MHT CET · Maths · Probability
If the probability density function of a continuous random variable is \(f(x)=\frac{x^{3}}{3}\) if \(-1 < x < 2\)
\(=0\), otherwise,
then the cumulative distribution function of \(X\) is
- A \(\frac{1}{14}\left[x^{4}-1\right]\)
- B \(\frac{1}{10}\left[x^{4}-1\right]\)
- C \(\frac{1}{16}\left[x^{4}-1\right]\)
- D \(\frac{1}{12}\left[x^{4}-1\right]\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{12}\left[x^{4}-1\right]\)
Step-by-step Solution
Detailed explanation
c.d.f. of \(x\) is given by
\(\begin{aligned}
\mathrm{f}(\mathrm{x}) &=\int_{-1}^{\mathrm{x}} \mathrm{f}(\mathrm{y}) \mathrm{dy} \quad=\int_{-1}^{\mathrm{x}} \frac{\mathrm{y}^{3}}{3} \mathrm{dy} \\
&=\left[\frac{\mathrm{y}^{4}}{12}\right]_{-1}^{\mathrm{x}}=\frac{1}{12}\left(\mathrm{x}^{4}-1\right)
\end{aligned}\)
\(\begin{aligned}
\mathrm{f}(\mathrm{x}) &=\int_{-1}^{\mathrm{x}} \mathrm{f}(\mathrm{y}) \mathrm{dy} \quad=\int_{-1}^{\mathrm{x}} \frac{\mathrm{y}^{3}}{3} \mathrm{dy} \\
&=\left[\frac{\mathrm{y}^{4}}{12}\right]_{-1}^{\mathrm{x}}=\frac{1}{12}\left(\mathrm{x}^{4}-1\right)
\end{aligned}\)
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