Question

Let 𝑋~𝐹_6,2 and 𝑌~𝐹_2,6 . If 𝑃(𝑋 ≤ 2) =\( \frac{216}{343}\) and 𝑃 (𝑌≤\(\frac{1}{2}\) )=𝛼, then 686𝛼 equals

• A. 246
• B. 254
• C. 260
• D. 264

Answer 1

The Correct Option is B

To solve this problem, we need to understand the properties and relationships of the F-distribution. The F-distribution is a continuous probability distribution that arises frequently in the analysis of variance, among other statistical tests.

We are given:

• \(X \sim F_{6,2}\), meaning \(X\) follows an F-distribution with 6 numerator degrees of freedom and 2 denominator degrees of freedom.
• \(Y \sim F_{2,6}\), meaning \(Y\) follows an F-distribution with 2 numerator degrees of freedom and 6 denominator degrees of freedom.
• \(P(X \leq 2) = \frac{216}{343}\)
• We want to find \(686\alpha\) where \(P(Y \leq \frac{1}{2}) = \alpha\).

Given the reciprocal relationship of F-distributions, where if \(X \sim F_{n,m}\), then \(\frac{1}{X} \sim F_{m,n}\), we can use this relationship between \(X\) and \(Y\).

For the specified distributions: \[ P(X \leq 2) = \frac{216}{343} \implies P\left( \frac{1}{X} \geq \frac{1}{2} \right) = \frac{216}{343} \]

Using the complement rule: \[ P\left( \frac{1}{X} \leq \frac{1}{2} \right) = 1 - \frac{216}{343} = \frac{127}{343} \]

Here, \(\frac{1}{X} \sim F_{2,6}\), which is the same distribution as \(Y\). Hence: \[ P(Y \leq \frac{1}{2}) = \frac{127}{343} \]

Therefore, \(\alpha = \frac{127}{343}\).

We need to find \(686 \alpha\): \[ 686 \alpha = 686 \times \frac{127}{343} = 2 \times 127 = 254 \]

Thus, the correct answer is 254.