Thread: Probability Question regarding determining the constant in the pdf

Question : A random variable X has a pdf

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

i) Determine $\displaystyle A$

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

Originally Posted by zorro

Question : A random variable X has a pdf

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

i) Determine $\displaystyle A$

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

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

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

Originally Posted by Chris L T521

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

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

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.

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?

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


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

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

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

Originally Posted by zorro

Is this correct?

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


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

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

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

Close... You need to add both of the values together and then set equal to 1. So $\displaystyle A=\ldots$

Originally Posted by Chris L T521

Close... You need to add both of the values together and then set equal to 1. So $\displaystyle A=\ldots$

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

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

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

$\displaystyle 25 A = \ 1$

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


Now for

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

Is this correct???

Originally Posted by zorro

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

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

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

$\displaystyle 25 A = \ 1$

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


Now for

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

Is this correct???

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

Originally Posted by Chris L T521

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

thanks Chris L T521 for helping me so much
Cheers mite