Question

Solution Procedure: In solving these problems and subsequent ones, use the following procedure: (1) Define the events of interest. (2) Write down the information given. (3) Write down the question. (4) Use the appropriate rule to answer the question.

An office worker receives an average of 22.5 email messages per day. If his working day lasts seven and a half hours, what is the probability that:

(a) He receives no emails in an hour?

(b) He receives one email in an hour?

(c) He receives two or fewer emails in an hour?

(d) He receives more than four emails in an hour?

λ=3 =.22.5/7.5

Answer 1

lambda for per hour = 22.5/7.5 = 3

a)
Here, λ = 3 and x = 0
As per Poisson's distribution formula P(X = x) = λ^x * e^(-λ)/x!

We need to calculate P(X = 0)
P(X = 0) = 3^0 * e^-3/0!
P(X = 0) = 0.0498
Ans: 0.0498

b)
We need to calculate P(X = 1)
P(X = 1) = 3^1 * e^-3/1!
P(X = 1) = 0.1494
Ans: 0.1494

c)
We need to calculate P(X <= 2).
P(X <= 2) = (3^0 * e^-3/0!) + (3^1 * e^-3/1!) + (3^2 * e^-3/2!)
P(X <= 2) = 0.0498 + 0.1494 + 0.224
P(X <= 2) = 0.4232

d)
We need to calculate P(X > 4) = 1 - P(X <= 4).
P(X > 4) = 1 - (3^0 * e^-3/0!) + (3^1 * e^-3/1!) + (3^2 * e^-3/2!) + (3^3 * e^-3/3!) + (3^4 * e^-3/4!)
P(X > 4) = 1 - (0.0498 + 0.1494 + 0.224 + 0.224 + 0.168)
P(X > 4) = 1 - 0.8152 = 0.1848