Question

Let \( X \) be a continuous random variable with the probability density function
\[ f(x) = \begin{cases} \frac{x}{8}, & \text{if} \, 0 < x < 2, \\ k, & \text{if} \, 2 \leq x \leq 4, \\ \frac{6 - x}{8}, & \text{if} \, 4 < x < 6, \\ 0, & \text{otherwise}. \end{cases} \] Then \( P(1 < X < 5) \) equals

Hint:

To calculate probabilities for continuous random variables, integrate the probability density function over the desired range.

Answer 1

Correct Answer: 0.87 - 0.88

Step 1: Finding \( k \).
We are given the probability density function \( f(x) \). To determine the constant \( k \), we use the fact that the total probability must sum to 1, i.e., \[ \int_0^2 \frac{x}{8} \, dx + \int_2^4 k \, dx + \int_4^6 \frac{6 - x}{8} \, dx = 1. \] Solving for \( k \), we find that \( k = \frac{1}{4} \).
Step 2: Finding \( P(1<X<5) \).
Now, we compute \( P(1<X<5) \) by integrating the probability density function over the appropriate range: \[ P(1<X<5) = \int_1^2 \frac{x}{8} \, dx + \int_2^4 \frac{1}{4} \, dx + \int_4^5 \frac{6 - x}{8} \, dx. \] Evaluating these integrals gives the result approximately between 0.87 and 0.88.

Step 3: Conclusion.
Thus, \( P(1<X<5) \) is approximately \( 0.87 \) to \( 0.88 \).