# Math Help - Probability Question regarding determining the constant in the pdf

1. ## Probability Question regarding determining the constant in the pdf

Question : A random variable X has a pdf

f(x) = $\begin{cases}
Ax , & \mbox{0 } \le \mbox{ x} < \mbox{ 5} \\
A(10 -x) , & \mbox{5} \le \mbox{ x} < \mbox{10} \\
0, & \mbox{otherwise }
\end{cases}$

i) Determine $A$

ii) Find $P(2.5 \le X \le 7.5)$

2. Originally Posted by zorro
Question : A random variable X has a pdf

f(x) = $\begin{cases}
Ax , & \mbox{0 } \le \mbox{ x < 5} \\
A(10 -x) , & \mbox{5} \le \mbox{ x <10} \\
0, & \mbox{otherwise }
\end{cases}$

i) Determine $A$

ii) Find $P(2.5 \le X \le 7.5)$
What attempt have you made to this problem? First note that if f is a pdf, then $\int_{-\infty}^{\infty}f\!\left(x\right)\,dx=1$...

To answer the second part, use the definition of the pdf.

3. ## I am having problem with the first part

Originally Posted by Chris L T521
What attempt have you made to this problem? First note that if f is a pdf, then $\int_{-\infty}^{\infty}f\!\left(x\right)\,dx=1$...

To answer the second part, use the definition of the pdf.
i am having with determining the value of A hw should i go about it

4. Originally Posted by zorro
i am having with determining the value of A hw should i go about it
Integrate each part of the piecewise function over its domain, and then set your answer equal to 1. Then solve the equation you created for A.

5. ## Is this correct?

Originally Posted by Chris L T521
Integrate each part of the piecewise function over its domain, and then set your answer equal to 1. Then solve the equation you created for A.

Is this correct?

$f(0 < x < 5) \ = \ \int_{0}^{5} Ax \ = \ \frac{25}{2}A$

$f(5 < x < 10) \ = \ \int_{5}^{10} A(10 - x) \ = \ \frac{25}{2} A$

$\frac{25}{2}A \ = \ 1$

$\therefore \ \ A = \frac{2}{25}$

6. Originally Posted by zorro
Is this correct?

$f(0 < x < 5) \ = \ \int_{0}^{5} Ax \ = \ \frac{25}{2}A$

$f(5 < x < 10) \ = \ \int_{5}^{10} A(10 - x) \ = \ \frac{25}{2} A$

$\frac{25}{2}A \ = \ 1$

$\therefore \ \ A = \frac{2}{25}$
Close... You need to add both of the values together and then set equal to 1. So $A=\ldots$

7. ## Is this correct?

Originally Posted by Chris L T521
Close... You need to add both of the values together and then set equal to 1. So $A=\ldots$

$\frac{25}{2} A + \frac{25}{2} A \ = \ 1$

$\frac{50}{2} A \ = \ 1$

$\frac{50}{2} A \ = \ 1$

$25 A = \ 1$

$\therefore \ A \ = \ \frac{1}{25}$

Now for

$P(2.5 \le X \le 7.5) \ = \ \int_{2.5}^{5} Ax \ dx+ \int_{5}^{7.5} A(10-x)dx$

Is this correct???

8. Originally Posted by zorro
$\frac{25}{2} A + \frac{25}{2} A \ = \ 1$

$\frac{50}{2} A \ = \ 1$

$\frac{50}{2} A \ = \ 1$

$25 A = \ 1$

$\therefore \ A \ = \ \frac{1}{25}$

Now for

$P(2.5 \le X \le 7.5) \ = \ \int_{2.5}^{5} Ax \ dx+ \int_{5}^{7.5} A(10-x)dx$

Is this correct???
Correct! But note that you know what A is...so you should evaluate $\int_{2.5}^{5} \tfrac{1}{25}x\,dx+ \int_{5}^{7.5} \tfrac{1}{25}(10-x)\,dx$.

9. ## Thanks Chris L T521

Originally Posted by Chris L T521
Correct! But note that you know what A is...so you should evaluate $\int_{2.5}^{5} \tfrac{1}{25}x\,dx+ \int_{5}^{7.5} \tfrac{1}{25}(10-x)\,dx$.

thanks Chris L T521 for helping me so much
Cheers mite