Question:
Let π1, π2, be a sequence of π.π.π. random variables each having π(0,1) distribution. Let π be a random variable having the distribution function πΊ. Suppose that
\(limπ_{nββ} \frac{( π-1+π_2+π_π}{4} β€ π₯)=πΊ(π₯)\), for all π₯ β β .
Then, πππ(π) equals
Let π1, π2, be a sequence of π.π.π. random variables each having π(0,1) distribution. Let π be a random variable having the distribution function πΊ. Suppose that
\(limπ_{nββ} \frac{( π-1+π_2+π_π}{4} β€ π₯)=πΊ(π₯)\), for all π₯ β β .
Then, πππ(π) equals
\(limπ_{nββ} \frac{( π-1+π_2+π_π}{4} β€ π₯)=πΊ(π₯)\), for all π₯ β β .
Then, πππ(π) equals
Updated On: Oct 1, 2024
- \(\frac{1}{12}\)
- \(\frac{1}{32}\)
- \(\frac{1}{48}\)
- \(\frac{1}{64}\)
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The Correct Option is D
Solution and Explanation
The correct option is (D): \(\frac{1}{64}\)
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