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