Probability - Mean & Variance?

A hat contains 21 identical tags numbered 1, 2, 3, .. 21. (The numbers are printed on the tags, and each number in the list occurs only once.) You select one tag at random from the hat, without looking. (Assume that each tag has an equal probability of being picked.) Let x = the number on


the tag you have picked


(a) Determine the expected value of x.


(b) Find the variance of x.


(c) What is the probability of selecting a tag with an odd number?

Probability - Mean %26amp; Variance?
This is a discrete uniform distribution over the interval (1,21):





a)


The mean of a discrete uniform distribution is (a+b)/2 where a and b are the endpoints. We have:


E(x) = (1+21)/2 = 11





b)


The variance of a discrete uniform distribution is ((n^2)-1)/12 where n is the number of terms in the range. We have:


Var(x) = ((21^2)-1)/12 = 36.67





c)


There are 11 odd numbered tags out of the 21 total (1,3,5,7,9,11,13,15,17,19, and 21). This means this probability is 11/21 = .5238 or 52.38%