Answer: 0.5493

## Explanation The interquartile range (IQR) is the difference between the upper quartile (Q₃) and lower quartile (Q₁). For a continuous random variable with PDF f_X(x), we need to find the values q₁ and q₃ such that: - F(q₁) = P(X ≤ q₁) = 0.25 - F(q₃) = P(X ≤ q₃) = 0.75 **Step 1: Find the CDF** Given PDF: f_X(x) = 2e^{-2x}, x > 0 The cumulative distribution function (CDF) is: $$ F(x) = \int_0^x 2e^{-2t} dt = \left[ -e^{-2t} \right]_0^x = 1 - e^{-2x} $$ **Step 2: Find Q₁ (lower quartile)** Set F(q₁) = 0.25: $$ 1 - e^{-2q₁} = 0.25 $$ $$ e^{-2q₁} = 0.75 $$ $$ -2q₁ = \ln(0.75) $$ $$ q₁ = -\frac{\ln(0.75)}{2} = -\frac{-0.287682}{2} = 0.143841 $$ **Step 3: Find Q₃ (upper quartile)** Set F(q₃) = 0.75: $$ 1 - e^{-2q₃} = 0.75 $$ $$ e^{-2q₃} = 0.25 $$ $$ -2q₃ = \ln(0.25) $$ $$ q₃ = -\frac{\ln(0.25)}{2} = -\frac{-1.386294}{2} = 0.693147 $$ **Step 4: Calculate IQR** $$ IQR = Q₃ - Q₁ = 0.693147 - 0.143841 = 0.549306 $$ **Note**: The solution in the text has a sign error in the calculations. The correct approach yields positive values for q₁ and q₃ since x > 0. The final IQR calculation is correct: 0.5493. **Key Points**: - The distribution is exponential with rate parameter λ = 2 - For exponential distribution, the CDF is F(x) = 1 - e^{-λx} - The interquartile range is Q₃ - Q₁ = (ln(4) - ln(4/3))/λ = ln(3)/λ = ln(3)/2 ≈ 0.5493

Author: Nikitesh Somanthe