.∣Xˉ−β∣=2.

  1. Find ∣α−β∣|\alpha-\beta|∣α−β∣

Since Xˉ=α+2,\bar X=\alpha+2,Xˉ=α+2, we get Xˉ−β=(α+2)−β=(α−β)+2.\bar X-\beta=(\alpha+2)-\beta=(\alpha-\beta)+2.Xˉ−β=(α+2)−β=(α−β)+2.

And from above, ∣Xˉ−β∣=2,|\bar X-\beta|=2,∣Xˉ−β∣=2, so ∣(α−β)+2∣=2.|(\alpha-\beta)+2|=2.∣(α−β)+2∣=2.

Thus, (α−β)+2=±2.(\alpha-\beta)+2=\pm 2.(α−β)+2=±2.

So either α−β=0\alpha-\beta=0α−β=0 or α−β=−4.\alpha-\beta=-4.α−β=−4.

But the question states that α\alphaα and β\betaβ are distinct, so α−β≠0\alpha-\beta\neq 0α−β=0. Hence, α−β=−4.\alpha-\beta=-4.α−β=−4.

Therefore, ∣α−β∣=4.|\alpha-\beta|=4.∣α−β∣=4.

  1. Final answer

4\boxed{4}4​

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