Answer to Question #327580 in Statistics and Probability for Khaita

Question

Question 3 [25]

Suppose that a mobile telecommunication company’s helpline receives five calls, on average, per

minute.

Required:

a) Discuss the difference between the Binomial probability distribution and the Poisson

probability distribution.

b) How many calls does the company expect to receive in a period of 30 minutes?

c) What is the probability that the company will receive at most four calls in a period of 4

minutes?

d) What is the probability that the company will receive at least three calls in a period of 5

minutes?

e) What is the probability that the company will receive between six and nine calls in a period

of 2 minutes?

Answer 1

$a:Binomial\,\,takes\,\,finite\,\,number\,\,of\,\,values, Poisson\,\,infinite\\b:\\En\left( t \right) =\lambda t=5\cdot 30=150\\c:\\P\left( n\left( 4 \right) \leqslant 4 \right) =\sum_{i=0}^4{P\left( n\left( 4 \right) =i \right)}=\sum_{i=0}^4{\frac{20^ie^{-20}}{i!}}=\\=e^{-20}\left( 1+\frac{20}{1}+\frac{20^2}{2}+\frac{20^3}{6}+\frac{20^4}{24} \right) =1.69447\times 10^{-5}\\d:\\P\left( n\left( 5 \right) \geqslant 3 \right) =1-P\left( n\left( 5 \right) <3 \right) =1-\sum_{i=0}^2{\frac{25^ie^{-25}}{i!}}=\\=1-e^{-25}\left( 1+\frac{25}{1}+\frac{25^2}{2} \right) =1-4.70107\times 10^{-9}$