Answer to Question #176952 in Statistics and Probability for Sean

Question

Question III [25]

Suppose that a Covid-19 testing centre receives two phones calls, on average, per minute 

concerning test results. 

Required:

a) What are the conditions for this experiment to be considered a Poisson experiment? 

Motivate. (4)

b) What is the expected number of phone calls regarding test results per hour? (2)

c) What is the probability that more zero but less than four phone calls regarding test results 

are received in any given period of two minutes? (9) 

d) What is the probability that more than one phone calls regarding test results are received 

in any given period of three minutes? (7)

e) What is the probability that no phone calls regarding test results are received in any given 

period of one minute?

Answer 1

a) Poisson experiment expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event. The Poisson experiment can also be used for the number of events in other specified intervals such as distance, area or volume.

It can be applied to systems with a large number of possible events, each of which is rare. The number of such events that occur during a fixed time interval is, under the right circumstances, a random number with a Poisson distribution.

b) $\lambda=2/60=1/30$

c)

$P(X=k)=\frac{\lambda^ke^{-\lambda}}{k!}$

for two minutes: $\lambda=1$

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

$P(X=1)=e^{-1},\ P(X=2)=e^{-1}/2,\ P(X=3)=e^{-1}/6$

$P(0<X<4)=e^{-1}(1+1/2+1/6)=0.61$

d) for three minutes: $\lambda=2/3$

$P(X>1)=1-P(X=0)-P(X=1)$

$P(X=0)=e^{-2/3},\ P(X=1)=2e^{-2/3}/3$

$P(X>1)=1-e^{-2/3}(1+2/3)=0.14$

e)

$P(X=0)=e^{-2}=0.135$