Is there any quick way to solve this question?












0












$begingroup$


Let $X$ be a random variable with the following cumulative distribution function
$$F(x) =
begin{cases}
{0}, & x<0 \[2ex]
x^2, & 0le x<dfrac{1}{2}\[2ex]
dfrac{3}{4}, &dfrac{1}{2}le x< 1 \[2ex]
1 & xge1
end{cases}$$



Find $P(dfrac{1}{4}<X<1)$



My input



Here I used $P(Xle x)= F(x) $ and $P(X< x)= F(x)^{*}$



At $dfrac{1}{2}$ our pdf is discontinuous and in the interval $0le x<dfrac{1}{2}$ $f(x)$ is continuous and in $dfrac{1}{2}le x< 1 $ it is discrete. So



$P(dfrac{1}{4}<X<dfrac{1}{2})+P( dfrac{1}{2}le x< 1 )=Fbigg(dfrac{1}{2}bigg)^{*}-Fbigg(dfrac{1}{4}bigg)+F(1)^{*}-Fbigg(dfrac{1}{2}bigg) + Pbigg(X=dfrac{1}{2}bigg)$



$= Fbigg(dfrac{1}{2}bigg)^{*}-Fbigg(dfrac{1}{4}bigg)+F(1)^{*}-Fbigg(dfrac{1}{2}bigg) + Fbigg(dfrac{1}{2}bigg)-Fbigg(dfrac{1}{2}bigg)^{*}$



$=F(1)^{*}-Fbigg(dfrac{1}{4}bigg)=dfrac{11}{16}$



So this question came in one mark. Is there any quick way to solve this ?






probability probability-distributions







share|cite|improve this question











$endgroup$












  • $begingroup$
    $P(1/4<X<1)=F(1)^{ast}-F(1/4)$ by your definitions. The rest is redundant.
    $endgroup$
    – metamorphy
    Dec 6 '18 at 12:41












  • $begingroup$
    @metamorphy So it's always true if we have mixed distribution then valid too?
    $endgroup$
    – Daman deep
    Dec 6 '18 at 12:45
















0












$begingroup$


Let $X$ be a random variable with the following cumulative distribution function
$$F(x) =
begin{cases}
{0}, & x<0 \[2ex]
x^2, & 0le x<dfrac{1}{2}\[2ex]
dfrac{3}{4}, &dfrac{1}{2}le x< 1 \[2ex]
1 & xge1
end{cases}$$



Find $P(dfrac{1}{4}<X<1)$



My input



Here I used $P(Xle x)= F(x) $ and $P(X< x)= F(x)^{*}$



At $dfrac{1}{2}$ our pdf is discontinuous and in the interval $0le x<dfrac{1}{2}$ $f(x)$ is continuous and in $dfrac{1}{2}le x< 1 $ it is discrete. So



$P(dfrac{1}{4}<X<dfrac{1}{2})+P( dfrac{1}{2}le x< 1 )=Fbigg(dfrac{1}{2}bigg)^{*}-Fbigg(dfrac{1}{4}bigg)+F(1)^{*}-Fbigg(dfrac{1}{2}bigg) + Pbigg(X=dfrac{1}{2}bigg)$



$= Fbigg(dfrac{1}{2}bigg)^{*}-Fbigg(dfrac{1}{4}bigg)+F(1)^{*}-Fbigg(dfrac{1}{2}bigg) + Fbigg(dfrac{1}{2}bigg)-Fbigg(dfrac{1}{2}bigg)^{*}$



$=F(1)^{*}-Fbigg(dfrac{1}{4}bigg)=dfrac{11}{16}$



So this question came in one mark. Is there any quick way to solve this ?






probability probability-distributions







share|cite|improve this question











$endgroup$












  • $begingroup$
    $P(1/4<X<1)=F(1)^{ast}-F(1/4)$ by your definitions. The rest is redundant.
    $endgroup$
    – metamorphy
    Dec 6 '18 at 12:41












  • $begingroup$
    @metamorphy So it's always true if we have mixed distribution then valid too?
    $endgroup$
    – Daman deep
    Dec 6 '18 at 12:45














0












0








0





$begingroup$


Let $X$ be a random variable with the following cumulative distribution function
$$F(x) =
begin{cases}
{0}, & x<0 \[2ex]
x^2, & 0le x<dfrac{1}{2}\[2ex]
dfrac{3}{4}, &dfrac{1}{2}le x< 1 \[2ex]
1 & xge1
end{cases}$$



Find $P(dfrac{1}{4}<X<1)$



My input



Here I used $P(Xle x)= F(x) $ and $P(X< x)= F(x)^{*}$



At $dfrac{1}{2}$ our pdf is discontinuous and in the interval $0le x<dfrac{1}{2}$ $f(x)$ is continuous and in $dfrac{1}{2}le x< 1 $ it is discrete. So



$P(dfrac{1}{4}<X<dfrac{1}{2})+P( dfrac{1}{2}le x< 1 )=Fbigg(dfrac{1}{2}bigg)^{*}-Fbigg(dfrac{1}{4}bigg)+F(1)^{*}-Fbigg(dfrac{1}{2}bigg) + Pbigg(X=dfrac{1}{2}bigg)$



$= Fbigg(dfrac{1}{2}bigg)^{*}-Fbigg(dfrac{1}{4}bigg)+F(1)^{*}-Fbigg(dfrac{1}{2}bigg) + Fbigg(dfrac{1}{2}bigg)-Fbigg(dfrac{1}{2}bigg)^{*}$



$=F(1)^{*}-Fbigg(dfrac{1}{4}bigg)=dfrac{11}{16}$



So this question came in one mark. Is there any quick way to solve this ?






probability probability-distributions







share|cite|improve this question











$endgroup$




Let $X$ be a random variable with the following cumulative distribution function
$$F(x) =
begin{cases}
{0}, & x<0 \[2ex]
x^2, & 0le x<dfrac{1}{2}\[2ex]
dfrac{3}{4}, &dfrac{1}{2}le x< 1 \[2ex]
1 & xge1
end{cases}$$



Find $P(dfrac{1}{4}<X<1)$



My input



Here I used $P(Xle x)= F(x) $ and $P(X< x)= F(x)^{*}$



At $dfrac{1}{2}$ our pdf is discontinuous and in the interval $0le x<dfrac{1}{2}$ $f(x)$ is continuous and in $dfrac{1}{2}le x< 1 $ it is discrete. So



$P(dfrac{1}{4}<X<dfrac{1}{2})+P( dfrac{1}{2}le x< 1 )=Fbigg(dfrac{1}{2}bigg)^{*}-Fbigg(dfrac{1}{4}bigg)+F(1)^{*}-Fbigg(dfrac{1}{2}bigg) + Pbigg(X=dfrac{1}{2}bigg)$



$= Fbigg(dfrac{1}{2}bigg)^{*}-Fbigg(dfrac{1}{4}bigg)+F(1)^{*}-Fbigg(dfrac{1}{2}bigg) + Fbigg(dfrac{1}{2}bigg)-Fbigg(dfrac{1}{2}bigg)^{*}$



$=F(1)^{*}-Fbigg(dfrac{1}{4}bigg)=dfrac{11}{16}$



So this question came in one mark. Is there any quick way to solve this ?






probability probability-distributions




probability probability-distributions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 6 '18 at 12:48







Daman deep

















asked Dec 6 '18 at 12:30









Daman deepDaman deep

749318




749318












  • $begingroup$
    $P(1/4<X<1)=F(1)^{ast}-F(1/4)$ by your definitions. The rest is redundant.
    $endgroup$
    – metamorphy
    Dec 6 '18 at 12:41












  • $begingroup$
    @metamorphy So it's always true if we have mixed distribution then valid too?
    $endgroup$
    – Daman deep
    Dec 6 '18 at 12:45


















  • $begingroup$
    $P(1/4<X<1)=F(1)^{ast}-F(1/4)$ by your definitions. The rest is redundant.
    $endgroup$
    – metamorphy
    Dec 6 '18 at 12:41












  • $begingroup$
    @metamorphy So it's always true if we have mixed distribution then valid too?
    $endgroup$
    – Daman deep
    Dec 6 '18 at 12:45
















$begingroup$
$P(1/4<X<1)=F(1)^{ast}-F(1/4)$ by your definitions. The rest is redundant.
$endgroup$
– metamorphy
Dec 6 '18 at 12:41






$begingroup$
$P(1/4<X<1)=F(1)^{ast}-F(1/4)$ by your definitions. The rest is redundant.
$endgroup$
– metamorphy
Dec 6 '18 at 12:41














$begingroup$
@metamorphy So it's always true if we have mixed distribution then valid too?
$endgroup$
– Daman deep
Dec 6 '18 at 12:45




$begingroup$
@metamorphy So it's always true if we have mixed distribution then valid too?
$endgroup$
– Daman deep
Dec 6 '18 at 12:45










1 Answer
1






active

oldest

votes


















1












$begingroup$

$$Pleft(frac14<X<1right)=P(X<1)-Pleft(Xleqfrac14right)=lim_{xto1-}F(x)-Fleft(frac14right)=frac34-frac1{16}=frac{11}{16}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Always valid no conditions about continuity, discontinuity, discrete or continuous?
    $endgroup$
    – Daman deep
    Dec 6 '18 at 12:47










  • $begingroup$
    Indeed always valid.
    $endgroup$
    – drhab
    Dec 6 '18 at 12:47










  • $begingroup$
    What will be the probability at $P(X=dfrac{1}{2})$?
    $endgroup$
    – Daman deep
    Dec 6 '18 at 12:49












  • $begingroup$
    $P(X=frac12)=P(Xleqfrac12)-P(X<frac12)=F(frac12)-lim_{xtofrac12-}F(x)=frac34-frac14=frac12$
    $endgroup$
    – drhab
    Dec 6 '18 at 12:51












  • $begingroup$
    In your answer you wrote $Pbigg( X le dfrac{1}{4}bigg)$ but $Pbigg( X < dfrac{1}{4}bigg) $ is valid too write?
    $endgroup$
    – Daman deep
    Dec 6 '18 at 12:53













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1 Answer
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oldest

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1 Answer
1






active

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active

oldest

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active

oldest

votes









1












$begingroup$

$$Pleft(frac14<X<1right)=P(X<1)-Pleft(Xleqfrac14right)=lim_{xto1-}F(x)-Fleft(frac14right)=frac34-frac1{16}=frac{11}{16}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Always valid no conditions about continuity, discontinuity, discrete or continuous?
    $endgroup$
    – Daman deep
    Dec 6 '18 at 12:47










  • $begingroup$
    Indeed always valid.
    $endgroup$
    – drhab
    Dec 6 '18 at 12:47










  • $begingroup$
    What will be the probability at $P(X=dfrac{1}{2})$?
    $endgroup$
    – Daman deep
    Dec 6 '18 at 12:49












  • $begingroup$
    $P(X=frac12)=P(Xleqfrac12)-P(X<frac12)=F(frac12)-lim_{xtofrac12-}F(x)=frac34-frac14=frac12$
    $endgroup$
    – drhab
    Dec 6 '18 at 12:51












  • $begingroup$
    In your answer you wrote $Pbigg( X le dfrac{1}{4}bigg)$ but $Pbigg( X < dfrac{1}{4}bigg) $ is valid too write?
    $endgroup$
    – Daman deep
    Dec 6 '18 at 12:53


















1












$begingroup$

$$Pleft(frac14<X<1right)=P(X<1)-Pleft(Xleqfrac14right)=lim_{xto1-}F(x)-Fleft(frac14right)=frac34-frac1{16}=frac{11}{16}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Always valid no conditions about continuity, discontinuity, discrete or continuous?
    $endgroup$
    – Daman deep
    Dec 6 '18 at 12:47










  • $begingroup$
    Indeed always valid.
    $endgroup$
    – drhab
    Dec 6 '18 at 12:47










  • $begingroup$
    What will be the probability at $P(X=dfrac{1}{2})$?
    $endgroup$
    – Daman deep
    Dec 6 '18 at 12:49












  • $begingroup$
    $P(X=frac12)=P(Xleqfrac12)-P(X<frac12)=F(frac12)-lim_{xtofrac12-}F(x)=frac34-frac14=frac12$
    $endgroup$
    – drhab
    Dec 6 '18 at 12:51












  • $begingroup$
    In your answer you wrote $Pbigg( X le dfrac{1}{4}bigg)$ but $Pbigg( X < dfrac{1}{4}bigg) $ is valid too write?
    $endgroup$
    – Daman deep
    Dec 6 '18 at 12:53
















1












1








1





$begingroup$

$$Pleft(frac14<X<1right)=P(X<1)-Pleft(Xleqfrac14right)=lim_{xto1-}F(x)-Fleft(frac14right)=frac34-frac1{16}=frac{11}{16}$$






share|cite|improve this answer











$endgroup$



$$Pleft(frac14<X<1right)=P(X<1)-Pleft(Xleqfrac14right)=lim_{xto1-}F(x)-Fleft(frac14right)=frac34-frac1{16}=frac{11}{16}$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 6 '18 at 12:48

























answered Dec 6 '18 at 12:45









drhabdrhab

100k544130




100k544130












  • $begingroup$
    Always valid no conditions about continuity, discontinuity, discrete or continuous?
    $endgroup$
    – Daman deep
    Dec 6 '18 at 12:47










  • $begingroup$
    Indeed always valid.
    $endgroup$
    – drhab
    Dec 6 '18 at 12:47










  • $begingroup$
    What will be the probability at $P(X=dfrac{1}{2})$?
    $endgroup$
    – Daman deep
    Dec 6 '18 at 12:49












  • $begingroup$
    $P(X=frac12)=P(Xleqfrac12)-P(X<frac12)=F(frac12)-lim_{xtofrac12-}F(x)=frac34-frac14=frac12$
    $endgroup$
    – drhab
    Dec 6 '18 at 12:51












  • $begingroup$
    In your answer you wrote $Pbigg( X le dfrac{1}{4}bigg)$ but $Pbigg( X < dfrac{1}{4}bigg) $ is valid too write?
    $endgroup$
    – Daman deep
    Dec 6 '18 at 12:53




















  • $begingroup$
    Always valid no conditions about continuity, discontinuity, discrete or continuous?
    $endgroup$
    – Daman deep
    Dec 6 '18 at 12:47










  • $begingroup$
    Indeed always valid.
    $endgroup$
    – drhab
    Dec 6 '18 at 12:47










  • $begingroup$
    What will be the probability at $P(X=dfrac{1}{2})$?
    $endgroup$
    – Daman deep
    Dec 6 '18 at 12:49












  • $begingroup$
    $P(X=frac12)=P(Xleqfrac12)-P(X<frac12)=F(frac12)-lim_{xtofrac12-}F(x)=frac34-frac14=frac12$
    $endgroup$
    – drhab
    Dec 6 '18 at 12:51












  • $begingroup$
    In your answer you wrote $Pbigg( X le dfrac{1}{4}bigg)$ but $Pbigg( X < dfrac{1}{4}bigg) $ is valid too write?
    $endgroup$
    – Daman deep
    Dec 6 '18 at 12:53


















$begingroup$
Always valid no conditions about continuity, discontinuity, discrete or continuous?
$endgroup$
– Daman deep
Dec 6 '18 at 12:47




$begingroup$
Always valid no conditions about continuity, discontinuity, discrete or continuous?
$endgroup$
– Daman deep
Dec 6 '18 at 12:47












$begingroup$
Indeed always valid.
$endgroup$
– drhab
Dec 6 '18 at 12:47




$begingroup$
Indeed always valid.
$endgroup$
– drhab
Dec 6 '18 at 12:47












$begingroup$
What will be the probability at $P(X=dfrac{1}{2})$?
$endgroup$
– Daman deep
Dec 6 '18 at 12:49






$begingroup$
What will be the probability at $P(X=dfrac{1}{2})$?
$endgroup$
– Daman deep
Dec 6 '18 at 12:49














$begingroup$
$P(X=frac12)=P(Xleqfrac12)-P(X<frac12)=F(frac12)-lim_{xtofrac12-}F(x)=frac34-frac14=frac12$
$endgroup$
– drhab
Dec 6 '18 at 12:51






$begingroup$
$P(X=frac12)=P(Xleqfrac12)-P(X<frac12)=F(frac12)-lim_{xtofrac12-}F(x)=frac34-frac14=frac12$
$endgroup$
– drhab
Dec 6 '18 at 12:51














$begingroup$
In your answer you wrote $Pbigg( X le dfrac{1}{4}bigg)$ but $Pbigg( X < dfrac{1}{4}bigg) $ is valid too write?
$endgroup$
– Daman deep
Dec 6 '18 at 12:53






$begingroup$
In your answer you wrote $Pbigg( X le dfrac{1}{4}bigg)$ but $Pbigg( X < dfrac{1}{4}bigg) $ is valid too write?
$endgroup$
– Daman deep
Dec 6 '18 at 12:53




















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