Show that $P(0<Xdfrac23$
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Show that $P(0<X<6)>dfrac23$, if $X$ has probability density function $f(x)=dfrace^-xx^22$, if $x>0$ and $0$, otherwise.
Now, $E(X)=3$ and $Var(X)=3$, so I can not use Chebyshev's inequality. Moreover I used the inequality, $e^-xgeq dfracx^22-x+1$, but also this not working here.
What inequality I should use here? Thanks for any help.
probability probability-theory statistics inequality
asked Aug 27 at 12:42
Stat_prob_001
276111
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up vote
1
down vote
favorite
Show that $P(0<X<6)>dfrac23$, if $X$ has probability density function $f(x)=dfrace^-xx^22$, if $x>0$ and $0$, otherwise.
Now, $E(X)=3$ and $Var(X)=3$, so I can not use Chebyshev's inequality. Moreover I used the inequality, $e^-xgeq dfracx^22-x+1$, but also this not working here.
What inequality I should use here? Thanks for any help.
probability probability-theory statistics inequality
asked Aug 27 at 12:42
Stat_prob_001
276111
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Show that $P(0<X<6)>dfrac23$, if $X$ has probability density function $f(x)=dfrace^-xx^22$, if $x>0$ and $0$, otherwise.
Now, $E(X)=3$ and $Var(X)=3$, so I can not use Chebyshev's inequality. Moreover I used the inequality, $e^-xgeq dfracx^22-x+1$, but also this not working here.
What inequality I should use here? Thanks for any help.
probability probability-theory statistics inequality
asked Aug 27 at 12:42
Stat_prob_001
276111
Show that $P(0<X<6)>dfrac23$, if $X$ has probability density function $f(x)=dfrace^-xx^22$, if $x>0$ and $0$, otherwise.
Now, $E(X)=3$ and $Var(X)=3$, so I can not use Chebyshev's inequality. Moreover I used the inequality, $e^-xgeq dfracx^22-x+1$, but also this not working here.
What inequality I should use here? Thanks for any help.
probability probability-theory statistics inequality
asked Aug 27 at 12:42
Stat_prob_001
276111
asked Aug 27 at 12:42
Stat_prob_001
276111
asked Aug 27 at 12:42
Stat_prob_001
276111
asked Aug 27 at 12:42
Stat_prob_001
276111
276111
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
$mathbbP(0<X<6) = 1 - mathbbP(|X-mathbbE(X)|geq 3)$.
Now you can try with Chebyshev's inequality.
answered Aug 27 at 12:49
LucaMac
1,20114
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up vote
4
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We have that
$$P(0<X<6)=int_0^6f(x)mathrmdx$$
$$=frac12int_0^6e^-xx^2mathrmdx$$
$$=frac12left.e^-x(-x^2-2x-2)right|_x=0^6$$
$$=1-25e^-6 approx 0.93803>frac23$$
answered Aug 27 at 12:48
Botond
4,0082632
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
$mathbbP(0<X<6) = 1 - mathbbP(|X-mathbbE(X)|geq 3)$.
Now you can try with Chebyshev's inequality.
answered Aug 27 at 12:49
LucaMac
1,20114
add a comment |Â
up vote
4
down vote
accepted
$mathbbP(0<X<6) = 1 - mathbbP(|X-mathbbE(X)|geq 3)$.
Now you can try with Chebyshev's inequality.
answered Aug 27 at 12:49
LucaMac
1,20114
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
$mathbbP(0<X<6) = 1 - mathbbP(|X-mathbbE(X)|geq 3)$.
Now you can try with Chebyshev's inequality.
answered Aug 27 at 12:49
LucaMac
1,20114
$mathbbP(0<X<6) = 1 - mathbbP(|X-mathbbE(X)|geq 3)$.
Now you can try with Chebyshev's inequality.
answered Aug 27 at 12:49
LucaMac
1,20114
answered Aug 27 at 12:49
LucaMac
1,20114
answered Aug 27 at 12:49
LucaMac
1,20114
answered Aug 27 at 12:49
LucaMac
1,20114
1,20114
add a comment |Â
add a comment |Â
up vote
4
down vote
We have that
$$P(0<X<6)=int_0^6f(x)mathrmdx$$
$$=frac12int_0^6e^-xx^2mathrmdx$$
$$=frac12left.e^-x(-x^2-2x-2)right|_x=0^6$$
$$=1-25e^-6 approx 0.93803>frac23$$
answered Aug 27 at 12:48
Botond
4,0082632
add a comment |Â
up vote
4
down vote
We have that
$$P(0<X<6)=int_0^6f(x)mathrmdx$$
$$=frac12int_0^6e^-xx^2mathrmdx$$
$$=frac12left.e^-x(-x^2-2x-2)right|_x=0^6$$
$$=1-25e^-6 approx 0.93803>frac23$$
answered Aug 27 at 12:48
Botond
4,0082632
add a comment |Â
up vote
4
down vote
up vote
4
down vote
We have that
$$P(0<X<6)=int_0^6f(x)mathrmdx$$
$$=frac12int_0^6e^-xx^2mathrmdx$$
$$=frac12left.e^-x(-x^2-2x-2)right|_x=0^6$$
$$=1-25e^-6 approx 0.93803>frac23$$
answered Aug 27 at 12:48
Botond
4,0082632
We have that
$$P(0<X<6)=int_0^6f(x)mathrmdx$$
$$=frac12int_0^6e^-xx^2mathrmdx$$
$$=frac12left.e^-x(-x^2-2x-2)right|_x=0^6$$
$$=1-25e^-6 approx 0.93803>frac23$$
answered Aug 27 at 12:48
Botond
4,0082632
answered Aug 27 at 12:48
Botond
4,0082632
answered Aug 27 at 12:48
Botond
4,0082632
answered Aug 27 at 12:48
Botond
4,0082632
4,0082632
add a comment |Â
add a comment |Â
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