A professor receives, on average, 8 emails from students on the day before an online quiz. If the distribution of the calls is Poisson, what is the chance that he received at least 6 and at most 10 emails on such a day?
Poisson distribution (m = mean):
P(x) = e^(-m) m^x / x!
Values:
x = 6,7,8,9,10
m = 8
Substitute and calculate for each x. Add all those calculations together for a total probability.
I hope this will help.
To solve this problem, we can use the Poisson distribution formula. The Poisson distribution is a probability distribution that models the number of events occurring in a fixed interval of time or space when these events occur with a known average rate and independently of the time since the last event.
The probability mass function (PMF) for the Poisson distribution is given by:
P(X = k) = (e^(-λ) * λ^k) / k!
Where:
- P(X = k) is the probability of exactly k events occurring,
- e is the mathematical constant (~2.71828),
- λ is the average rate of events (in this case, the average number of emails received by the professor on the day before an online quiz), and
- k is the number of events we are interested in.
In this case, λ is given as 8 (the average number of emails received).
Now, we need to find the probability that the professor received at least 6 and at most 10 emails. To do this, we calculate the cumulative probability from 6 to 10.
P(6 ≤ X ≤ 10) = P(X ≤ 10) - P(X ≤ 5)
To find P(X ≤ 10), we sum the probabilities for k = 0 to 10:
P(X ≤ 10) = P(X = 0) + P(X = 1) + P(X = 2) + ... + P(X = 10)
To find P(X ≤ 5), we sum the probabilities for k = 0 to 5:
P(X ≤ 5) = P(X = 0) + P(X = 1) + P(X = 2) + ... + P(X = 5)
To calculate each individual probability, we can use the Poisson distribution formula mentioned earlier.
Alternatively, we can use a computer software or calculator that has the capability to calculate the cumulative Poisson probability for a range of values. This would save us from manually calculating and summing each individual probability.
Once we have the values for P(X ≤ 10) and P(X ≤ 5), we can subtract P(X ≤ 5) from P(X ≤ 10) to find P(6 ≤ X ≤ 10).