Question

For a uniform distribution in the range \([0, k]\), the mean and the variance are equal if:

• A. \(k = 1\)
• B. \(k = 2\)
• C. \(k = 4\)
• D. \(k = 6\)

Hint:

Recall mean \(=k/2\) and variance \(=k^2/12\) for \(U(0,k)\); equate them and solve for \(k\).

Answer 1

The Correct Option is D

Step 1: For a continuous uniform distribution on \([0, k]\), the mean is \[\text{Mean} = \frac{k}{2}\] and the variance is \[\text{Variance} = \frac{k^2}{12}\]

Step 2: Setting mean equal to variance: \[\frac{k}{2} = \frac{k^2}{12}\]

Step 3: Multiply both sides by 12: \[6k = k^2 \implies k^2 - 6k = 0 \implies k(k-6) = 0\]

Step 4: This gives \(k = 0\) or \(k = 6\). Since \(k=0\) gives a degenerate (zero-width) distribution, the valid answer is \(k = 6\).

Final answer: \(\boxed{k = 6}\)