JEE Main 2024 — Mathematics Probability & Statistics

Given observations are ${x}_{1},{x}_{2},...,{x}_{10}$x1​,x2​,...,x10​

$$\Rightarrow \sum _{i=1}^{10}({x}_{i}-\alpha )=2$$⇒i=1∑10​(xi​−α)=2

$$\Rightarrow \sum _{i=1}^{10}{x}_{i}-10\alpha =2$$⇒i=1∑10​xi​−10α=2

Mean, $\displaystyle \mu =\frac{6}{5}=\frac{\sum {x}_{i}}{10}$μ=56​=10∑xi​​

$$\Rightarrow \sum {x}_{i}=12$$⇒∑xi​=12

$$\Rightarrow 10\alpha +2=12$$⇒10α+2=12

$$\Rightarrow \alpha =1$$⇒α=1

Now, $\displaystyle \sum _{i=1}^{10}{({x}_{i}-\beta )}^{2}=40$i=1∑10​(xi​−β)2=40. Let ${y}_{i}={x}_{i}-\beta$yi​=xi​−β

$$\Rightarrow {\sigma }_{y}^{2}=\frac{1}{10}\sum {y}_{i}^{2}-{(\bar{y})}^{2}$$⇒σy2​=101​∑yi2​−(yˉ​)2

$$\Rightarrow \frac{84}{25}=\frac{1}{10}\sum {({x}_{i}-\beta )}^{2}-{(\frac{\sum _{i=1}^{10}({x}_{i}-\beta )}{10})}^{2}$$⇒2584​=101​∑(xi​−β)2−(10∑i=110​(xi​−β)​)2

$$\Rightarrow \frac{84}{25}=4-{(\frac{12-10\beta }{10})}^{2}$$⇒2584​=4−(1012−10β​)2

$$\Rightarrow {(\frac{6-5\beta }{5})}^{2}=4-\frac{84}{25}$$⇒(56−5β​)2=4−2584​

$$\Rightarrow {(\frac{6-5\beta }{5})}^{2}=\frac{16}{25}$$⇒(56−5β​)2=2516​

$$\Rightarrow 6-5\beta =\pm 4$$⇒6−5β=±4

$\Rightarrow \beta =\frac{2}{5}$⇒β=52​ (not possible) or $\beta =2$β=2

Hence, $\frac{\beta }{\alpha }=2$αβ​=2