JEE Main 2023 — Mathematics Probability & Statistics

Given that,

For set $1:$1:

Mean $=12$=12 and Variance $=14$=14

For set $2:$2:

Mean $=14$=14 and Variance $={\sigma }^{2}$=σ2

Now, For set $1$1 we have,

$$\Rightarrow \frac{\sum _{i=1}^{i=15}{x}_{i}}{15}=12$$⇒15∑i=1i=15​xi​​=12

$$\Rightarrow \sum _{i=1}^{i=15}{x}_{i}=15\times 12$$⇒i=1∑i=15​xi​=15×12

Also, $\displaystyle \frac{\sum _{i=1}^{i=15}{{x}_{i}}^{2}}{15}-{(\frac{\sum _{i=1}^{i=15}{x}_{i}}{15})}^{2}=14$15∑i=1i=15​xi​2​−(15∑i=1i=15​xi​​)2=14

$$\Rightarrow \frac{\sum _{i=1}^{i=15}{{x}_{i}}^{2}}{15}-{12}^{2}=14$$⇒15∑i=1i=15​xi​2​−122=14

$$\Rightarrow \sum _{i=1}^{i=15}{{x}_{i}}^{2}=(14+144)\times 15$$⇒i=1∑i=15​xi​2=(14+144)×15

Now for set $2:$2:

$$\Rightarrow \frac{\sum _{i=1}^{i=15}{y}_{i}}{15}=14$$⇒15∑i=1i=15​yi​​=14

and $\displaystyle \Rightarrow \sum _{i=1}^{i=15}{y}_{i}=15\times 14$⇒i=1∑i=15​yi​=15×14

Also, $\displaystyle \frac{\sum _{i=1}^{i=15}{{y}_{i}}^{2}}{15}-{(\frac{\sum _{i=1}^{i=15}{y}_{i}}{15})}^{2}={\sigma }^{2}$15∑i=1i=15​yi​2​−(15∑i=1i=15​yi​​)2=σ2

$$\Rightarrow \frac{\sum _{i=1}^{i=15}{{y}_{i}}^{2}}{15}-{14}^{2}={\sigma }^{2}$$⇒15∑i=1i=15​yi​2​−142=σ2

$$\Rightarrow \sum _{i=1}^{i=15}{{y}_{i}}^{2}=({\sigma }^{2}+196)\times 15$$⇒i=1∑i=15​yi​2=(σ2+196)×15

Now,

The combined variance will be

$$\Rightarrow 13=\frac{(\sum _{i=1}^{i=15}{{x}_{i}}^{2})\times 15+(\sum _{i=1}^{i=15}{{y}_{i}}^{2})\times 15}{15+15}-{(\frac{\sum _{i=1}^{i=15}{x}_{i}+\sum _{i=1}^{i=15}{y}_{i}}{15+15})}^{2}$$⇒13=15+15(∑i=1i=15​xi​2)×15+(∑i=1i=15​yi​2)×15​−(15+15∑i=1i=15​xi​+∑i=1i=15​yi​​)2

$$\Rightarrow 13=\frac{(14+144)\times 15+({\sigma }^{2}+196)\times 15}{30}-{13}^{2}$$⇒13=30(14+144)×15+(σ2+196)×15​−132

$$\therefore {\sigma }^{2}=10$$∴σ2=10

Hence, ${\sigma }^{2}=10$σ2=10