Part BCSIR NET June 2025integrating-the-squared-density-raises-the-exponent-so-the-tails-are-cubes

Integrating the squared density raises the exponent so the tails are cubes

Let X be a random sample of size 1 from the probability density function f(x| $θ) = (3/ θ^{3}) (θ - x)^{2}$ if $0 < x < θ$ , and 0 otherwise. If $(X / (1 - λ_{1}), X / (1 - λ_{2}))$ is a confidence interval for $θ$ with confidence coefficient $1 - α$ , where $λ_{i} \in (0, 1), i = 1, 2, λ_{1} < λ_{2}$ , and $α \in (0, 1)$ , then which of the following statements is true?

A. $λ_{2}^{2} - λ_{1}^{2} = 1 - α$
B. $λ_{2}^{3} - λ_{1}^{3} = 1 - α$ ✓
C. $λ_{2}^{2} - λ_{1}^{2} = 4 (1 - α)$
D. $λ_{2}^{3} - λ_{1}^{3} = 9 (1 - α)$

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests execution slip.

See pricing

50 are analysed free — try those first.

The trap it tests

Execution slip

The idea was right. The computation was not.

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Integrating the squared density raises the exponent so the tails are cubes

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