# CDF and PDF Calculations

1. Apr 13, 2015

### _N3WTON_

1. The problem statement, all variables and given/known data
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)

2. Relevant equations

3. 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

Last edited: Apr 13, 2015
2. Apr 13, 2015

### 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].

3. Apr 13, 2015

### _N3WTON_

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)?

4. Apr 13, 2015

### RUber

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.

5. Apr 14, 2015

### Ray Vickson

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, $G = 1-F$.

Last edited: Apr 15, 2015
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