Suppose the probability density function is given by $f\left(x\right)=\frac{1}{2}$ for $0\le x\le 2$ and $f\left(x\right)=0$ otherwise. What is the probability that $x$ is between $1$ and $3$?

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $0$
• D. $1$

Answer: (B)

Verified step by step guidance

1. Identify the probability density function (pdf) given: \(f(x) = \frac{1}{2}\) for \$0 \leq x \leq 2\(, and \)f(x) = 0$ otherwise.
2. Understand that the probability that \(x\) lies between two values \(a\) and \(b\) is found by integrating the pdf over that interval: \(P(a \leq x \leq b) = \int_a^b f(x) \, dx\).
3. Set the limits of integration according to the problem: since we want \(P(1 \leq x \leq 3)\), but the pdf is zero for \(x > 2\), adjust the upper limit to 2, so the integral becomes \(\int_1^2 \frac{1}{2} \, dx\).
4. Perform the integration: \(\int_1^2 \frac{1}{2} \, dx = \frac{1}{2} \times (2 - 1)\).
5. Interpret the result of the integral as the probability that \(x\) is between 1 and 3, noting that values beyond 2 contribute zero probability.