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

The density function is proportional to \(x^2\) on the interval \([1,4]\). Hence, probabilities over subintervals are proportional to the increase in \(x^3\) over those intervals.

Over the full interval \([1,4]\), the total contribution is:

\(4^3 - 1^3 = 64 - 1 = 63\)

This represents the entire probability mass.

Over the interval \([2,3]\), the corresponding contribution is:

\(3^3 - 2^3 = 27 - 8 = 19\)

Therefore, the probability that \(x\) lies between \(2\) and \(3\) is simply the ratio:

\(\dfrac{19}{63}\)

Numerically,

\(\dfrac{19}{63} \approx 0.3016\)

Hence, the required probability is \(\boxed{0.302}\) (rounded to three decimal places).