QJEEM24Ma54f2: JEE Main 2024 Mathematics - Class 12 JEEM24Ma54f2 Solution: If the mean of the following probability distribution of a…

$\begin{aligned}& \text { Mean }=\sum x_i P\left(x_i\right) \\& \frac{46}{9}=4 a+4 a+4 b+12 b+24 b \\& \frac{46}{9}=8 a+40 b \\& \frac{23}{9}=4 a+20 b \\& 36 a+180 b=23 \quad \text{.... (1)}\end{aligned}$

Sum of probability is 1

$\Rightarrow 4 a+6 b=1 \quad \text{... (2)}$

$\begin{aligned}& \text { Solving (1) and (2) } \\& a=\frac{1}{12}, b=\frac{1}{9} \\& \sigma^2=\sum x_i^2 P\left(x_i\right)-\left(\sum x_i P\left(x_i\right)\right)^2 \\& =4 \times 2 a+16(a+b)+36(2 b)+64(3 b)-\left(\frac{46}{9}\right)^2 \\& =8(a+2(a+b)+9 b+24 b)-\left(\frac{46}{9}\right)^2 \\& =8(3 a+35 b)-\left(\frac{46}{9}\right)^2 \\& =8\left(\frac{3}{12}+\frac{35}{9}\right)-\left(\frac{46}{9}\right)^2 \\& =8\left(\frac{149}{36}\right)-\left(\frac{46}{9}\right)^2=\frac{566}{81} \\\end{aligned}$