Answer:
0.145
Step-by-step explanation:
Suppose λ = average number of an event occurring in an interval of a presented length; &
x = no. of events occurring in an interval length t
Thus, x has a Poisson distribution with parameter λt.
i.e.
[tex]P[X=k] = \dfrac{(\lambda t)^k\times e^{-\lambda t}}{k!} \ \ \ where; k = 0,1,2 ...[/tex]
Given that:
The average number of call every 8 hours = 11
Then, the parameter for the Poisson process [tex]\lambda =\dfrac{11}{8}[/tex]
where;
t = 2 hours
[tex]\lambda t = \dfrac{11}{8}\times 2[/tex]
[tex]\lambda t = \dfrac{11}{4}[/tex]
Also,
x = no. of calls occurring at an interval of length 2
The required probability is as follows:
P(x > 4) = 1 - [P ( x≤ 4) ]
[tex]=1 - [ P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) ][/tex]
[tex]= 1 - \bigg [ \dfrac{(\dfrac{11}{4})^0 \times e^{-11/4}}{0!}+\dfrac{(\dfrac{11}{4})^1 \times e^{-11/4}}{1!}+ \dfrac{(\dfrac{11}{4})^2 \times e^{-11/4}}{2!}+ \dfrac{(\dfrac{11}{4})^3 \times e^{-11/4}}{3!}+ \dfrac{(\dfrac{11}{4})^4 \times e^{-11/4}}{4!} \bigg ][/tex]
[tex]= 1 - \begin {bmatrix} 1 + \dfrac{11}{4} + \dfrac{\bigg (\dfrac{11}{4}\bigg)^2}{2}+\dfrac{\bigg (\dfrac{11}{4}\bigg)^3}{6}+\dfrac{\bigg (\dfrac{11}{4}\bigg)^4}{24} \end {bmatrix} e^{-11/4}[/tex]
[tex]= 1 - [ 1 + 2.75 + 3.78125 + 3.466145833 + 2.3829] \ e^{-11/4}[/tex]
= 1 - [13.38037]× 0.06393
= 1 - 0.855407
= 0.14459
≅ 0.145 ( to 3 decimal place)
