Question

Consider a probability distribution given by the density function \( P(x) \).

\[ P(x) = \begin{cases} Cx^2, & \text{for } 1 \leq x \leq 4, \\ 0, & \text{for } x<1 \text{ or } x>4. \end{cases} \]
The probability that \( x \) lies between 2 and 3, i.e., \( P(2 \leq x \leq 3) \), is ______. (rounded off to three decimal places)

Hint:

To find the probability over an interval for a continuous probability distribution, integrate the probability density function over that interval and normalize the result if necessary.

Answer 1

We are given the probability density function: \[ P(x) = \begin{cases} Cx^2, & \text{for } 1 \leq x \leq 4, \\ 0, & \text{for } x<1 \text{ or } x>4. \end{cases} \]
- The total probability over the interval \( [1, 4] \) must be 1, since the total probability in any probability distribution is 1. Therefore, we can find the value of the constant \( C \) by integrating \( P(x) \) over the interval \( [1, 4] \): \[ \int_1^4 Cx^2 \, dx = 1. \]
Compute the integral: \[ \int_1^4 x^2 \, dx = \left[ \frac{x^3}{3} \right]_1^4 = \frac{4^3}{3} - \frac{1^3}{3} = \frac{64}{3} - \frac{1}{3} = \frac{63}{3} = 21. \] Therefore, we have: \[ C \times 21 = 1 \quad \Rightarrow \quad C = \frac{1}{21}. \]
Step 1: Calculate the probability \( P(2 \leq x \leq 3) \)
Now that we have the value of \( C \), we can calculate the probability that \( x \) lies between 2 and 3: \[ P(2 \leq x \leq 3) = \int_2^3 \frac{1}{21} x^2 \, dx. \]
Compute the integral: \[ \int_2^3 x^2 \, dx = \left[ \frac{x^3}{3} \right]_2^3 = \frac{3^3}{3} - \frac{2^3}{3} = \frac{27}{3} - \frac{8}{3} = \frac{19}{3}. \] Therefore: \[ P(2 \leq x \leq 3) = \frac{1}{21} \times \frac{19}{3} = \frac{19}{63}. \]
Step 2: Round the answer
Now, let's calculate the value of \( \frac{19}{63} \): \[ P(2 \leq x \leq 3) = \frac{19}{63} \approx 0.3016. \]
Thus, the probability that \( x \) lies between 2 and 3 is \( \boxed{0.302} \) (rounded to three decimal places).