Answer: Here's my answer! <3 This took me soooo long!
Explanation:
(a) To find P(0.5 < X < 1.5) using the probability density function (pdf) fx(x), we need to integrate the pdf over the interval [0.5, 1.5].
Since the pdf is given as fx(x) = cx for 0 < x < 2, and 0 otherwise, we can find the value of c by integrating the pdf over its entire range and setting it equal to 1 (since the total probability of the random variable X must be 1).
Integrating the pdf over the range [0, 2], we get:
∫[0,2] cx dx = 1
Integrating cx, we get:
[cx^2/2] from 0 to 2 = 1
(c(2)^2/2) - (c(0)^2/2) = 1
2c = 1
c = 1/2
Now that we know the value of c, we can calculate P(0.5 < X < 1.5) by integrating the pdf over the interval [0.5, 1.5] using the value of c.
∫[0.5,1.5] (1/2)x dx = (1/2) * ∫[0.5,1.5] x dx
Integrating x, we get:
[(1/2)(x^2/2)] from 0.5 to 1.5
[(1/2)((1.5)^2/2)] - [(1/2)((0.5)^2/2)]
(1/2)(2.25/2) - (1/2)(0.25/2)
(1/2)(2.25/2 - 0.25/2)
(1/2)(2 - 0.25)
(1/2)(1.75)
0.875
Therefore, P(0.5 < X < 1.5) = 0.875.
(b) To find P(1 < X < 2) using the cumulative distribution function (cdf) Fx(x), we can integrate the pdf fx(x) over the interval [1, 2] using the value of c.
Fx(x) is the integral of fx(x) from negative infinity to x. Since the pdf is 0 for x < 0 and x > 2, we only need to consider the integral over the interval [0, x]. Therefore, the cdf Fx(x) is:
Fx(x) = ∫[0,x] fx(t) dt
For x < 0, Fx(x) = 0.
For 0 ≤ x ≤ 2, Fx(x) = ∫[0,x] (1/2)t dt
For x > 2, Fx(x) = 1.
To find P(1 < X < 2), we need to evaluate Fx(2) - Fx(1).
Fx(2) = ∫[0,2] (1/2)t dt
= (1/2) * ∫[0,2] t dt
= (1/2) * (t^2/2) | from 0 to 2
= (1/2) * (2^2/2 - 0^2/2)
= (1/2) * (4/2 - 0)
= (1/2) * 2
= 1
Fx(1) = ∫[0,1] (1/2)t dt
= (1/2) * ∫[0,1] t dt
= (1/2) * (t^2/2) | from 0 to 1
= (1/2) * (1^2/2 - 0^2/2)
= (1/2) * (1/2 - 0)
= (1/2) * (1/2)
= 1/4
Therefore, P(1 < X < 2) = Fx(2) - Fx(1) = 1 - 1/4 = 3/4.