Q57 / 150 · MHT CET 2023 (10 May Shift 1) · Mathematics

The p.d.f. of a discrete random variable is defined as $f(x)= \begin{cases}\mathrm{k} x^2, & 0 \leq x \leq 6 \\ 0, & \text { otherwise }\end{cases}$ Then the value of $F(4)$ (c.d.f) is

  • Correct answer
    A
    $\frac{30}{91}$
  • B
    $\frac{30}{97}$
  • C
    $\frac{15}{47}$
  • D
    $\frac{15}{97}$

Solution

$\begin{aligned} \mathrm{f}(x) & =\mathrm{k} x^2, 0 \leq x \leq 6 \\ \mathrm{k}(0)^2 & +\mathrm{k}(1)^2+\mathrm{k}(2)^2+\mathrm{k}(3)^2+\mathrm{k}(4)^2 \\ \Rightarrow \mathrm{k} & +4 \mathrm{k}(5)^2+\mathrm{k}(6)^2=1 \\ \Rightarrow & 91 \mathrm{k}=1 \\ \Rightarrow \mathrm{k} & =\frac{1}{91} \\ \mathrm{~F}(4) & =\mathrm{P}(\mathrm{X} \leq 4)=\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1) \\ & =\mathrm{k}(0)^2+\mathrm{k}(\mathrm{k})^2+\mathrm{k}(2)^2+\mathrm{k}(3)^2+\mathrm{k}(4)^2 \\ & =\mathrm{k}+4 \mathrm{k}+9 \mathrm{k}+16 \mathrm{k} \\ & =30 \mathrm{k} \\ & =30\left(\frac{1}{91}\right) \\ & =\frac{30}{91}\end{aligned}$

Hence, the correct answer is option: 'A'.

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