Answer:
(a) The probability that a text message user receives or sends three messages per hour is 0.2180.
(b) The probability that a text message user receives or sends more than three messages per hour is 0.2667.
Step-by-step explanation:
Let <em>X</em> = number of text messages receive or send in an hour.
The random variable <em>X</em> follows a Poisson distribution with parameter <em>λ</em>.
It is provided that users receive or send 62.7 text messages in 24 hours.
Then the average number of text messages received or sent in an hour is:
.
The probability of a random variable can be computed using the formula:

(a)
Compute the probability that a text message user receives or sends three messages per hour as follows:

Thus, the probability that a text message user receives or sends three messages per hour is 0.2180.
(b)
Compute the probability that a text message user receives or sends more than three messages per hour as follows:
P (X > 3) = 1 - P (X ≤ 3)
= 1 - P (X = 0) - P (X = 1) - P (X = 2) - P (X = 3)

Thus, the probability that a text message user receives or sends more than three messages per hour is 0.2667.