MHT CET · Maths · Probability
The p.d.f. of a continuous r.v. \(\mathrm{X}\) is given by \(\mathrm{f}(x)=\frac{x}{8}, 0 < x < 4\) \(=0\), otherwise, then \(\mathrm{P}(\mathrm{X} \leq 2)\) is
- A \(\frac{5}{16}\)
- B \(\frac{9}{16}\)
- C \(\frac{7}{16}\)
- D \(\frac{1}{4}\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{4}\)
Step-by-step Solution
Detailed explanation
(C)
\(\mathrm{P}(\mathrm{x} \leq 2) =\int_{0}^{2} \mathrm{f}(\mathrm{x}) \mathrm{dx}\)
\(=\int_{0}^{2} \frac{\mathrm{x}}{8} \mathrm{dx}=\frac{1}{8}\left[\frac{\mathrm{x}^{2}}{2}\right]_{0}^{2} \Rightarrow \frac{2^{2}}{16}-0=\frac{4}{16}=\frac{1}{4}\)
\(\mathrm{P}(\mathrm{x} \leq 2) =\int_{0}^{2} \mathrm{f}(\mathrm{x}) \mathrm{dx}\)
\(=\int_{0}^{2} \frac{\mathrm{x}}{8} \mathrm{dx}=\frac{1}{8}\left[\frac{\mathrm{x}^{2}}{2}\right]_{0}^{2} \Rightarrow \frac{2^{2}}{16}-0=\frac{4}{16}=\frac{1}{4}\)
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