Question

Suppose that X has the density function \( f(x) = cx^2 \) for \( 0 \le x \le 1 \) and \( f(x) = 0 \) otherwise. What is \( P(0.1 \le X \le 0.5) \)?

• A. \( \frac{1}{25} \)
• B. \( \frac{6}{25} \)
• C. \( \frac{4}{250} \)
• D. \( \frac{31}{250} \)

Hint:

Always normalize the PDF to find any unknown constants before computing interval probabilities. For power-law distributions like \(x^n\), the normalization constant over \([0,1]\) is always \(n+1\).

Answer 1

The Correct Option is D

Concept: For a continuous random variable \( X \) defined by a Probability Density Function (PDF) \( f(x) \):
• Normalization Condition: The total area under the PDF curve over the entire support domain must equal 1: \[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \] This condition allows us to determine unknown normalization constants like \( c \).
• Probability Calculation: The probability that \( X \) falls within a specific interval \([a, b]\) is given by the definite integral: \[ P(a \le X \le b) = \int_{a}^{b} f(x) \, dx \]

Step 1: Find the normalization constant \( c \)
Apply the normalization condition to the non-zero interval of the PDF: \[ \int_{0}^{1} cx^2 \, dx = 1 \] Evaluate the definite integral: \[ c \left[ \frac{x^3}{3} \right]_{0}^{1} = 1 \implies c \left( \frac{1^3}{3} - 0 \right) = 1 \implies \frac{c}{3} = 1 \implies c = 3 \] Thus, the complete probability density function is: \[ f(x) = 3x^2 \quad \text{for } 0 \le x \le 1 \]

Step 2: Calculate the target interval probability
We want to find \( P(0.1 \le X \le 0.5) \). Set up the definite integral over these limits using our fully determined PDF: \[ P(0.1 \le X \le 0.5) = \int_{0.1}^{0.5} 3x^2 \, dx \] Evaluate the integration: \[ \int_{0.1}^{0.5} 3x^2 \, dx = \left[ x^3 \right]_{0.1}^{0.5} = (0.5)^3 - (0.1)^3 \]

Step 3: Convert decimals to fractional forms
Expressing the values as fractions simplifies the arithmetic: \[ 0.5 = \frac{1}{2} \implies (0.5)^3 = \frac{1}{8} \] \[ 0.1 = \frac{1}{10} \implies (0.1)^3 = \frac{1}{1000} \] Subtract these fractional values: \[ P(0.1 \le X \le 0.5) = \frac{1}{8} - \frac{1}{1000} \] Find a common denominator, which is 1000: \[ \frac{1}{8} = \frac{125}{1000} \] \[ P(0.1 \le X \le 0.5) = \frac{125}{1000} - \frac{1}{1000} = \frac{124}{1000} \] Simplify the fraction by dividing the numerator and denominator by 4: \[ \frac{124 \div 4}{1000 \div 4} = \frac{31}{250} \] This matches Option (D).