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I have done this problem in two ways and I get two different answer.Which one is correct.

I provided the link to the image below.

https://drive.google.com/file/d/1acToL8QBVq05mTQ9t7TgHgedremJdydA/view?usp=drivesdk

enter image description here

For the probability density function $$ f(x)=\begin{cases} 20x(1-x)^3, & 0<x<1 \\ 0, & \text{elsewhere} \end{cases} $$ find $P\bigl(x<\frac{1}{2}\bigr)$.

Method A

\begin{align} P\Bigl(x<\frac{1}{2}\Bigr) &= \int_0^{1/2} 20x(1-x^3)\,dx \\ &= 20\int_0^{1/2} (1-x)x^3\,dx \\ &= \frac{13}{16} \end{align}

Method B

For continuous distribution $P\bigl(x<\frac{1}{2}\bigr)=P\bigl(x\le\frac{1}{2}\bigr)$ so \begin{align} P\Bigl(x<\frac{1}{2}\Bigr) &= f\Bigl(\frac{1}{2}\Bigr) \\ &= 20\Bigl(\frac{1}{2}\Bigr)\Bigl(1-\frac{1}{2}\Bigr)^3 \\ &= \frac{10}{8} \end{align}

egreg
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user218102
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1 Answers1

2

First method is the right one. In the second method you have confused the density function $f$ with the cumulative distribution function $F$. $P\{X\leq \frac 1 2\}=F(\frac 1 2)$ which is not the same as $f(\frac 1 2)$.

Kavi Rama Murthy
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  • Thanks for your answer.But can you tell me about its difference.It be great help. – user218102 Dec 31 '18 at 08:49
  • @user218102 Consider, for example, the uniform distribution on $(0,1)$. Here $f(x)=1$ for $x \leq 0$ and $0$ for other values of $x$. But $F(x)$ is $0$ for $x \leq 0$, $x$ for $0<x<1$ and $1$ for $x \geq 1$. In general, $f$ and $F$ are related by the equation $F(x)=\int_{-\infty}^{x} f(t), dt$. If $X$ is a random variable with this distribution then $P{X\leq x}=F(x)=\int_{-\infty}^{x} f(t), dt$. – Kavi Rama Murthy Dec 31 '18 at 08:57