I really need help with this problem:
Suppose that the 5% of population is left-handed.
In a party there is a random number of guests, given by a r.v. of Poisson with parameter 1000. Find:
a) The expected value of left-handed people in the party
b) The distribution of left-handed people in the party
c) The probability that one random person in the party is left-handed.
My problem is that I have only "classic examples" of Poisson distribution (i.e. "A machine produces an item A... The number of calls... ") where the sample size is given (i.e. "a box of 100 items... 300 calls in a week..."), so I don't know how to approach this problem.
Can someone help me?
Thanks in advance.
So, considering guests were chosen randomly (i.e. without handedness-related bias), we have the following:
a) The E.V. of left-handed people is just the E.V. of total guests multiplied by 5%. I guess you can easily calculate it having a parameter.
b) Again, just take CDF of Poisson distribution with parameter 1000 and multiply that function by 5%.
c) It's 5%, as it's the same if you were randomly choosing a person from the whole population