Calculating Probability of a given X range if X is a continuous random variable with a given PDF. The 2019 Stack Overflow Developer Survey Results Are Inprobability and random variable with joint pdfCalculating PDF of $Z$ from $X,Y$ when $Z=X+Y$, given the PDFs of $X$ and $Y$Suppose that $U$ is uniformly distributed on $[0,1]$. Given its p.d.f. and c.d.f, find $P(U<a|U<b)$ for $0<a<b<1$.Continuous random variable pdf question.Find the cumulative probability function given a probability density functionTrouble with sum of two independent random variables continuousFind all the points that will qualify as being median of XFinding the PDF of a function if X is exponentially distributed with a given parameter $lambda$ = 3Calculate $f_X(x)$ and $f_Y(y)$ given a pair $(X,Y)$ of continuous random variables with a joint PDF of…Calculating probability given joint probability density of a pair given by $f_X,Y(x,y)$
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Calculating Probability of a given X range if X is a continuous random variable with a given PDF.
The 2019 Stack Overflow Developer Survey Results Are Inprobability and random variable with joint pdfCalculating PDF of $Z$ from $X,Y$ when $Z=X+Y$, given the PDFs of $X$ and $Y$Suppose that $U$ is uniformly distributed on $[0,1]$. Given its p.d.f. and c.d.f, find $P(U<a|U<b)$ for $0<a<b<1$.Continuous random variable pdf question.Find the cumulative probability function given a probability density functionTrouble with sum of two independent random variables continuousFind all the points that will qualify as being median of XFinding the PDF of a function if X is exponentially distributed with a given parameter $lambda$ = 3Calculate $f_X(x)$ and $f_Y(y)$ given a pair $(X,Y)$ of continuous random variables with a joint PDF of…Calculating probability given joint probability density of a pair given by $f_X,Y(x,y)$
$begingroup$
This question was given to me as a review for an upcoming exam.
Find $P(2 leq X leq 3)$ if $X$ is a continous random variable with pdf:
$$f_X(x) =
begincases
xe^-x & xgeq 0 \
0 & x lt 0
endcases
$$
My work:
$$P(2 leq X leq 3) = int_2^3xe^-xdx$$
Integration by parts:
$u=x,u'=1,v'=e^-x,v=-e^-x$
$$int_2^3xe^-x dx = -xe^-xBig|^3_2-int_2^3-e^-xdx\
=frac-4e^3 + frac3e^2 approx 0.2069$$
Did I miss anything or do something incorrectly?
Also, I had a quick question regarding a different case, let's say problem was asking for $P(-1 leq X leq 3)$ instead, would we do this instead?:
$$P(-1 leq X leq 3)= int_-1^00 dx + int_0^3xe^-xdx$$
Obviously the left integral will be 0 in this case, I'm just wondering if we split up the integrals with addition if the range we are looking for is in two different domains of the PDF.
probability statistics
edited Mar 23 at 10:36
zhoraster
16k21853
asked Mar 23 at 1:48
JoeJoe
596
$endgroup$
add a comment |
$begingroup$
This question was given to me as a review for an upcoming exam.
Find $P(2 leq X leq 3)$ if $X$ is a continous random variable with pdf:
$$f_X(x) =
begincases
xe^-x & xgeq 0 \
0 & x lt 0
endcases
$$
My work:
$$P(2 leq X leq 3) = int_2^3xe^-xdx$$
Integration by parts:
$u=x,u'=1,v'=e^-x,v=-e^-x$
$$int_2^3xe^-x dx = -xe^-xBig|^3_2-int_2^3-e^-xdx\
=frac-4e^3 + frac3e^2 approx 0.2069$$
Did I miss anything or do something incorrectly?
Also, I had a quick question regarding a different case, let's say problem was asking for $P(-1 leq X leq 3)$ instead, would we do this instead?:
$$P(-1 leq X leq 3)= int_-1^00 dx + int_0^3xe^-xdx$$
Obviously the left integral will be 0 in this case, I'm just wondering if we split up the integrals with addition if the range we are looking for is in two different domains of the PDF.
probability statistics
edited Mar 23 at 10:36
zhoraster
16k21853
asked Mar 23 at 1:48
JoeJoe
596
$endgroup$
1
$begingroup$
To your quick question -- yes
$endgroup$
– Minus One-Twelfth
Mar 23 at 1:55
$begingroup$
Both parts are correct.
$endgroup$
– Kavi Rama Murthy
Mar 23 at 5:26
$begingroup$
Perhaps see Wikipedia. This is the PDF of a RV with dist'n $mathsfGamma(shape=2, rate=1).$ In R: codediff(pgamma(c(2,3), 2, 1))returns 0.2068576. Also,pgamma(3, 2, 1)returns 0.8008517. [pgammais the CDF of a gamma distribution.] Because the support is the positive half line, you could just find $P(0 le X le 3).$
$endgroup$
– BruceET
Mar 23 at 17:07
add a comment |
$begingroup$
This question was given to me as a review for an upcoming exam.
Find $P(2 leq X leq 3)$ if $X$ is a continous random variable with pdf:
$$f_X(x) =
begincases
xe^-x & xgeq 0 \
0 & x lt 0
endcases
$$
My work:
$$P(2 leq X leq 3) = int_2^3xe^-xdx$$
Integration by parts:
$u=x,u'=1,v'=e^-x,v=-e^-x$
$$int_2^3xe^-x dx = -xe^-xBig|^3_2-int_2^3-e^-xdx\
=frac-4e^3 + frac3e^2 approx 0.2069$$
Did I miss anything or do something incorrectly?
Also, I had a quick question regarding a different case, let's say problem was asking for $P(-1 leq X leq 3)$ instead, would we do this instead?:
$$P(-1 leq X leq 3)= int_-1^00 dx + int_0^3xe^-xdx$$
Obviously the left integral will be 0 in this case, I'm just wondering if we split up the integrals with addition if the range we are looking for is in two different domains of the PDF.
probability statistics
edited Mar 23 at 10:36
zhoraster
16k21853
asked Mar 23 at 1:48
JoeJoe
596
$endgroup$
This question was given to me as a review for an upcoming exam.
Find $P(2 leq X leq 3)$ if $X$ is a continous random variable with pdf:
$$f_X(x) =
begincases
xe^-x & xgeq 0 \
0 & x lt 0
endcases
$$
My work:
$$P(2 leq X leq 3) = int_2^3xe^-xdx$$
Integration by parts:
$u=x,u'=1,v'=e^-x,v=-e^-x$
$$int_2^3xe^-x dx = -xe^-xBig|^3_2-int_2^3-e^-xdx\
=frac-4e^3 + frac3e^2 approx 0.2069$$
Did I miss anything or do something incorrectly?
Also, I had a quick question regarding a different case, let's say problem was asking for $P(-1 leq X leq 3)$ instead, would we do this instead?:
$$P(-1 leq X leq 3)= int_-1^00 dx + int_0^3xe^-xdx$$
Obviously the left integral will be 0 in this case, I'm just wondering if we split up the integrals with addition if the range we are looking for is in two different domains of the PDF.
probability statistics
probability statistics
edited Mar 23 at 10:36
zhoraster
16k21853
asked Mar 23 at 1:48
JoeJoe
596
edited Mar 23 at 10:36
zhoraster
16k21853
asked Mar 23 at 1:48
JoeJoe
596
edited Mar 23 at 10:36
zhoraster
16k21853
edited Mar 23 at 10:36
zhoraster
16k21853
edited Mar 23 at 10:36
zhoraster
16k21853
16k21853
asked Mar 23 at 1:48
JoeJoe
596
asked Mar 23 at 1:48
JoeJoe
596
asked Mar 23 at 1:48
JoeJoe
596
596
1
$begingroup$
To your quick question -- yes
$endgroup$
– Minus One-Twelfth
Mar 23 at 1:55
$begingroup$
Both parts are correct.
$endgroup$
– Kavi Rama Murthy
Mar 23 at 5:26
$begingroup$
Perhaps see Wikipedia. This is the PDF of a RV with dist'n $mathsfGamma(shape=2, rate=1).$ In R: codediff(pgamma(c(2,3), 2, 1))returns 0.2068576. Also,pgamma(3, 2, 1)returns 0.8008517. [pgammais the CDF of a gamma distribution.] Because the support is the positive half line, you could just find $P(0 le X le 3).$
$endgroup$
– BruceET
Mar 23 at 17:07
add a comment |
1
$begingroup$
To your quick question -- yes
$endgroup$
– Minus One-Twelfth
Mar 23 at 1:55
$begingroup$
Both parts are correct.
$endgroup$
– Kavi Rama Murthy
Mar 23 at 5:26
$begingroup$
Perhaps see Wikipedia. This is the PDF of a RV with dist'n $mathsfGamma(shape=2, rate=1).$ In R: codediff(pgamma(c(2,3), 2, 1))returns 0.2068576. Also,pgamma(3, 2, 1)returns 0.8008517. [pgammais the CDF of a gamma distribution.] Because the support is the positive half line, you could just find $P(0 le X le 3).$
$endgroup$
– BruceET
Mar 23 at 17:07
1
1
$begingroup$
To your quick question -- yes
$endgroup$
– Minus One-Twelfth
Mar 23 at 1:55
$begingroup$
To your quick question -- yes
$endgroup$
– Minus One-Twelfth
Mar 23 at 1:55
$begingroup$
Both parts are correct.
$endgroup$
– Kavi Rama Murthy
Mar 23 at 5:26
$begingroup$
Both parts are correct.
$endgroup$
– Kavi Rama Murthy
Mar 23 at 5:26
$begingroup$
Perhaps see Wikipedia. This is the PDF of a RV with dist'n $mathsfGamma(shape=2, rate=1).$ In R: code
diff(pgamma(c(2,3), 2, 1)) returns 0.2068576. Also, pgamma(3, 2, 1) returns 0.8008517. [pgamma is the CDF of a gamma distribution.] Because the support is the positive half line, you could just find $P(0 le X le 3).$$endgroup$
– BruceET
Mar 23 at 17:07
$begingroup$
Perhaps see Wikipedia. This is the PDF of a RV with dist'n $mathsfGamma(shape=2, rate=1).$ In R: code
diff(pgamma(c(2,3), 2, 1)) returns 0.2068576. Also, pgamma(3, 2, 1) returns 0.8008517. [pgamma is the CDF of a gamma distribution.] Because the support is the positive half line, you could just find $P(0 le X le 3).$$endgroup$
– BruceET
Mar 23 at 17:07
add a comment |
0
active
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1
$begingroup$
To your quick question -- yes
$endgroup$
– Minus One-Twelfth
Mar 23 at 1:55
$begingroup$
Both parts are correct.
$endgroup$
– Kavi Rama Murthy
Mar 23 at 5:26
$begingroup$
Perhaps see Wikipedia. This is the PDF of a RV with dist'n $mathsfGamma(shape=2, rate=1).$ In R: code
diff(pgamma(c(2,3), 2, 1))returns 0.2068576. Also,pgamma(3, 2, 1)returns 0.8008517. [pgammais the CDF of a gamma distribution.] Because the support is the positive half line, you could just find $P(0 le X le 3).$$endgroup$
– BruceET
Mar 23 at 17:07