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

Correct Answer: 0.3

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. Therefore, we find the constant \( C \) by integrating \( P(x) \) over \( [1,4] \):
\[ \int_1^4 Cx^2 \, dx = 1. \] Compute the integral:
\[ \int_1^4 x^2 \, dx = \left[ \frac{x^3}{3} \right]_1^4 = \frac{64}{3} - \frac{1}{3} = \frac{63}{3} = 21. \] Hence,
\[ C \times 21 = 1 \quad \Rightarrow \quad C = \frac{1}{21}. \] Step 1: Calculate the probability \( P(2 \leq x \leq 3) \)
\[ P(2 \leq x \leq 3) = \int_2^3 \frac{1}{21} x^2 \, dx. \] Compute the integral:
\[ \int_2^3 x^2 \, dx = \left[ \frac{x^3}{3} \right]_2^3 = \frac{27}{3} - \frac{8}{3} = \frac{19}{3}. \] Therefore,
\[ P(2 \leq x \leq 3) = \frac{1}{21} \times \frac{19}{3} = \frac{19}{63}. \] Step 2: Round the answer
\[ \frac{19}{63} \approx 0.3016. \] Thus, the required probability is
\[ \boxed{0.302} \] (rounded to three decimal places).