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Probability Distributions — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsProbability Distributions2 Marks
MCQ
If a c.r.v. X has the probability density function
$\begin{aligned}\mathrm{f}(x) & =\mathrm{C}\left(9-x^2\right) ; 0<x<3 \\& =0 \quad ; \text { otherwise }\end{aligned}$
Then, the value of C is
  • A
    $\frac{1}{16}$
  • B
    $\frac{1}{15}$
  • C
    $\frac{2}{18}$
  • ✓
    $\frac{1}{18}$
✓

Answer

Correct option: D.
$\frac{1}{18}$
(D)
Since, $\mathrm{f}(x)$ is the p.d.f. of X
$\therefore \quad \int_{-\infty}^{\infty} \mathrm{f}(x) \mathrm{d} x=1$
$\begin{aligned} & \Rightarrow \int_0^3 \mathrm{C}\left(9-x^2\right) \mathrm{d} x=1 \\ & \Rightarrow \mathrm{C}\left[9 x-\frac{x^3}{3}\right]_0^3=1 \\ & \Rightarrow \mathrm{C}(27-9)=1 \Rightarrow \mathrm{C}=\frac{1}{18}\end{aligned}$

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