Question

Stephanie gets an average of 11 calls during her 8 hour work day. What is the probability that Stephanie will get more than 4 calls in a 2 hour portion of her work day

Answer 1

0.145

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.

`P[X=k] = ((\lambda t)^k* e^(-\lambda t))/(k!) \ \ \ where; k = 0,1,2 ...`

Given that:

The average number of call every 8 hours = 11

Then, the parameter for the Poisson process
`\lambda =(11)/(8)`

where;

t = 2 hours

`\lambda t = (11)/(8)* 2`

`\lambda t = (11)/(4)`

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) ]

`=1 - [ P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) ]`

`= 1 - \bigg [ (((11)/(4))^0 * e^(-11/4))/(0!)+(((11)/(4))^1 * e^(-11/4))/(1!)+ (((11)/(4))^2 * e^(-11/4))/(2!)+ (((11)/(4))^3 * e^(-11/4))/(3!)+ (((11)/(4))^4 * e^(-11/4))/(4!) \bigg ]`

`= 1 - \begin {bmatrix} 1 + (11)/(4) + (\bigg ((11)/(4)\bigg)^2)/(2)+(\bigg ((11)/(4)\bigg)^3)/(6)+(\bigg ((11)/(4)\bigg)^4)/(24) \end {bmatrix} e^(-11/4)`

`= 1 - [ 1 + 2.75 + 3.78125 + 3.466145833 + 2.3829] \ e^(-11/4)`

= 1 - [13.38037]× 0.06393

= 1 - 0.855407

= 0.14459

≅ 0.145 ( to 3 decimal place)