Question

If the p. d. f. of a continuous random variable X is given by \(f(x) = \{\begin{array}{cc}k(9+8x-x^2), & \text{for }-1\leq x\leq 4 \\ 0, & \text{otherwise}\end{array}\) then the value of \(k\) is \(\ldots\)

• A. \(\frac{1}{125}\)
• B. \(\frac{3}{250}\)
• C. \(\frac{3}{125}\)
• D. \(\frac{1}{250}\)

Hint:

Total probability must be 1, so integrate the pdf from -1 to 4 and set it equal to 1.

Answer 1

The Correct Option is B

Step 1: Factor to make the integral easier:
$9 + 8x - x^2 = (9 - x)(1 + x)$. Substitute $u = x + 1$ to get $u(10 - u)$ with $u$ from 0 to 5.

Step 2: Integrate:
$\int_0^5 (10u - u^2)\,du = \left[5u^2 - \frac{u^3}{3}\right]_0^5 = 125 - \frac{125}{3} = \frac{250}{3}$.

Step 3: Solve for k:
$k\cdot\frac{250}{3} = 1$, so $k = \frac{3}{250}$.

Step 4: Check:
The same value of $\frac{250}{3}$ results from the direct antiderivative, so the substitution is consistent.

Final Answer:
Option (B). \[ \boxed{\frac{3}{250} \text{ (B)}} \]