Question

Let \( f(x) \) be the probability distribution function of a random variable \( X \), where \[ f(x) = \frac{5x^4}{64} \quad \text{if} \quad -2<x<2 \] \[ = 0 \quad \text{otherwise} \] Let \( | \cdot | \) denote the modulus function. Then, \( P(|X|<1) \) and \( P(X^2<3) \) are given by (respectively):

• A. \( \frac{1}{32} \) and \( \frac{9}{\sqrt{32}} \)
• B. \( \frac{5}{64} \) and \( \frac{5}{\sqrt{32}} \)
• C. \( \frac{1}{64} \) and \( \frac{\sqrt{3}}{32} \)
• D. \( \frac{2\sqrt{3}}{32} \) and \( \frac{1}{32} \)

Hint:

For probability distributions, integrate the probability density function over the desired range to find the probability. Remember to handle modulus and other transformations appropriately.

Answer 1

The Correct Option is A

Step 1: Calculating \( P(|X|<1) \).
We are asked to find the probability that \( |X|<1 \), which is equivalent to finding the integral of \( f(x) \) over the interval \( -1 \leq x \leq 1 \): \[ P(|X|<1) = \int_{-1}^{1} \frac{5x^4}{64} \, dx \] We compute this integral: \[ P(|X|<1) = \frac{5}{64} \int_{-1}^{1} x^4 \, dx = \frac{5}{64} \left[ \frac{x^5}{5} \right]_{-1}^{1} = \frac{5}{64} \left( \frac{1^5}{5} - \frac{(-1)^5}{5} \right) \] \[ P(|X|<1) = \frac{5}{64} \cdot \frac{2}{5} = \frac{1}{32} \] Step 2: Calculating \( P(X^2<3) \).
Next, we need to calculate the probability that \( X^2<3 \), which is equivalent to finding the integral of \( f(x) \) over the interval \( -\sqrt{3} \leq x \leq \sqrt{3} \): \[ P(X^2<3) = \int_{-\sqrt{3}}^{\sqrt{3}} \frac{5x^4}{64} \, dx \] We compute this integral: \[ P(X^2<3) = \frac{5}{64} \int_{-\sqrt{3}}^{\sqrt{3}} x^4 \, dx = \frac{5}{64} \left[ \frac{x^5}{5} \right]_{-\sqrt{3}}^{\sqrt{3}} = \frac{5}{64} \left( \frac{(\sqrt{3})^5}{5} - \frac{(-\sqrt{3})^5}{5} \right) \] \[ P(X^2<3) = \frac{5}{64} \cdot \frac{18\sqrt{3}}{5} = \frac{9}{\sqrt{32}} \] Step 3: Conclusion.
Thus, the correct answer is (A) \( \frac{1}{32} \) and \( \frac{9}{\sqrt{32}} \).