How. to find the c.d.f from a continuous random variable?
$begingroup$
Given the following density function for continuous random variable x
f(x)=2/3($x^4$+5$x^3$+2$x^2$+3x-2) for x between [2,3]
Determine the c.df and use the c.df to determine p(2.25 < x < 2.5)
So first I realize I can't take the integral between 2 and 3 since the area under the curve would be 1 for a continuous random variable. Therefore I'm going to integrate between my lower limit 2 and my upper limit x. Would that be a good first step? Then from there I would just take my c.d.f and plug in the lower bounds and upper bounds. Not asking anyone to actually work through the problem just want to understand c.d.f better and how to tackle this sort of problem. Advice would be very helpful.
probability probability-distributions random-variables
asked Dec 13 '18 at 14:07
LilLil
96032544
$endgroup$
add a comment |
$begingroup$
Given the following density function for continuous random variable x
f(x)=2/3($x^4$+5$x^3$+2$x^2$+3x-2) for x between [2,3]
Determine the c.df and use the c.df to determine p(2.25 < x < 2.5)
So first I realize I can't take the integral between 2 and 3 since the area under the curve would be 1 for a continuous random variable. Therefore I'm going to integrate between my lower limit 2 and my upper limit x. Would that be a good first step? Then from there I would just take my c.d.f and plug in the lower bounds and upper bounds. Not asking anyone to actually work through the problem just want to understand c.d.f better and how to tackle this sort of problem. Advice would be very helpful.
probability probability-distributions random-variables
asked Dec 13 '18 at 14:07
LilLil
96032544
$endgroup$
1
$begingroup$
seems good, just carry out the plan.
$endgroup$
– Siong Thye Goh
Dec 13 '18 at 14:09
$begingroup$
i don't think you need more derivatives thereafter.
$endgroup$
– Siong Thye Goh
Dec 13 '18 at 15:28
add a comment |
$begingroup$
Given the following density function for continuous random variable x
f(x)=2/3($x^4$+5$x^3$+2$x^2$+3x-2) for x between [2,3]
Determine the c.df and use the c.df to determine p(2.25 < x < 2.5)
So first I realize I can't take the integral between 2 and 3 since the area under the curve would be 1 for a continuous random variable. Therefore I'm going to integrate between my lower limit 2 and my upper limit x. Would that be a good first step? Then from there I would just take my c.d.f and plug in the lower bounds and upper bounds. Not asking anyone to actually work through the problem just want to understand c.d.f better and how to tackle this sort of problem. Advice would be very helpful.
probability probability-distributions random-variables
asked Dec 13 '18 at 14:07
LilLil
96032544
$endgroup$
Given the following density function for continuous random variable x
f(x)=2/3($x^4$+5$x^3$+2$x^2$+3x-2) for x between [2,3]
Determine the c.df and use the c.df to determine p(2.25 < x < 2.5)
So first I realize I can't take the integral between 2 and 3 since the area under the curve would be 1 for a continuous random variable. Therefore I'm going to integrate between my lower limit 2 and my upper limit x. Would that be a good first step? Then from there I would just take my c.d.f and plug in the lower bounds and upper bounds. Not asking anyone to actually work through the problem just want to understand c.d.f better and how to tackle this sort of problem. Advice would be very helpful.
probability probability-distributions random-variables
probability probability-distributions random-variables
asked Dec 13 '18 at 14:07
LilLil
96032544
asked Dec 13 '18 at 14:07
LilLil
96032544
edited Dec 13 '18 at 15:16
Lil
asked Dec 13 '18 at 14:07
LilLil
96032544
asked Dec 13 '18 at 14:07
LilLil
96032544
asked Dec 13 '18 at 14:07
LilLil
96032544
96032544
1
$begingroup$
seems good, just carry out the plan.
$endgroup$
– Siong Thye Goh
Dec 13 '18 at 14:09
$begingroup$
i don't think you need more derivatives thereafter.
$endgroup$
– Siong Thye Goh
Dec 13 '18 at 15:28
add a comment |
1
$begingroup$
seems good, just carry out the plan.
$endgroup$
– Siong Thye Goh
Dec 13 '18 at 14:09
$begingroup$
i don't think you need more derivatives thereafter.
$endgroup$
– Siong Thye Goh
Dec 13 '18 at 15:28
1
1
$begingroup$
seems good, just carry out the plan.
$endgroup$
– Siong Thye Goh
Dec 13 '18 at 14:09
$begingroup$
seems good, just carry out the plan.
$endgroup$
– Siong Thye Goh
Dec 13 '18 at 14:09
$begingroup$
i don't think you need more derivatives thereafter.
$endgroup$
– Siong Thye Goh
Dec 13 '18 at 15:28
$begingroup$
i don't think you need more derivatives thereafter.
$endgroup$
– Siong Thye Goh
Dec 13 '18 at 15:28
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I notice that
$$int_2^3frac23(x^4+5x^3+2x^2+3x-2) ne 1$$
Hence, there could be a typo in the question.
After you fix the typo.
suppsoe the pdf is $f(x)$ from $a$ to b.
If $x>b$, then $F(x)=1$.
if $x<a$, then $F(x) = 0$.
If $a<x<b$, then $F(x) = int_a^x f(x) , dx$.
After you computed $F(x)$,$$P(2.25<x<2.5)=F(2.5)-F(2.25).$$
answered Dec 13 '18 at 15:27
Siong Thye GohSiong Thye Goh
101k1466118
$endgroup$
$begingroup$
yeah I realized I need a constant so it actually equals 1. One more question, If I want the probability that x is between 2.25 and 2.5 I can find the probability x is less than 2.5 and then subtract the probability that x is less than 2.25
$endgroup$
– Lil
Dec 13 '18 at 15:35
$begingroup$
so to get the probability that x is less than 2.5 I would evaluate the integral of the density function provided from 2 to 2.5?
$endgroup$
– Lil
Dec 13 '18 at 15:36
$begingroup$
After you compute $F$, you can use that function, no more integration is needed.
$endgroup$
– Siong Thye Goh
Dec 13 '18 at 15:38
$begingroup$
Alright so the already integrated function I'm going to substitute those bounds, correct?
$endgroup$
– Lil
Dec 13 '18 at 15:39
$begingroup$
meaning the probability from 2 to 2.5 minus the probability of 2 to 2.25
$endgroup$
– Lil
Dec 13 '18 at 15:40
|
show 2 more comments
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1 Answer
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active
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1 Answer
1
active
oldest
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$begingroup$
I notice that
$$int_2^3frac23(x^4+5x^3+2x^2+3x-2) ne 1$$
Hence, there could be a typo in the question.
After you fix the typo.
suppsoe the pdf is $f(x)$ from $a$ to b.
If $x>b$, then $F(x)=1$.
if $x<a$, then $F(x) = 0$.
If $a<x<b$, then $F(x) = int_a^x f(x) , dx$.
After you computed $F(x)$,$$P(2.25<x<2.5)=F(2.5)-F(2.25).$$
answered Dec 13 '18 at 15:27
Siong Thye GohSiong Thye Goh
101k1466118
$endgroup$
$begingroup$
yeah I realized I need a constant so it actually equals 1. One more question, If I want the probability that x is between 2.25 and 2.5 I can find the probability x is less than 2.5 and then subtract the probability that x is less than 2.25
$endgroup$
– Lil
Dec 13 '18 at 15:35
$begingroup$
so to get the probability that x is less than 2.5 I would evaluate the integral of the density function provided from 2 to 2.5?
$endgroup$
– Lil
Dec 13 '18 at 15:36
$begingroup$
After you compute $F$, you can use that function, no more integration is needed.
$endgroup$
– Siong Thye Goh
Dec 13 '18 at 15:38
$begingroup$
Alright so the already integrated function I'm going to substitute those bounds, correct?
$endgroup$
– Lil
Dec 13 '18 at 15:39
$begingroup$
meaning the probability from 2 to 2.5 minus the probability of 2 to 2.25
$endgroup$
– Lil
Dec 13 '18 at 15:40
|
show 2 more comments
$begingroup$
I notice that
$$int_2^3frac23(x^4+5x^3+2x^2+3x-2) ne 1$$
Hence, there could be a typo in the question.
After you fix the typo.
suppsoe the pdf is $f(x)$ from $a$ to b.
If $x>b$, then $F(x)=1$.
if $x<a$, then $F(x) = 0$.
If $a<x<b$, then $F(x) = int_a^x f(x) , dx$.
After you computed $F(x)$,$$P(2.25<x<2.5)=F(2.5)-F(2.25).$$
answered Dec 13 '18 at 15:27
Siong Thye GohSiong Thye Goh
101k1466118
$endgroup$
$begingroup$
yeah I realized I need a constant so it actually equals 1. One more question, If I want the probability that x is between 2.25 and 2.5 I can find the probability x is less than 2.5 and then subtract the probability that x is less than 2.25
$endgroup$
– Lil
Dec 13 '18 at 15:35
$begingroup$
so to get the probability that x is less than 2.5 I would evaluate the integral of the density function provided from 2 to 2.5?
$endgroup$
– Lil
Dec 13 '18 at 15:36
$begingroup$
After you compute $F$, you can use that function, no more integration is needed.
$endgroup$
– Siong Thye Goh
Dec 13 '18 at 15:38
$begingroup$
Alright so the already integrated function I'm going to substitute those bounds, correct?
$endgroup$
– Lil
Dec 13 '18 at 15:39
$begingroup$
meaning the probability from 2 to 2.5 minus the probability of 2 to 2.25
$endgroup$
– Lil
Dec 13 '18 at 15:40
|
show 2 more comments
$begingroup$
I notice that
$$int_2^3frac23(x^4+5x^3+2x^2+3x-2) ne 1$$
Hence, there could be a typo in the question.
After you fix the typo.
suppsoe the pdf is $f(x)$ from $a$ to b.
If $x>b$, then $F(x)=1$.
if $x<a$, then $F(x) = 0$.
If $a<x<b$, then $F(x) = int_a^x f(x) , dx$.
After you computed $F(x)$,$$P(2.25<x<2.5)=F(2.5)-F(2.25).$$
answered Dec 13 '18 at 15:27
Siong Thye GohSiong Thye Goh
101k1466118
$endgroup$
I notice that
$$int_2^3frac23(x^4+5x^3+2x^2+3x-2) ne 1$$
Hence, there could be a typo in the question.
After you fix the typo.
suppsoe the pdf is $f(x)$ from $a$ to b.
If $x>b$, then $F(x)=1$.
if $x<a$, then $F(x) = 0$.
If $a<x<b$, then $F(x) = int_a^x f(x) , dx$.
After you computed $F(x)$,$$P(2.25<x<2.5)=F(2.5)-F(2.25).$$
answered Dec 13 '18 at 15:27
Siong Thye GohSiong Thye Goh
101k1466118
answered Dec 13 '18 at 15:27
Siong Thye GohSiong Thye Goh
101k1466118
answered Dec 13 '18 at 15:27
Siong Thye GohSiong Thye Goh
101k1466118
answered Dec 13 '18 at 15:27
Siong Thye GohSiong Thye Goh
101k1466118
101k1466118
$begingroup$
yeah I realized I need a constant so it actually equals 1. One more question, If I want the probability that x is between 2.25 and 2.5 I can find the probability x is less than 2.5 and then subtract the probability that x is less than 2.25
$endgroup$
– Lil
Dec 13 '18 at 15:35
$begingroup$
so to get the probability that x is less than 2.5 I would evaluate the integral of the density function provided from 2 to 2.5?
$endgroup$
– Lil
Dec 13 '18 at 15:36
$begingroup$
After you compute $F$, you can use that function, no more integration is needed.
$endgroup$
– Siong Thye Goh
Dec 13 '18 at 15:38
$begingroup$
Alright so the already integrated function I'm going to substitute those bounds, correct?
$endgroup$
– Lil
Dec 13 '18 at 15:39
$begingroup$
meaning the probability from 2 to 2.5 minus the probability of 2 to 2.25
$endgroup$
– Lil
Dec 13 '18 at 15:40
|
show 2 more comments
$begingroup$
yeah I realized I need a constant so it actually equals 1. One more question, If I want the probability that x is between 2.25 and 2.5 I can find the probability x is less than 2.5 and then subtract the probability that x is less than 2.25
$endgroup$
– Lil
Dec 13 '18 at 15:35
$begingroup$
so to get the probability that x is less than 2.5 I would evaluate the integral of the density function provided from 2 to 2.5?
$endgroup$
– Lil
Dec 13 '18 at 15:36
$begingroup$
After you compute $F$, you can use that function, no more integration is needed.
$endgroup$
– Siong Thye Goh
Dec 13 '18 at 15:38
$begingroup$
Alright so the already integrated function I'm going to substitute those bounds, correct?
$endgroup$
– Lil
Dec 13 '18 at 15:39
$begingroup$
meaning the probability from 2 to 2.5 minus the probability of 2 to 2.25
$endgroup$
– Lil
Dec 13 '18 at 15:40
$begingroup$
yeah I realized I need a constant so it actually equals 1. One more question, If I want the probability that x is between 2.25 and 2.5 I can find the probability x is less than 2.5 and then subtract the probability that x is less than 2.25
$endgroup$
– Lil
Dec 13 '18 at 15:35
$begingroup$
yeah I realized I need a constant so it actually equals 1. One more question, If I want the probability that x is between 2.25 and 2.5 I can find the probability x is less than 2.5 and then subtract the probability that x is less than 2.25
$endgroup$
– Lil
Dec 13 '18 at 15:35
$begingroup$
so to get the probability that x is less than 2.5 I would evaluate the integral of the density function provided from 2 to 2.5?
$endgroup$
– Lil
Dec 13 '18 at 15:36
$begingroup$
so to get the probability that x is less than 2.5 I would evaluate the integral of the density function provided from 2 to 2.5?
$endgroup$
– Lil
Dec 13 '18 at 15:36
$begingroup$
After you compute $F$, you can use that function, no more integration is needed.
$endgroup$
– Siong Thye Goh
Dec 13 '18 at 15:38
$begingroup$
After you compute $F$, you can use that function, no more integration is needed.
$endgroup$
– Siong Thye Goh
Dec 13 '18 at 15:38
$begingroup$
Alright so the already integrated function I'm going to substitute those bounds, correct?
$endgroup$
– Lil
Dec 13 '18 at 15:39
$begingroup$
Alright so the already integrated function I'm going to substitute those bounds, correct?
$endgroup$
– Lil
Dec 13 '18 at 15:39
$begingroup$
meaning the probability from 2 to 2.5 minus the probability of 2 to 2.25
$endgroup$
– Lil
Dec 13 '18 at 15:40
$begingroup$
meaning the probability from 2 to 2.5 minus the probability of 2 to 2.25
$endgroup$
– Lil
Dec 13 '18 at 15:40
|
show 2 more comments
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$begingroup$
seems good, just carry out the plan.
$endgroup$
– Siong Thye Goh
Dec 13 '18 at 14:09
$begingroup$
i don't think you need more derivatives thereafter.
$endgroup$
– Siong Thye Goh
Dec 13 '18 at 15:28