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How do I myself compute the median of the following pdf? [closed]
Clash Royale CLAN TAG#URR8PPP
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I know how to calculate the median and the third quartile of a set of data but if it’s continuous I don’t know how to compute it
statistics
edited Jul 14 at 16:19
Alex Francisco
15.7k92047
asked Jul 14 at 15:31
Roy Rizk
887
closed as off-topic by Alex Francisco, StubbornAtom, Jyrki Lahtonen, Shaun, Chris Custer Jul 14 at 18:01
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, StubbornAtom, Jyrki Lahtonen, Shaun, Chris Custer
add a comment |Â
up vote
0
down vote
favorite
I know how to calculate the median and the third quartile of a set of data but if it’s continuous I don’t know how to compute it
statistics
edited Jul 14 at 16:19
Alex Francisco
15.7k92047
asked Jul 14 at 15:31
Roy Rizk
887
closed as off-topic by Alex Francisco, StubbornAtom, Jyrki Lahtonen, Shaun, Chris Custer Jul 14 at 18:01
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, StubbornAtom, Jyrki Lahtonen, Shaun, Chris Custer
math.stackexchange.com/questions/397563/…
– saulspatz
Jul 14 at 15:42
Your question was put on hold, the message above (and possibly comments) should give an explanation why. (In particular, this link might be useful.) You might try to edit your question to address these issues. Note that the next edit puts your post in the review queue, where users can vote whether to reopen it or leave it closed. (Therefore it would be good to avoid minor edits and improve your question as much as possible with the next edit.)
– Martin Sleziak
Jul 15 at 9:28
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up vote
0
down vote
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up vote
0
down vote
favorite
I know how to calculate the median and the third quartile of a set of data but if it’s continuous I don’t know how to compute it
statistics
edited Jul 14 at 16:19
Alex Francisco
15.7k92047
asked Jul 14 at 15:31
Roy Rizk
887
I know how to calculate the median and the third quartile of a set of data but if it’s continuous I don’t know how to compute it
statistics
edited Jul 14 at 16:19
Alex Francisco
15.7k92047
asked Jul 14 at 15:31
Roy Rizk
887
edited Jul 14 at 16:19
Alex Francisco
15.7k92047
edited Jul 14 at 16:19
Alex Francisco
15.7k92047
edited Jul 14 at 16:19
Alex Francisco
15.7k92047
15.7k92047
asked Jul 14 at 15:31
Roy Rizk
887
asked Jul 14 at 15:31
Roy Rizk
887
asked Jul 14 at 15:31
Roy Rizk
887
887
closed as off-topic by Alex Francisco, StubbornAtom, Jyrki Lahtonen, Shaun, Chris Custer Jul 14 at 18:01
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, StubbornAtom, Jyrki Lahtonen, Shaun, Chris Custer
closed as off-topic by Alex Francisco, StubbornAtom, Jyrki Lahtonen, Shaun, Chris Custer Jul 14 at 18:01
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, StubbornAtom, Jyrki Lahtonen, Shaun, Chris Custer
math.stackexchange.com/questions/397563/…
– saulspatz
Jul 14 at 15:42
Your question was put on hold, the message above (and possibly comments) should give an explanation why. (In particular, this link might be useful.) You might try to edit your question to address these issues. Note that the next edit puts your post in the review queue, where users can vote whether to reopen it or leave it closed. (Therefore it would be good to avoid minor edits and improve your question as much as possible with the next edit.)
– Martin Sleziak
Jul 15 at 9:28
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math.stackexchange.com/questions/397563/…
– saulspatz
Jul 14 at 15:42
Your question was put on hold, the message above (and possibly comments) should give an explanation why. (In particular, this link might be useful.) You might try to edit your question to address these issues. Note that the next edit puts your post in the review queue, where users can vote whether to reopen it or leave it closed. (Therefore it would be good to avoid minor edits and improve your question as much as possible with the next edit.)
– Martin Sleziak
Jul 15 at 9:28
math.stackexchange.com/questions/397563/…
– saulspatz
Jul 14 at 15:42
math.stackexchange.com/questions/397563/…
– saulspatz
Jul 14 at 15:42
Your question was put on hold, the message above (and possibly comments) should give an explanation why. (In particular, this link might be useful.) You might try to edit your question to address these issues. Note that the next edit puts your post in the review queue, where users can vote whether to reopen it or leave it closed. (Therefore it would be good to avoid minor edits and improve your question as much as possible with the next edit.)
– Martin Sleziak
Jul 15 at 9:28
Your question was put on hold, the message above (and possibly comments) should give an explanation why. (In particular, this link might be useful.) You might try to edit your question to address these issues. Note that the next edit puts your post in the review queue, where users can vote whether to reopen it or leave it closed. (Therefore it would be good to avoid minor edits and improve your question as much as possible with the next edit.)
– Martin Sleziak
Jul 15 at 9:28
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2 Answers
2
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oldest
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0
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accepted
You have to integrate from $0$ to $z$ and see where $int_0^z f(x) , dx =0.5$. You are looking for the value of $X=z$. Here you have to do it stepwise.
$underlinetextHere are the steps:$
1) First you to check if $int_0^1 frac34x^2 , dx geq 0.5$. If this is the case then the equation is $int_0^z frac34x^2 , dx=0.5$. Calculate the integral and then solve the equation for $z$.
2) If $int_0^1 frac34x^2 , dx<0.5$ then you have to use the following equation to evaluate the corresponing value of $X=z$:
$int_0^1 frac34x^2 , dx+int_e^z frac1x , dx =0.5$
answered Jul 14 at 15:48
callculus
16.4k31427
Thanks you anlot for your answer, But I think that the 1) and 2) here must be reversed
– Roy Rizk
Jul 14 at 16:22
@RoyRizk Why do you think that?
– callculus
Jul 14 at 16:23
Because if the integral of 3/4*x2 was sufficient to give me a value more than 0.5 I’d be then sure that the median is located within this area. PS: how can I write the symbol of integral and power and others.. with my pc
– Roy Rizk
Jul 14 at 16:25
@RoyRizk Good catch. I´ve mixed up the equality signs. I´ve made an edit. You can evaluate the codes if you click on the edit links. In my case it is this link In general see here for more formatting tips.
– callculus
Jul 14 at 16:47
add a comment |Â
up vote
1
down vote
In general, the median of a pdf $f:[a,b]$ is calculated by $int_a^mf(x) dx=0.5$.
The integral of $3/4x^2$ from $0$ to $1$ is $1/4$.
So we need $int_e^mfrac 1 x, dx = 1/4$.
So we need $log_e(m)-log_e(e)=1/4$.
So $log(m)=1.25$ so $m=e^1.25$. Which is.....drumroll....3.49 :)
answered Jul 14 at 15:48
Simon Terrington
1466
Made minor change in formatting of second integral for clarification.
– BruceET
Jul 14 at 16:05
Thank you, that is better.
– Simon Terrington
Jul 14 at 17:24
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
You have to integrate from $0$ to $z$ and see where $int_0^z f(x) , dx =0.5$. You are looking for the value of $X=z$. Here you have to do it stepwise.
$underlinetextHere are the steps:$
1) First you to check if $int_0^1 frac34x^2 , dx geq 0.5$. If this is the case then the equation is $int_0^z frac34x^2 , dx=0.5$. Calculate the integral and then solve the equation for $z$.
2) If $int_0^1 frac34x^2 , dx<0.5$ then you have to use the following equation to evaluate the corresponing value of $X=z$:
$int_0^1 frac34x^2 , dx+int_e^z frac1x , dx =0.5$
answered Jul 14 at 15:48
callculus
16.4k31427
Thanks you anlot for your answer, But I think that the 1) and 2) here must be reversed
– Roy Rizk
Jul 14 at 16:22
@RoyRizk Why do you think that?
– callculus
Jul 14 at 16:23
Because if the integral of 3/4*x2 was sufficient to give me a value more than 0.5 I’d be then sure that the median is located within this area. PS: how can I write the symbol of integral and power and others.. with my pc
– Roy Rizk
Jul 14 at 16:25
@RoyRizk Good catch. I´ve mixed up the equality signs. I´ve made an edit. You can evaluate the codes if you click on the edit links. In my case it is this link In general see here for more formatting tips.
– callculus
Jul 14 at 16:47
add a comment |Â
up vote
0
down vote
accepted
You have to integrate from $0$ to $z$ and see where $int_0^z f(x) , dx =0.5$. You are looking for the value of $X=z$. Here you have to do it stepwise.
$underlinetextHere are the steps:$
1) First you to check if $int_0^1 frac34x^2 , dx geq 0.5$. If this is the case then the equation is $int_0^z frac34x^2 , dx=0.5$. Calculate the integral and then solve the equation for $z$.
2) If $int_0^1 frac34x^2 , dx<0.5$ then you have to use the following equation to evaluate the corresponing value of $X=z$:
$int_0^1 frac34x^2 , dx+int_e^z frac1x , dx =0.5$
answered Jul 14 at 15:48
callculus
16.4k31427
Thanks you anlot for your answer, But I think that the 1) and 2) here must be reversed
– Roy Rizk
Jul 14 at 16:22
@RoyRizk Why do you think that?
– callculus
Jul 14 at 16:23
Because if the integral of 3/4*x2 was sufficient to give me a value more than 0.5 I’d be then sure that the median is located within this area. PS: how can I write the symbol of integral and power and others.. with my pc
– Roy Rizk
Jul 14 at 16:25
@RoyRizk Good catch. I´ve mixed up the equality signs. I´ve made an edit. You can evaluate the codes if you click on the edit links. In my case it is this link In general see here for more formatting tips.
– callculus
Jul 14 at 16:47
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
You have to integrate from $0$ to $z$ and see where $int_0^z f(x) , dx =0.5$. You are looking for the value of $X=z$. Here you have to do it stepwise.
$underlinetextHere are the steps:$
1) First you to check if $int_0^1 frac34x^2 , dx geq 0.5$. If this is the case then the equation is $int_0^z frac34x^2 , dx=0.5$. Calculate the integral and then solve the equation for $z$.
2) If $int_0^1 frac34x^2 , dx<0.5$ then you have to use the following equation to evaluate the corresponing value of $X=z$:
$int_0^1 frac34x^2 , dx+int_e^z frac1x , dx =0.5$
answered Jul 14 at 15:48
callculus
16.4k31427
You have to integrate from $0$ to $z$ and see where $int_0^z f(x) , dx =0.5$. You are looking for the value of $X=z$. Here you have to do it stepwise.
$underlinetextHere are the steps:$
1) First you to check if $int_0^1 frac34x^2 , dx geq 0.5$. If this is the case then the equation is $int_0^z frac34x^2 , dx=0.5$. Calculate the integral and then solve the equation for $z$.
2) If $int_0^1 frac34x^2 , dx<0.5$ then you have to use the following equation to evaluate the corresponing value of $X=z$:
$int_0^1 frac34x^2 , dx+int_e^z frac1x , dx =0.5$
answered Jul 14 at 15:48
callculus
16.4k31427
edited Jul 14 at 16:47
answered Jul 14 at 15:48
callculus
16.4k31427
answered Jul 14 at 15:48
callculus
16.4k31427
answered Jul 14 at 15:48
callculus
16.4k31427
16.4k31427
Thanks you anlot for your answer, But I think that the 1) and 2) here must be reversed
– Roy Rizk
Jul 14 at 16:22
@RoyRizk Why do you think that?
– callculus
Jul 14 at 16:23
Because if the integral of 3/4*x2 was sufficient to give me a value more than 0.5 I’d be then sure that the median is located within this area. PS: how can I write the symbol of integral and power and others.. with my pc
– Roy Rizk
Jul 14 at 16:25
@RoyRizk Good catch. I´ve mixed up the equality signs. I´ve made an edit. You can evaluate the codes if you click on the edit links. In my case it is this link In general see here for more formatting tips.
– callculus
Jul 14 at 16:47
add a comment |Â
Thanks you anlot for your answer, But I think that the 1) and 2) here must be reversed
– Roy Rizk
Jul 14 at 16:22
@RoyRizk Why do you think that?
– callculus
Jul 14 at 16:23
Because if the integral of 3/4*x2 was sufficient to give me a value more than 0.5 I’d be then sure that the median is located within this area. PS: how can I write the symbol of integral and power and others.. with my pc
– Roy Rizk
Jul 14 at 16:25
@RoyRizk Good catch. I´ve mixed up the equality signs. I´ve made an edit. You can evaluate the codes if you click on the edit links. In my case it is this link In general see here for more formatting tips.
– callculus
Jul 14 at 16:47
Thanks you anlot for your answer, But I think that the 1) and 2) here must be reversed
– Roy Rizk
Jul 14 at 16:22
Thanks you anlot for your answer, But I think that the 1) and 2) here must be reversed
– Roy Rizk
Jul 14 at 16:22
@RoyRizk Why do you think that?
– callculus
Jul 14 at 16:23
@RoyRizk Why do you think that?
– callculus
Jul 14 at 16:23
Because if the integral of 3/4*x2 was sufficient to give me a value more than 0.5 I’d be then sure that the median is located within this area. PS: how can I write the symbol of integral and power and others.. with my pc
– Roy Rizk
Jul 14 at 16:25
Because if the integral of 3/4*x2 was sufficient to give me a value more than 0.5 I’d be then sure that the median is located within this area. PS: how can I write the symbol of integral and power and others.. with my pc
– Roy Rizk
Jul 14 at 16:25
@RoyRizk Good catch. I´ve mixed up the equality signs. I´ve made an edit. You can evaluate the codes if you click on the edit links. In my case it is this link In general see here for more formatting tips.
– callculus
Jul 14 at 16:47
@RoyRizk Good catch. I´ve mixed up the equality signs. I´ve made an edit. You can evaluate the codes if you click on the edit links. In my case it is this link In general see here for more formatting tips.
– callculus
Jul 14 at 16:47
add a comment |Â
up vote
1
down vote
In general, the median of a pdf $f:[a,b]$ is calculated by $int_a^mf(x) dx=0.5$.
The integral of $3/4x^2$ from $0$ to $1$ is $1/4$.
So we need $int_e^mfrac 1 x, dx = 1/4$.
So we need $log_e(m)-log_e(e)=1/4$.
So $log(m)=1.25$ so $m=e^1.25$. Which is.....drumroll....3.49 :)
answered Jul 14 at 15:48
Simon Terrington
1466
Made minor change in formatting of second integral for clarification.
– BruceET
Jul 14 at 16:05
Thank you, that is better.
– Simon Terrington
Jul 14 at 17:24
add a comment |Â
up vote
1
down vote
In general, the median of a pdf $f:[a,b]$ is calculated by $int_a^mf(x) dx=0.5$.
The integral of $3/4x^2$ from $0$ to $1$ is $1/4$.
So we need $int_e^mfrac 1 x, dx = 1/4$.
So we need $log_e(m)-log_e(e)=1/4$.
So $log(m)=1.25$ so $m=e^1.25$. Which is.....drumroll....3.49 :)
answered Jul 14 at 15:48
Simon Terrington
1466
Made minor change in formatting of second integral for clarification.
– BruceET
Jul 14 at 16:05
Thank you, that is better.
– Simon Terrington
Jul 14 at 17:24
add a comment |Â
up vote
1
down vote
up vote
1
down vote
In general, the median of a pdf $f:[a,b]$ is calculated by $int_a^mf(x) dx=0.5$.
The integral of $3/4x^2$ from $0$ to $1$ is $1/4$.
So we need $int_e^mfrac 1 x, dx = 1/4$.
So we need $log_e(m)-log_e(e)=1/4$.
So $log(m)=1.25$ so $m=e^1.25$. Which is.....drumroll....3.49 :)
answered Jul 14 at 15:48
Simon Terrington
1466
In general, the median of a pdf $f:[a,b]$ is calculated by $int_a^mf(x) dx=0.5$.
The integral of $3/4x^2$ from $0$ to $1$ is $1/4$.
So we need $int_e^mfrac 1 x, dx = 1/4$.
So we need $log_e(m)-log_e(e)=1/4$.
So $log(m)=1.25$ so $m=e^1.25$. Which is.....drumroll....3.49 :)
answered Jul 14 at 15:48
Simon Terrington
1466
edited Jul 14 at 17:52
answered Jul 14 at 15:48
Simon Terrington
1466
answered Jul 14 at 15:48
Simon Terrington
1466
answered Jul 14 at 15:48
Simon Terrington
1466
1466
Made minor change in formatting of second integral for clarification.
– BruceET
Jul 14 at 16:05
Thank you, that is better.
– Simon Terrington
Jul 14 at 17:24
add a comment |Â
Made minor change in formatting of second integral for clarification.
– BruceET
Jul 14 at 16:05
Thank you, that is better.
– Simon Terrington
Jul 14 at 17:24
Made minor change in formatting of second integral for clarification.
– BruceET
Jul 14 at 16:05
Made minor change in formatting of second integral for clarification.
– BruceET
Jul 14 at 16:05
Thank you, that is better.
– Simon Terrington
Jul 14 at 17:24
Thank you, that is better.
– Simon Terrington
Jul 14 at 17:24
add a comment |Â
math.stackexchange.com/questions/397563/…
– saulspatz
Jul 14 at 15:42
Your question was put on hold, the message above (and possibly comments) should give an explanation why. (In particular, this link might be useful.) You might try to edit your question to address these issues. Note that the next edit puts your post in the review queue, where users can vote whether to reopen it or leave it closed. (Therefore it would be good to avoid minor edits and improve your question as much as possible with the next edit.)
– Martin Sleziak
Jul 15 at 9:28