(1)
Let’s define some variables: n=12, p=0.18, q=1-p=0.82, where n is the number of calls and p is the probability of a long-distance (LD) call based on observation.
(p+q)¹²=p¹²+12p¹¹q+66p¹⁰q²+220p⁹q³+...+220p³q⁹+66p²q¹⁰+12pq¹¹+q¹²=1 is the binomial expansion. B(X) is the binomial term for exactly X LD calls. For example, if X=2, B(X)=66p²q¹⁰.
p¹² is the probability of all calls being LD, and q¹² the probability of none. 12pq¹¹ is the probability of 1 out of 12 being LD, etc. To find out the probability of 3 or more, we subtract the accumulated probability of 0, 1 and 2 from 1, that is, 1-(B(0)+B(1)+B(2))=0.3702 approx or about 37%.
(2)
To use the normal distribution we use mean=np=2.16 and SD=1.7712. The Z score for 3 calls is:
(3-np)/(npq)=0.4743 approx. Using tables for this value of Z we get 0.6822 as the probability that there will be 3 LD calls or fewer. So 1-0.6822=0.3178 is the probability for at least 3 LD calls.
(3)
So the actual probability using the binomial distribution was about 0.3702, and the normal distribution gave us 0.3178 which is 0.3178/0.3702=0.8585, an error of about 14% (too low) which we could write -14%.