Question

Given the probability density function $ f(X=x)= \begin{cases} 0, & x<2 \\ \frac{3+2x}{18}, & 2\leq x\leq4 \\ 0, & x>4 \end{cases} $ then the probability that X lies between 2 and 3 is

• A. \( \frac{4}{9} \)
• B. \( \frac{2}{3} \)
• C. \( \frac{5}{8} \)
• D. \( \frac{3}{8} \)

Hint:

When dealing with piecewise probability density functions, always ensure that the integration limits for finding probability match the specific definition of \( f(x) \) for that interval. If the interval spans multiple definitions, you might need to break the integral into multiple parts. Remember that the total probability over the entire range of X must equal 1.

Answer 1

The Correct Option is A

Step 1: Understand the definition of probability for a continuous random variable.
For a continuous random variable \(X\), its probability distribution is described by a Probability Density Function (PDF), denoted as \( f(x) \). The probability that the random variable \(X\) takes a value within a certain interval, say from \(a\) to \(b\), is calculated by integrating the PDF over that interval. The formula for this probability is: \[ P(a \leq X \leq b) = \int_{a}^{b} f(x) \, dx \] In this problem, we are given the piecewise PDF: \[ f(X=x)= \begin{cases} 0, & x<2 \\ \frac{3+2x}{18}, & 2\leq x\leq4 \\ 0, & x>4 \end{cases} \] We need to find the probability that \(X\) lies between 2 and 3, which can be written as \( P(2 \leq X \leq 3) \).
Step 2: Identify the relevant part of the PDF and set up the definite integral.
Looking at the given PDF, for the interval \( 2 \leq x \leq 4 \), the function is defined as \( f(x) = \frac{3+2x}{18} \).
Since the interval of interest for finding the probability is \( [2, 3] \), which falls entirely within the \( 2 \leq x \leq 4 \) range, we will use the function \( f(x) = \frac{3+2x}{18} \) for our integration.
The integral will be set up with lower limit \(a = 2\) and upper limit \(b = 3\): \[ P(2 \leq X \leq 3) = \int_{2}^{3} \frac{3+2x}{18} \, dx \] 
Step 3: Evaluate the definite integral.
To evaluate the integral, we can first pull the constant \( \frac{1}{18} \) outside the integral sign: \[ P(2 \leq X \leq 3) = \frac{1}{18} \int_{2}^{3} (3+2x) \, dx \] Now, integrate the expression \( (3+2x) \) with respect to \(x\): The integral of a constant, \(3\), is \(3x\). The integral of \(2x\) is \(2 \cdot \frac{x^{1+1}}{1+1} = 2 \cdot \frac{x^2}{2} = x^2\). So, the antiderivative of \( (3+2x) \) is \( 3x + x^2 \). Next, apply the limits of integration (from 2 to 3) using the Fundamental Theorem of Calculus: \( \int_{a}^{b} g(x) dx = G(b) - G(a) \), where \(G(x)\) is the antiderivative of \(g(x)\). \[ \int_{2}^{3} (3x + x^2)\Big|_{2}^{3} \] Substitute the upper limit \(x=3\): \[ (3(3) + (3)^2) = (9 + 9) = 18 \] Substitute the lower limit \(x=2\): \[ (3(2) + (2)^2) = (6 + 4) = 10 \] Subtract the value at the lower limit from the value at the upper limit: \[ 18 - 10 = 8 \] Finally, multiply this result by the constant \( \frac{1}{18} \) that we pulled out earlier: \[ P(2 \leq X \leq 3) = \frac{1}{18} \times 8 \] \[ P(2 \leq X \leq 3) = \frac{8}{18} \] Simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 2: \[ P(2 \leq X \leq 3) = \frac{8 \div 2}{18 \div 2} = \frac{4}{9} \] The final answer is \( \boxed{\frac{4}{9}} \).