Probability Question regarding determining the constant in the pdf

**zorro** · December 13th 2009, 06:16 PM

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

**Chris L T521** · December 13th 2009, 06:38 PM

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.

**zorro** · December 13th 2009, 06:40 PM

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

**Chris L T521** · December 13th 2009, 06:47 PM

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.

**zorro** · December 13th 2009, 07:01 PM

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?

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

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

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

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

**Chris L T521** · December 13th 2009, 07:08 PM

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 $A=\ldots$

**zorro** · December 13th 2009, 07:27 PM

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???

**Chris L T521** · December 13th 2009, 07:29 PM

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 $\int_{2.5}^{5} \tfrac{1}{25}x\,dx+ \int_{5}^{7.5} \tfrac{1}{25}(10-x)\,dx$ .

**zorro** · December 18th 2009, 09:41 PM

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

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Probability Question regarding determining the constant in the pdf

I am having problem with the first part

Is this correct?

Is this correct?

Thanks Chris L T521

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