ExamBro
ExamBro
MHT CET · Maths · Probability

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

  1. A \(\frac{30}{91}\)
  2. B \(\frac{30}{97}\)
  3. C \(\frac{15}{47}\)
  4. D \(\frac{15}{97}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{30}{91}\)

Step-by-step Solution

Detailed explanation

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