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_N3WTON_

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## 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:

[itex] f(x) = 0.057x + 0.272 [/itex] 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:

[itex] \int_{3}^{5} (0.057x+0.272)dx = 1 [/itex]

So the pdf is confirmed

For part b:

[itex] F(x) = \int_{3}^{x}(0.057y+0.272)dy [/itex]

[itex] = \frac{0.057}{2}x^{2}+0.272x - (\frac{0.057}{2}3^{2}+0.272(3)) [/itex]

[itex]F(x) = \frac{0.057}{2}x^{2} + 0.272x - 1.0725 [/itex]

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:

[itex] F(4)-F(3) = 0.4715 - 0 = 0.4715 [/itex]

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:

[itex] 1 - F(5.4) [/itex]. 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

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