Answer to Question #191361 in Statistics and Probability for NANGULA ABIA

Question

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? (3) 

Answer 1

(a). The rate of events is constant, since there are two phone calls per minute, each occurring independently. As a result, we classify it as a Poisson distribution.

(b). The average number of phone calls per min is 2

Hence, the average number of phone calls per hour will be:

$2\times 60=120$

Therefore, expected number of phone calls per hour is 120 phone calls.

(c). Let X be the number of phone calls received in 2 minutes

$λ=2×2=4$

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

$P\left(0<X<4\right)=e^{-\lambda }\left(\frac{4^1}{1!}+\frac{4^2}{2!}+\frac{4^3}{3!}\right)$

$P\left(0<X<4\right)=e^{-4\:}\times 22.66667\:\:$

$P\left(0<X<4\right)=0.4152$

(d). In a given period of 3 minutes

$\lambda =2\times 3=6$

$X=Number\:of\:phone\:calls\:received\:in\:3\:minutes$

$P\left(X>1\right)=1-P\left(X\le 1\right)$

$P\left(X>1\right)=1-\left[P\left(X=0\right)+P\left(X=1\right)\right]$

$P\left(X>1\right)=1-\left[e^{-6}\left(\frac{6^0}{0!}+\frac{6^1}{1!}\right)\right]$

$P\left(X>1\right)=1-0.01736$

$P\left(X>1\right)=0.9827$

(e). Given that: $\lambda =2$

$\:P\left(X=0\right)=\frac{e^{-\lambda \:}\lambda \:^0}{0!}$

$P\left(X=0\right)=e^{-2}$

$\:P\left(X=0\right)=0.1353$