MCQ
X is a continuous random variable with probability density function
$f(x)=\left\{\begin{array}{cc}\frac{x}{6}+k ; & 0 \leq x \leq 3 \\0 ; & \text { otherwise }\end{array}\right.$
The value of k is equal to
  • $\frac{1}{12}$
  • B
    $\frac{1}{3}$
  • C
    $\frac{1}{4}$
  • D
    $\frac{1}{6}$

Answer

Correct option: A.
$\frac{1}{12}$
(A)
Since, $\mathrm{f}(x)$ is the p.d.f. of X .
$\therefore \quad \int_{-\infty}^{\infty} \mathrm{f}(x) \mathrm{d} x=1$
$\begin{aligned} & \Rightarrow \int_0^3\left(\frac{x}{6}+\mathrm{k}\right) \mathrm{d} x=1 \\ & \Rightarrow\left[\frac{x^2}{12}+\mathrm{k} x\right]_0^3=1 \Rightarrow \frac{3}{4}+3 \mathrm{k}=1 \\ & \Rightarrow 3 \mathrm{k}=\frac{1}{4} \Rightarrow \mathrm{k}=\frac{1}{12}\end{aligned}$

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