Find (i) k (ii) mean (iii) variance for the p.d.f

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Find (i) k (ii) mean (iii) variance for the p.d.f

written 7.0 years ago by

teamques10 ★ 69k

modified 3.0 years ago by

RakeshBhuse • 3.2k

Subject: Applied Mathematics 4

Topic: Probability

Difficulty: Medium

$f(x) = k (x−x^2 ) , 0 \leq x \leq 1. k \gt 0$ = 0 ,otherwise

engineering mathematics

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1 Answer

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written 3.0 years ago by

RakeshBhuse • 3.2k

For continuous random variable

$\int_{\infty}^{\infty} f(x)=1$

For given problem

$\begin {aligned}\int_{0}^{1} f(x) d x=1\\ \therefore \int_{0}^{1} k\left(x-x^{2}\right)=1\\ \therefore k\left(\frac{x^{2}}{2}-\frac{x^{3}}{3}\right)_{0}^{1}=1\\ \therefore k\left(\frac{1}{2}- \frac{1}{3}\right)=1\\ \therefore k\left(\frac{3-2}{6}\right)=1\\ \therefore k=6 \end{aligned}$

Mean

$$\begin{aligned} E(X) &=\int_{\infty}^{-\infty} x f(x) d x\\ &=\int_{0}^{1} x k\left(x-x^{2}\right) d x\\ &=k \int_{0}^{1}\left(x^{2}-x^{3}\right) d x\\ &=k\left(\frac{x^{3}}{3} - \frac{x^{4}}{4}\right)_{0}^{1}\\ &=k\left(\frac{1}{3}-\frac{1}{4}\right)\\ &=6\left(\frac{4-3}{12}\right)\\ &=\frac{1}{2}=0.5 …

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written 3.0 years ago by

RakeshBhuse • 3.2k

For continuous random variable

$\int_{\infty}^{\infty} f(x)=1$

For given problem

$\begin {aligned}\int_{0}^{1} f(x) d x=1\\ \therefore \int_{0}^{1} k\left(x-x^{2}\right)=1\\ \therefore k\left(\frac{x^{2}}{2}-\frac{x^{3}}{3}\right)_{0}^{1}=1\\ \therefore k\left(\frac{1}{2}- \frac{1}{3}\right)=1\\ \therefore k\left(\frac{3-2}{6}\right)=1\\ \therefore k=6 \end{aligned}$

Mean

$$\begin{aligned} E(X) &=\int_{\infty}^{-\infty} x f(x) d x\\ &=\int_{0}^{1} x k\left(x-x^{2}\right) d x\\ &=k \int_{0}^{1}\left(x^{2}-x^{3}\right) d x\\ &=k\left(\frac{x^{3}}{3} - \frac{x^{4}}{4}\right)_{0}^{1}\\ &=k\left(\frac{1}{3}-\frac{1}{4}\right)\\ &=6\left(\frac{4-3}{12}\right)\\ &=\frac{1}{2}=0.5 …

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