For a random sample size n drawn from uniform distribution on[0,1], the excepted value of the maximum order stastic X(n) is:

a. 1/n
b. n+1/n
c. 1/n+1
d. n/n+1

The problem involves finding the expected value of the maximum order statistic \( X_{(n)} \) from a random sample of size \( n \) drawn from a uniform distribution on the interval \([0,1]\).

For a sample of size \( n \) from the uniform distribution \( U(0,1) \), the expected value of the maximum \( X_{(n)} \) is given by:

\[
\mathbb{E}[X_{(n)}] = \frac{n}{n+1}
\]

This formula arises from the properties of the uniform distribution and order statistics.

Therefore, the correct answer is:

**d. \(\frac{n}{n+1}\)**

Would you like more details or have any questions?

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Here are 5 relative questions that expand on this concept:

1. How do you derive the expected value of the minimum order statistic \( X_{(1)} \) from a uniform distribution on \([0,1]\)?
2. What is the variance of the maximum order statistic \( X_{(n)} \) in a uniform distribution?
3. How does the expected value of \( X_{(n)} \) change if the uniform distribution is on \([a,b]\) instead of \([0,1]\)?
4. Can you explain how order statistics are used in statistical estimation?
5. How does the expected value of \( X_{(n)} \) compare to the expected value of \( X_{(n-1)} \)?

**Tip:** The expected value of order statistics can often be derived using integration techniques and properties of the cumulative distribution function (CDF).

Explore the expected value of the maximum order statistic from a random sample drawn from a uniform distribution on [0,1]. Understand the formula derived from probability theory and properties of order statistics.

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