What is the expected value of k and variance of k if the total time it takes a tourist to visit k locations is distributed according to the pdf f(t)=1/(2k)*e^-t/(2k)?

The tourist must visit n locations, and the probability that the tourist visits a location on a given day is 0.5.

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I'm not sure the question really makes sense as it is asked, though, to be honest.

If you're saying that the tourist will visit n locations, with a probability of 0.5 that he will visit any location on a given day (the probability density function given indicates that he is limited to one location per day), then our expected value and variance for the time taken on the trip match up from the above analysis, except for our needing to tweak the answers to be 2n and 4n^2.

If we're trying to find the expected value and variance of k, the number of locations visited, the answer is simple:

k = n

Var(k) = 0

This would hold because we already **know** that the tourist will visit n locations.

Now, if we were trying to find the expected number of locations visited in a day given that distribution, the answer could be based on our above result:

E[visits in a day] = 1/2

Variance is slightly trickier, but pretty easy, considering we only have two situations with equal probability (p = 0.5): visit (x = 1) or not visit (x = 0) where x simply denotes how many places were visited in a given day. Again, we only have two possibilities based on the probability density function and our expected trip time.

Var (visits in a day) = `sum_(n=1)^2p_(n)(x_(n) - mu)^2`

`= 0.5(0 -1/2)^2 + 0.5(1 - 1/2)^2 = 1/4`

Therefore, the variance on the number of locations we will be visiting in a day is 1/4.

I hope that answers your question!

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