Probaility density function Determine A

[TomJerry]

Problem :
A random variable X has probability density function
$$f(n) = <br /> \begin{cases} <br /> Ax, & \mbox{0 lessthan equal x less than 5 } \\<br /> A(10-x), & \mbox{5 less than equal x less than 10 }\\ <br /> 0, & \mbox{otherwise } <br /> \end{cases}<br />$$

i)Determine A
ii)Find P(2.5 <= X <= 7.5)

Solutions
i) let P(A) = Ax and P(b)= A(10-x)

P(A) + P(B) = 1

Therefore on calculating i get A = 1/25 [IS THIS CORRECT]

ii)
integrate from [0-5] for Ax and substract it by [0-2.5] A(x) --> I will get [2.5-5] A(x)
then
integrate from [5-10] for A(10-x) and substract it by [7.5-10] A(10-x) --> I will get [5-7.5] A(10-x)

Add them both to get answer for ii)

IS THIS CORRECT

 

[tiny-tim]

Hi TomJerry!

Therefore on calculating i get A = 1/25 [IS THIS CORRECT]

Yes!

integrate from [0-5] for Ax and substract it by [0-2.5] A(x) --> I will get [2.5-5] A(x)
then
integrate from [5-10] for A(10-x) and substract it by [7.5-10] A(10-x) --> I will get [5-7.5] A(10-x)

Add them both to get answer for ii)

IS THIS CORRECT

shouldn't there be squares here?

(and i'd just do one of the integrals, and double it … the distribution is obviously syymmetric about 5 )

 

[TomJerry]

shouldn't there be squares here?

(and i'd just do one of the integrals, and double it … the distribution is obviously syymmetric about 5 )

I didnt get it... what squares

 

[tiny-tim]

If you're integrating Ax, won't you get an x² ?