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Compute \(P\left(Y_{3}<\xi_{0.5}

Short Answer

Expert verified
The probability \(P\left(Y_{3}<\xi_{0.5}<Y_{7}\right)\) is 0.3333.

Step by step solution

01

Formulate the Probability Expression

To solve this, it's necessary to express the probability just using the order statistics. Given that \(Y_{3}<\xi_{0.5}<Y_{7}\) with the ordered set \(Y_{1}<\cdots<Y_{9}\), we can look at it as the median \(\xi_{0.5}\) being anywhere between the 3rd and 7th element in the set. This can be formulated as \( P\left(Y_{3}<\xi_{0.5}<Y_{7}\right) = P\left(3<n<7\right) \) where n represents the position of the median in the ordered set.
02

Calculate the Probability for the Specified Range

Since the median can fall between any of the nine positions with equal probability in a random sample, we consider each position as an equally likely event. Thus, the probability of each event can be calculated as 1 divided by the total number of events. In this case that would be \(1/9 = 0.1111.\) Now, to find the overall probability of the median falling between the 3rd and 7th order statistic (both exclusive), simply multiply this probability with number of positions ranging from 4 to 6, which is 3. This gives us \(0.1111 * 3 = 0.3333.\)
03

Final Probability

So, the likelihood of the median of a random sample from a distribution of continuous type falling between the third and seventh order statistic is 0.3333. Hence, we have \(P\left(Y_{3}<\xi_{0.5}<Y_{7}\right)=0.3333\).

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