JEE Main 2026 — Statistics Question with Solution

Given the mean of the seven observations is $8$:

$\frac{2+4+\alpha +8+\beta +12+14}{7}=8$

$\frac{40+\alpha +\beta }{7}=8$

$\alpha +\beta =16$

Given the variance of the observations is $16$:

$\frac{\sum x_i^2}{n}-(\text{Mean}{)}^2=16$

$\frac{2^2+4^2+{\alpha }^2+8^2+{\beta }^2+12^2+14^2}{7}-8^2=16$

$\frac{4+16+{\alpha }^2+64+{\beta }^2+144+196}{7}-64=16$

$\frac{424+{\alpha }^2+{\beta }^2}{7}=80$

$424+{\alpha }^2+{\beta }^2=560$

${\alpha }^2+{\beta }^2=136$

Using the identity $(\alpha +\beta {)}^2={\alpha }^2+{\beta }^2+2\alpha \beta$:

$16^2=136+2\alpha \beta$

$256=136+2\alpha \beta$

$2\alpha \beta =120\Rightarrow \alpha \beta =60$

Solving for $\alpha$ and $\beta$, they are the roots of the equation $t^2-16t+60=0$, which are $6$ and $10$.

Since $\alpha <\beta$, we have $\alpha =6$ and $\beta =10$.

The roots of the required quadratic equation are $3\alpha +2$ and $2\beta +1$:

$3\alpha +2=3(6)+2=20$

$2\beta +1=2(10)+1=21$

Sum of the roots $=20+21=41$

Product of the roots $=20\times 21=420$

The required quadratic equation is $x^2-(\text{Sum})x+(\text{Product})=0$:

$x^2-41x+420=0$

Answer: $x^2-41x+420=0$