Probaility density function Determine A

[TomJerry]

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

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 ?

 

[blue_raver22]

Your solution for part i) is correct. To determine A, you need to use the fact that the total area under the probability density function must equal 1. Therefore, you can set up the following equation:
∫f(x)dx = 1
∫Ax dx + ∫A(10-x) dx = 1
[Ax^2/2] from 0 to 5 + [A(10x-x^2/2)] from 5 to 10 = 1
(25A/2) + (25A/2) = 1
50A/2 = 1
A = 1/25

For part ii), your approach is correct. You need to use the probability density function to calculate the probability of the random variable falling within the given range. Therefore, the solution would be:
P(2.5 ≤ X ≤ 7.5) = ∫f(x)dx from 2.5 to 7.5
= ∫Ax dx from 2.5 to 5 + ∫A(10-x) dx from 5 to 7.5
= [Ax^2/2] from 2.5 to 5 + [A(10x-x^2/2)] from 5 to 7.5
= (25A/2) - (6.25A/2) + (12.5A/2) - (9.375A/2)
= 8.75A/2
= 8.75(1/25)/2
= 0.175 or 17.5%
Therefore, the probability of the random variable falling between 2.5 and 7.5 is 17.5%.