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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}

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    $\begingroup$ You should upload images with the provided interface, rather than linking other sites that could disappear. In this particular case it wouldn't be too difficult to add the work using MathJax. $\endgroup$ – egreg Dec 31 '18 at 14:46
  • $\begingroup$ @egreg they linked using google docs. It can't/won't disappear unless the OP makes it disappear. Although I would say that for new users such as in this case mathjax is not easy to use. Not even remotely. $\endgroup$ – The Great Duck Jan 1 at 9:05
  • $\begingroup$ Honestly I don't know how to use mathjax.It seems complicated to me. $\endgroup$ – user218102 Jan 1 at 9:41
  • $\begingroup$ @user218102 I added it for you. $\endgroup$ – egreg Jan 1 at 9:55
  • $\begingroup$ Thanks sir. It means alot. $\endgroup$ – user218102 Jan 1 at 10:00
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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 $f(\frac 1 2)$.

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  • $\begingroup$ Thanks for your answer.But can you tell me about its difference.It be great help. $\endgroup$ – user218102 Dec 31 '18 at 8:49
  • $\begingroup$ @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 $P\{X\leq x\}=F(x)=\int_{-\infty}^{x} f(t)\, dt$. $\endgroup$ – Kavi Rama Murthy Dec 31 '18 at 8:57

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