A statistics question

Question

It is known that 18% of all calls coming into a telephone exchange are long-distance calls. During a certain period of time, 12 calls came into the exchange.

1) What is the probability that at least 3 are long-distance calls?

2) Find the probability of part (1) by using the normal approximation to binomial probabilities.

3) Find the percentage error of the normal approximation.

Thanks

Answer 1

(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%.