Question

The following is p.d.f. of continuous random variable X: $f(x) = \frac{x}{8}$ for $0 < x < 4$. Then $F(0.5)$, $F(1.7)$ and $F(5)$ is respectively ______.

• A. $\frac{1}{64}$, 1, 0.18
• B. 0.0156, 0.18, 1
• C. 0.18, 0.0156, 1
• D. 1, 0.0156, 0.18

Hint:

Always check the domain bounds first! If you are asked to find the CDF $F(x)$ for an $x$ that is larger than the maximum possible value in the probability space, the answer is instantly 1.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given a probability density function (p.d.f.) $f(x)$. We must calculate values for its corresponding Cumulative Distribution Function (C.D.F.), denoted as $F(x)$, at three specific points.

Step 2: Detailed Explanation:
By definition, the Cumulative Distribution Function $F(x)$ is the integral of the p.d.f. from the lowest bound up to $x$:
$F(x) = \int_{0}^{x} f(t) dt$ (for $0 \le x \le 4$)
Let's find the general formula for $F(x)$ in this active domain:
$F(x) = \int_{0}^{x} \frac{t}{8} dt = \left[ \frac{t^2}{16} \right]_0^x = \frac{x^2}{16}$
Now, evaluate $F(x)$ at the requested points:
1. Evaluate $F(0.5)$:
Since $0.5$ is within $[0, 4]$, use the formula:
$F(0.5) = \frac{(0.5)^2}{16} = \frac{0.25}{16} = \frac{1/4}{16} = \frac{1}{64}$
As a decimal: $\frac{1}{64} = 0.015625 \approx 0.0156$
2. Evaluate $F(1.7)$:
Since $1.7$ is within $[0, 4]$, use the formula:
$F(1.7) = \frac{(1.7)^2}{16} = \frac{2.89}{16}$
As a decimal: $\frac{2.89}{16} = 0.180625 \approx 0.18$
3. Evaluate $F(5)$:
The value $x=5$ lies completely outside and strictly above the upper bound of the active domain ($x=4$). In continuous probability, once the variable passes the maximum possible value, the cumulative probability maxes out at exactly 100%.
$F(x) = 1$ for all $x \ge 4$.
Therefore, $F(5) = 1$.
The values are $0.0156$, $0.18$, and $1$ respectively.

Step 3: Final Answer:
The values are 0.0156, 0.18, 1, matching option (b).