Question

1. Suppose you get on average 60 text messages per day between 8am and 8pm.  Compute the probability that in a particular hour between 8am and 8pm, you will get exactly 10 text messages. Give your answer to five decimal places with a leading zero.

Note: ON AVERAGE you get 60 text messages between 8am and 8pm...on average, how many text messages would you expect in one hour??

2.

Suppose you get on average 60 text messages per day between 8am and 8pm.  Compute the probability that in a particular hour between 8am and 8pm, you will get no more than 4 text messages. Give your answer to five decimal places with a leading zero.

Note: ON AVERAGE you get 60 text messages between 8am and 8pm...on average, how many text messages would you expect in one hour??

Answer 1

Q1) As the average number of messages in a 12 hour period is 60, therefore the average number of messages per hour is computed here as:
= 60/12 = 5

Therefore the probability that in a particular hour between 8am and 8pm, you will get exactly 10 text messages is computed using poisson probability function as:

$= \frac{10^5}{5!}e^{-10} = 0.01813$

Therefore 0.01813 is the required probability here.

Q2) As the average number of messages in a 12 hour period is 60, therefore the average number of messages per hour is computed here as:
= 60/12 = 5

Therefore the probability that in a particular hour between 8am and 8pm, you will get no more than 4 text messages is computed using poisson probability function as:

$P(X \leq 4) = P(X = 0) + P(X = 1) + ... + P( X = 4)$

$P(X \leq 4) = e^{-5} + 5e^{-5} + \frac{5^2}{2}e^{-5} + \frac{5^3}{3!}e^{-5} + \frac{5^4}{4!}e^{-5} = 0.44049$

Therefore 0.44049 is the required probability here.