How calculate probability of function of random variable
$begingroup$
I’m trying to solve an academic exercise in the form of “compute $P[f(X)>0]$ where $f$ is a polynomial(second degree) and $X$ is normally distributed with known mean and stdv.”
I’m familiar with the $Z$ reduction of the normal distribution but can’t figure out the solution neither on my school material.
In particular $f(x)$ is $X^2-3X-4$, $u=3.5$ and $d=2$.
I tried $(1-p[x-1])*(1-p[x<4])$ and applying z=(mean-a)/stdv but this leads to the wrong result using the tables
probability
edited Jan 2 at 5:02
twnly
1,2011214
asked Jan 2 at 2:59
user8047478user8047478
6
$endgroup$
add a comment |
$begingroup$
I’m trying to solve an academic exercise in the form of “compute $P[f(X)>0]$ where $f$ is a polynomial(second degree) and $X$ is normally distributed with known mean and stdv.”
I’m familiar with the $Z$ reduction of the normal distribution but can’t figure out the solution neither on my school material.
In particular $f(x)$ is $X^2-3X-4$, $u=3.5$ and $d=2$.
I tried $(1-p[x-1])*(1-p[x<4])$ and applying z=(mean-a)/stdv but this leads to the wrong result using the tables
probability
edited Jan 2 at 5:02
twnly
1,2011214
asked Jan 2 at 2:59
user8047478user8047478
6
$endgroup$
add a comment |
$begingroup$
I’m trying to solve an academic exercise in the form of “compute $P[f(X)>0]$ where $f$ is a polynomial(second degree) and $X$ is normally distributed with known mean and stdv.”
I’m familiar with the $Z$ reduction of the normal distribution but can’t figure out the solution neither on my school material.
In particular $f(x)$ is $X^2-3X-4$, $u=3.5$ and $d=2$.
I tried $(1-p[x-1])*(1-p[x<4])$ and applying z=(mean-a)/stdv but this leads to the wrong result using the tables
probability
edited Jan 2 at 5:02
twnly
1,2011214
asked Jan 2 at 2:59
user8047478user8047478
6
$endgroup$
I’m trying to solve an academic exercise in the form of “compute $P[f(X)>0]$ where $f$ is a polynomial(second degree) and $X$ is normally distributed with known mean and stdv.”
I’m familiar with the $Z$ reduction of the normal distribution but can’t figure out the solution neither on my school material.
In particular $f(x)$ is $X^2-3X-4$, $u=3.5$ and $d=2$.
I tried $(1-p[x-1])*(1-p[x<4])$ and applying z=(mean-a)/stdv but this leads to the wrong result using the tables
probability
probability
edited Jan 2 at 5:02
twnly
1,2011214
asked Jan 2 at 2:59
user8047478user8047478
6
edited Jan 2 at 5:02
twnly
1,2011214
asked Jan 2 at 2:59
user8047478user8047478
6
edited Jan 2 at 5:02
twnly
1,2011214
edited Jan 2 at 5:02
twnly
1,2011214
edited Jan 2 at 5:02
twnly
1,2011214
1,2011214
asked Jan 2 at 2:59
user8047478user8047478
6
asked Jan 2 at 2:59
user8047478user8047478
6
asked Jan 2 at 2:59
user8047478user8047478
6
6
add a comment |
add a comment |
1 Answer
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$begingroup$
Guide:
$$x^2-3x-4=(x+1)(x-4)$$
It is positive if $x<-1$ or $x>4$.
Hence you should compute $P(X < -1)+P(X>4).$
answered Jan 2 at 3:10
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Guide:
$$x^2-3x-4=(x+1)(x-4)$$
It is positive if $x<-1$ or $x>4$.
Hence you should compute $P(X < -1)+P(X>4).$
answered Jan 2 at 3:10
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
add a comment |
$begingroup$
Guide:
$$x^2-3x-4=(x+1)(x-4)$$
It is positive if $x<-1$ or $x>4$.
Hence you should compute $P(X < -1)+P(X>4).$
answered Jan 2 at 3:10
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
add a comment |
$begingroup$
Guide:
$$x^2-3x-4=(x+1)(x-4)$$
It is positive if $x<-1$ or $x>4$.
Hence you should compute $P(X < -1)+P(X>4).$
answered Jan 2 at 3:10
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
Guide:
$$x^2-3x-4=(x+1)(x-4)$$
It is positive if $x<-1$ or $x>4$.
Hence you should compute $P(X < -1)+P(X>4).$
answered Jan 2 at 3:10
Siong Thye GohSiong Thye Goh
103k1468119
answered Jan 2 at 3:10
Siong Thye GohSiong Thye Goh
103k1468119
answered Jan 2 at 3:10
Siong Thye GohSiong Thye Goh
103k1468119
answered Jan 2 at 3:10
Siong Thye GohSiong Thye Goh
103k1468119
103k1468119
add a comment |
add a comment |
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