JEE Main 2023 — Mathematics Probability & Statistics

Let the mean is denoted by $\bar{{x}_{i}}$xi​ˉ​, variance by ${{\sigma }_{i}}^{2}$σi​2, so standard deviation will be ${\sigma }_{i}$σi​, now as per given data we get,

$$\begin{matrix}A & & B & & A+B \\ {\bar{x}}_{1}=40 & & {\bar{x}}_{2}=55 & & \bar{x}=50 \\ {\sigma }_{1}=\alpha & & {\sigma }_{2}=30-\alpha & & {\sigma }_{2}=350 \\ {n}_{1}=100 & & {n}_{2}=n & & 100+n\end{matrix}$$Axˉ1​=40σ1​=αn1​=100​​Bxˉ2​=55σ2​=30−αn2​=n​​A+Bxˉ=50σ2​=350100+n​

So, using the formula of mean in $A+B$A+B we get,

$$\bar{x}=\frac{100\times 40+55n}{100+n}=50$$xˉ=100+n100×40+55n​=50

$$\Rightarrow n=200$$⇒n=200

Now using the formula of variance we get,

$${\sigma }_{1}^{2}=\frac{\sum {x}_{i}^{2}}{100}-{40}^{2}={\alpha }^{2}$$σ12​=100∑xi2​​−402=α2

And $\displaystyle {\sigma }_{2}^{2}=\frac{\sum {y}_{i}^{2}}{200}-{55}^{2}={(30-\alpha )}^{2}$σ22​=200∑yi2​​−552=(30−α)2

So, variance of $A+B$A+B will be,

$${\sigma }^{2}=\frac{\sum {x}_{i}^{2}+\sum {y}_{i}^{2}}{300}-{50}^{2}={350}^{2}$$σ2=300∑xi2​+∑yi2​​−502=3502

$$\Rightarrow {350}^{2}=\frac{(1600+{\alpha }^{2})\times 100+[3025+{(30-\alpha )}^{2}]\times 200}{300}-{50}^{2}$$⇒3502=300(1600+α2)×100+[3025+(30−α)2]×200​−502

$$\Rightarrow {\alpha }^{2}-40\alpha +300=0$$⇒α2−40α+300=0

$\Rightarrow \alpha =10$⇒α=10 or $30$30

And $\alpha =30$α=30 is not possible

So, ${\sigma }_{1}+2{\sigma }_{2}=2{10}^{2}+{20}^{2}=500$σ1​+2σ2​=2102+202=500