P(smallest observation is < 0.5) = p(at least one < 0.5) = 1 - P(all ovservations >= 0.5)
can you evaluate the RHS?
The other question is similar
A sample of 5 iid is taken from an exponential distribution with λ =1.
1) What is the P(smallest observation is < 0.5)?
2) What is the P(largest observation is < 0.5)?
I am thrown off by the smallest/largest condition.
Many thanks in advance for any feedback.
I did not check your arithmatic. Assuming you calculated properly 0.3934 would be the chance of one obvservation being greater than 0.5.
the chance of 5 being great than 0.5 would be 0.3934 ^ 5
so answer = 1-(0.3934^5)
I dont think that's right.P(largest observation is < 0.5) = 1 - P(all ovservations >= 0.5)
because of the fact that the exponential distribution is memoryless??
p(largest < 0.5) = p(all are < 0.5)
This would be true for any distribution. I dont think the memorylessness property is relevent here, but i could be wrong.
Is it possible for you to elaborate?
So I am gathering the answer is
so answer = 1-(0.6065^5)If that is the case would this calculation then be:p(largest < 0.5) = p(all are < 0.5)
This would be true for any distribution. I don't think the memoryless property is relevant here, but i could be wrong.
Thank you for your patience