CDF and PDF Calculations

[_N3WTON_]

Homework Statement

The current in a certain circuit as measured by an ammeter is a continuous random variable X with the following probability density function:
$f(x) = 0.057x + 0.272$ if 3 <= x <= 5. It is 0 otherwise.
a) Verify that the total area under the density curve is 1.
b) Obtain the CDF
c) Calculate P(x<= 4)
d) Is P(3<=x<=5) strictly less than, strictly greater than, or exactly the same as P(x<=4)
e) Calculate P(5.4<X)

Homework Equations

The Attempt at a Solution

For part a:
$\int_{3}^{5} (0.057x+0.272)dx = 1$
So the pdf is confirmed
For part b:
$F(x) = \int_{3}^{x}(0.057y+0.272)dy$
$= \frac{0.057}{2}x^{2}+0.272x - (\frac{0.057}{2}3^{2}+0.272(3))$
$F(x) = \frac{0.057}{2}x^{2} + 0.272x - 1.0725$
This is from 3<=X<=5
For part c:
I said that P(X<=4) is exactly the same as computing P(3<=X<=5). So From the CDF above I did:
$F(4)-F(3) = 0.4715 - 0 = 0.4715$
For part d:
I said that it would be exactly the same
For part e:
This is where I am getting confused, for P(5.4<X), I thought the formula was:
$1 - F(5.4)$. However, when I do this calculation, I get back a negative value, which doesn't make any sense. So I was wondering if my computed CDF is wrong? Another thought I had was since 5.4 is out of the range of the pdf, would the probability just be equal to 1? Any help is appreciated. Thanks

 

[RUber]

For part d, I am not sure I understand the wording...to me P(3<x<5) is 1, since it covers the whole range, so P(anything else) would be strictly less than that.
for part e, P(5.4<x) is the same as 1-P(x<5.4). Be careful not to plug an x value into your CDF which is not in the valid domain. I think all allowable x values are in [3,5].

 

[_N3WTON_]

For part d, I am not sure I understand the wording...to me P(3<x<5) is 1, since it covers the whole range, so P(anything else) would be strictly less than that.
for part e, P(5.4<x) is the same as 1-P(x<5.4). Be careful not to plug an x value into your CDF which is not in the valid domain. I think all allowable x values are in [3,5].

Ok I cleaned things up a bit, I realized it was a little hard to follow. So what you are saying is my calculation for P(X<=4) is incorrect? This is NOT equal to P(3<=X<=5)? Also, for the last part I would compute 1-F(5)?

 

[RUber]

Ok I cleaned things up a bit, I realized it was a little hard to follow. So what you are saying is my calculation for P(X<=4) is incorrect? This is NOT equal to P(3<=X<=5)? Also, for the last part I would compute 1-F(5)?

It looks like your calculation for P(x<=4) is correct. But it is NOT equal to P(3<=x<=5), since the latter is 1.
And, yes, for the last part 1-F(5) would work, or just look at the (piecewise) definition of the pdf and say what $\int_{5.4}^{\infty} f(x) dx$ would have to be. Either way you get the same answer.

 

[Ray Vickson]

Homework Statement

The current in a certain circuit as measured by an ammeter is a continuous random variable X with the following probability density function:
$f(x) = 0.057x + 0.272$ if 3 <= x <= 5. It is 0 otherwise.
a) Verify that the total area under the density curve is 1.
b) Obtain the CDF
c) Calculate P(x<= 4)
d) Is P(3<=x<=5) strictly less than, strictly greater than, or exactly the same as P(x<=4)
e) Calculate P(5.4<X)

Homework Equations

The Attempt at a Solution

For part a:
$\int_{3}^{5} (0.057x+0.272)dx = 1$
So the pdf is confirmed
For part b:
$F(x) = \int_{3}^{x}(0.057y+0.272)dy$
$= \frac{0.057}{2}x^{2}+0.272x - (\frac{0.057}{2}3^{2}+0.272(3))$
$F(x) = \frac{0.057}{2}x^{2} + 0.272x - 1.0725$
This is from 3<=X<=5
For part c:
I said that P(X<=4) is exactly the same as computing P(3<=X<=5). So From the CDF above I did:
$F(4)-F(3) = 0.4715 - 0 = 0.4715$
For part d:
I said that it would be exactly the same
For part e:
This is where I am getting confused, for P(5.4<X), I thought the formula was:
$1 - F(5.4)$. However, when I do this calculation, I get back a negative value, which doesn't make any sense. So I was wondering if my computed CDF is wrong? Another thought I had was since 5.4 is out of the range of the pdf, would the probability just be equal to 1? Any help is appreciated. Thanks

Since the density vanishes outside the interval [3,5], the CDF $F(x)= P(X \leq x)$ must satisfy $F(x) = 0$ for $x < 3$ and $F(x) = 1$ for $x \geq 5$--no calculations needed. For $x$ between 3 and 5, $F(x)$ is given by some formula involving $x$, and you have found it (but I have not checked it).

Similarly, the complementary cumulative $G(x) = P(X > x)$ satisfies $G(x) = 1$ if $x \leq 3$ and $G(x) = 0$ if $x \geq 5$---again, without calculations needed. Of course, $G = 1-F$.