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Question
The probability density function of a continuous random variable X is
f(x) = `{{:("a" + "b"x^2",", 0 ≤ x ≤ 1),(0",", "otherwise"):}`
where a and b are some constants. Find a and b if E(X) = `3/5`
Solution
Let X be due continuous variable of density function
`int_(-oo)^oo "f"(x) "d"x` = 1
Here `int_0^1 ("a" + "b" x^2) "d"x` = 1
`["a"x + ("b"x^3)/3]` = 1
`["a"(1) + ("b"(1)^3)/3] - ["a"(0) + ("b"(0))/3]` = 1
`"a" + "b"/3` = 1
⇒ 3a + b = 3 → (1)
Given that E(x) = `3/5`
`int_0^1 x "f"(x) "d"x = 1 = 3/5`
`int_0^1 x("a" + "b"x^2) "d"x = 3/5`
`["a" x^2/2 + "b" x^4/4]_0^1 = 3/5`
`["a"(1/2) + "b"(1/4)] - [0]] = 3/5`
`"a"/2 + "b"/4 = 3/5`
`(2"a"+ "b")/4 = 3/5`
⇒ `2"a" + "b" = 12/5` → (2)
Equation (1) - Equation (2)
⇒ 3a + b = 3
2a + b = 12/5
a = 3 - 12/5
a = `(15 - 12)/5`
∴ a = `3/5`
Substitute the value of a = `3/5` in equation
`3(3/5) + "b"` = 3
`9/5 + "b"` = 3
⇒ b = `3 - 9/5`
b = `(15 - 9)/5`
∴ b = `6/5`
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