In: Statistics and Probability
The number of calls that come into a small mail-order company
follows a Poisson distribution. Currently, these calls are serviced
by a single operator. The manager knows from past experience that
an additional operator will be needed if the rate of calls exceeds
15 per hour. The manager observes that 6 calls came into the
mail-order company during a randomly selected 15-minute
period.
a. If the rate of calls is actually 20 per hour, what is the
probability that 9 or more calls will come in during a given
15-minute period?
b. If the rate of calls is really 30 per hour, what is the
probability that 9 or more calls will come in during a given
15-minute period?
c. Based on the calculations in parts a and b, do you think that
the rate of incoming calls is more likely to be 20 or 30 per
hour?
d. Would you advise the manager to hire a second operator ?
Explain.
POISSON DISTRIBUTION (linear adjustment)
a) You have a random variable X(the number of calls which come in service mail order ), which has Poisson distribution, with parameter λ=20. The probability that there will be k "events" in time t, where tis measured in hours, is
e-20t (20t)k/k!
Let Y be a new random variable, the number of events in time s, where s is time measured in units of 15/60 = 1/4hour(for 15minutes). I changed the letter used for time, so as to avoid confusion with time as measured in hours. And I am using a new letter for the random variable, to prevent confusion with X
The random variable Y has Poisson distribution, with parameter θ=20/4=5, In other words, the parameter scales as one would expect. After all, for the Poisson, the parameter is the mean.
Then for example the probability that there will be k events in time interval of length s=1(quarter-hours) is given by
e-5(5 )k/ k!
(We did not need to invent the new Y to find this probability. In the original X, we could simply let t=1/4t=1/4.)
The probability of k events in time interval s=4s=4 (aka 1 hour) is also easily computed. Here θs=4θs=20, and you get probability
e-20t (20t)k/k!
now let's compute the probability
P(X>=9 ) = = 1-(P(X=8)+P(X=7)+....P(X=0) = 0.06810 ~ 6.8%
B) AVG RATE OF INCOMING CALLS = 30 PER HOUR
30 PER HOUR = 30/60 PER MINUTE = 1/2 PER MINUTE * 15 = 15/2 PER 15 MINUTES
P(X>=9) =. = 1-(P(X=8)+P(X=7)+....P(X=0) = 0.33803~33.803%
c) avg rate of incoming calls can be 20 per hours or 30 per hour as the observed rate of incoming calls in random 15minutes by manager was 6 per 15min or 24 per hour which lies in this interval , so yes incoming call rate can be in between but we cannot conclude it on the basis of one random 15 min observation , we need more observations to explore it further
d) i would suggest manager to hire a second operator as the rate of incoming can be more than 15calls per hour but we cannot conclude it because of only 1 random 15 minute observation , we need more observations like this to explore it further