To find the probability density function












3












$begingroup$


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 $Pbigl(x<frac{1}{2}bigr)$.




Method A



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



Method B



For continuous distribution $Pbigl(x<frac{1}{2}bigr)=Pbigl(xlefrac{1}{2}bigr)$ so
begin{align}
PBigl(x<frac{1}{2}Bigr)
&= fBigl(frac{1}{2}Bigr) \
&= 20Bigl(frac{1}{2}Bigr)Bigl(1-frac{1}{2}Bigr)^3 \
&= frac{10}{8}
end{align}






probability-distributions







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    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
















3












$begingroup$


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 $Pbigl(x<frac{1}{2}bigr)$.




Method A



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



Method B



For continuous distribution $Pbigl(x<frac{1}{2}bigr)=Pbigl(xlefrac{1}{2}bigr)$ so
begin{align}
PBigl(x<frac{1}{2}Bigr)
&= fBigl(frac{1}{2}Bigr) \
&= 20Bigl(frac{1}{2}Bigr)Bigl(1-frac{1}{2}Bigr)^3 \
&= frac{10}{8}
end{align}






probability-distributions







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    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














3












3








3


0



$begingroup$


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 $Pbigl(x<frac{1}{2}bigr)$.




Method A



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



Method B



For continuous distribution $Pbigl(x<frac{1}{2}bigr)=Pbigl(xlefrac{1}{2}bigr)$ so
begin{align}
PBigl(x<frac{1}{2}Bigr)
&= fBigl(frac{1}{2}Bigr) \
&= 20Bigl(frac{1}{2}Bigr)Bigl(1-frac{1}{2}Bigr)^3 \
&= frac{10}{8}
end{align}






probability-distributions







share|cite|improve this question











$endgroup$




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 $Pbigl(x<frac{1}{2}bigr)$.




Method A



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



Method B



For continuous distribution $Pbigl(x<frac{1}{2}bigr)=Pbigl(xlefrac{1}{2}bigr)$ so
begin{align}
PBigl(x<frac{1}{2}Bigr)
&= fBigl(frac{1}{2}Bigr) \
&= 20Bigl(frac{1}{2}Bigr)Bigl(1-frac{1}{2}Bigr)^3 \
&= frac{10}{8}
end{align}






probability-distributions




probability-distributions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 31 '18 at 14:54









egreg

186k1486209




186k1486209










asked Dec 31 '18 at 8:44









user218102user218102

234




234








  • 2




    $begingroup$
    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














  • 2




    $begingroup$
    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








2




2




$begingroup$
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




$begingroup$
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






$begingroup$
@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




$begingroup$
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




$begingroup$
@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




$begingroup$
Thanks sir. It means alot.
$endgroup$
– user218102
Jan 1 at 10:00










1 Answer
1






active

oldest

votes


















2












$begingroup$

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






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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 $P{Xleq x}=F(x)=int_{-infty}^{x} f(t), dt$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 31 '18 at 8:57














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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

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






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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 $P{Xleq x}=F(x)=int_{-infty}^{x} f(t), dt$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 31 '18 at 8:57


















2












$begingroup$

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






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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 $P{Xleq x}=F(x)=int_{-infty}^{x} f(t), dt$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 31 '18 at 8:57
















2












2








2





$begingroup$

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






share|cite|improve this answer









$endgroup$



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







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 31 '18 at 8:47









Kavi Rama MurthyKavi Rama Murthy

75.9k53270




75.9k53270












  • $begingroup$
    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 $P{Xleq x}=F(x)=int_{-infty}^{x} f(t), dt$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 31 '18 at 8:57




















  • $begingroup$
    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 $P{Xleq x}=F(x)=int_{-infty}^{x} f(t), dt$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 31 '18 at 8:57


















$begingroup$
Thanks for your answer.But can you tell me about its difference.It be great help.
$endgroup$
– user218102
Dec 31 '18 at 8:49




$begingroup$
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 $P{Xleq x}=F(x)=int_{-infty}^{x} f(t), dt$.
$endgroup$
– Kavi Rama Murthy
Dec 31 '18 at 8:57






$begingroup$
@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 $P{Xleq x}=F(x)=int_{-infty}^{x} f(t), dt$.
$endgroup$
– Kavi Rama Murthy
Dec 31 '18 at 8:57




















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