91Ó°ÊÓ

It is given that the density function of $X$is

localid="1646816654969" $f\left(x\right)=\left\{\begin{array}{l}a+b{x}^2;0≤x≤1\\ 0;\text{Otherwise}\end{array}\right.$

and

$E\left(X\right)=\frac{3}{5}$

The given density function is

$f\left(x\right)=\left\{\begin{array}{l}a+b{x}^2;0≤x≤1\\ 0;\text{Otherwise}\end{array}\right.$

According to the given information, the area under the density curve is $1$.

So,

$$\begin{gathered} {∫}_{-∞}^{∞}f\left(x\right)dx=1 \\ ⇒{∫}_0^1\left(a+b{x}^2\right)dx=1 \\ ⇒{\left[ax+\frac{b{x}^3}{3}\right]}_0^1=1 \\ ⇒a\left(1-0\right)+\frac{b\left(1^3-0^3\right)}{3}=1 \\ ⇒a+\frac{b}{3}=1 \\ ⇒3a+b=3..................\left(1\right) \end{gathered}$$

It is given that $E\left[X\right]=\frac{3}{5}$.

According to the given information, the expectation for a continuous random variable is $E\left[X\right]={∫}_{-∞}^{∞}f\left(x\right)dx=1$

$$\begin{gathered} ⇒\frac{3}{5}={∫}_0^{1}x\left(a+b{x}^2\right)dx \\ ⇒\frac{3}{5}={∫}_0^1\left(ax+b{x}^3\right)dx \\ ⇒\frac{3}{5}={\left[\frac{a{x}^2}{2}+\frac{b{x}^4}{4}\right]}_0^1 \\ ⇒\frac{3}{5}=\left[\frac{a\left(1^2-0^2\right)}{2}+\frac{b\left(1^4-0^4\right)}{4}\right] \\ ⇒\frac{3}{5}=\frac{a}{2}+\frac{b}{4} \\ ⇒12=10a+5b......................\left(2\right) \end{gathered}$$

Multiplying equation $\left(1\right)$by $5$we get,

$15a+5b=15$

Subtracting equation $\left(2\right)$from $\left(3\right)$we get,

$$\begin{gathered} 5a=3 \\ ⇒a=\frac{3}{5} \end{gathered}$$

Substituting the value of $a$in equation $\left(1\right)$we get,

$$\begin{gathered} 3\left(\frac{3}{5}\right)+b=3 \\ ⇒b=3-\frac{9}{5} \\ ⇒b=\frac{15-9}{5}=\frac{6}{5} \end{gathered}$$

Therefore, the value of $a=\frac{3}{5}\text{and}b=\frac{6}{5}$and

$$\begin{gathered} f\left(x\right)=\frac{3}{5}+\frac{6}{5}{x}^2 \\ =\frac{3}{5}\left(1+2x^2\right) \end{gathered}$$

So, the density function of $X$will be

$f\left(x\right)=\left\{\begin{array}{l}\frac{3}{5}\left(1+2x^2\right),0≤x≤1\\ 0,\text{Otherwise}\end{array}\right.$