Suppose that the number e-mail messages received by your instructor during an 8-hour period follows a Poisson distribution with a mean of four messages per hour. Find the probability that less than three messages are received by your instructor during the next 1-hour period.
To find the probability that less than three messages are received by your instructor during the next 1-hour period, we can use the Poisson distribution.
The Poisson distribution formula is:
P(x; λ) = (e^(-λ) * λ^x) / x!
Where P(x; λ) is the probability of getting x events in a certain time period, λ is the average number of events per time period, e is Euler's number (approximately 2.71828), and x is the number of events we are interested in.
In this case, λ is the mean number of messages received per hour, which is 4.
So, λ = 4.
To find the probability of receiving less than three messages, we need to calculate:
P(x < 3) = P(x = 0) + P(x = 1) + P(x = 2)
Let's calculate each term separately:
P(x = 0):
P(0; 4) = (e^(-4) * 4^0) / 0!
= (e^(-4) * 1) / 1
= e^(-4)
P(x = 1):
P(1; 4) = (e^(-4) * 4^1) / 1!
= (e^(-4) * 4) / 1
= 4e^(-4)
P(x = 2):
P(2; 4) = (e^(-4) * 4^2) / 2!
= (e^(-4) * 16) / 2
= 8e^(-4)
Now, we can calculate the probability of receiving less than three messages:
P(x < 3) = e^(-4) + 4e^(-4) + 8e^(-4)
= 13e^(-4)
Therefore, the probability that less than three messages will be received by your instructor during the next 1-hour period is approximately 0.0025 (or 0.25%).