SOLUTION: Let X be a random variable with pdf f (x) = ax^2 + b, 0 ≤ x ≤ 1. Find a and b such that E(X) = 5/8 . Thank You

Question 1177296: Let X be a random variable with pdf f (x) = ax^2 + b, 0 ≤ x ≤ 1. Find a and b such that E(X) = 5/8 .
Thank You
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
**1. Determine the Relationship Between a and b**
Since f(x) is a probability density function (pdf), the integral of f(x) over its domain must equal 1:
* ∫[0, 1] (ax² + b) dx = 1
Integrate:
* [ (ax³/3) + bx ] from 0 to 1 = 1
* (a/3 + b) - (0) = 1
* a/3 + b = 1
* a + 3b = 3
* a = 3 - 3b
**2. Calculate the Expected Value E(X)**
The expected value of X is given by:
* E(X) = ∫[0, 1] x * f(x) dx
* E(X) = ∫[0, 1] x * (ax² + b) dx
* E(X) = ∫[0, 1] (ax³ + bx) dx
Integrate:
* E(X) = [ (ax⁴/4) + (bx²/2) ] from 0 to 1
* E(X) = (a/4 + b/2) - (0)
* E(X) = a/4 + b/2
**3. Use the Given E(X) = 5/8**
We are given that E(X) = 5/8. So:
* a/4 + b/2 = 5/8
Multiply both sides by 8 to eliminate fractions:
* 2a + 4b = 5
**4. Substitute a = 3 - 3b**
Substitute a = 3 - 3b into the equation 2a + 4b = 5:
* 2(3 - 3b) + 4b = 5
* 6 - 6b + 4b = 5
* 6 - 2b = 5
* -2b = -1
* b = 1/2
**5. Find a**
Substitute b = 1/2 back into a = 3 - 3b:
* a = 3 - 3(1/2)
* a = 3 - 3/2
* a = 6/2 - 3/2
* a = 3/2
**Therefore, a = 3/2 and b = 1/2.**

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