Question

Let X be the random variable representing the number of calls received in an hour by a 911 emergency service. A portion of the probability distribution of X is given below. Value of X 0 1 2 3 4 Probability P(x) 0.25 ___ ___ 0.10 0.05 Suppose the probability that X = 1 and the probability that X = 2 are the same. What are these probabilities? What is the expected number of 911 calls in an hour?

Answer 1

For any distribution, the sum of the probabilities of all possible outcomes must be 1. In this case, we have to have

$P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)=1$

We're told that $p=P(X=1)=P(X=2)$, and we're given other probabilities, so we have

$0.25+2p+0.10+0.05=1\implies2p=0.6\implies p=0.3$

The expected number of calls would be

$E[X]=\displaystyle\sum_xx\,P(X=x)$

$E[X]=0\,P(X=0)+1\,P(X=1)+\cdots+4\,P(X=4)$

$E[X]=1.4$