A history professor finds that when he schedules an office hour at the 1030 am time slot an average of 3 students arrive. Use the poisson distribution to find the probability that in a randomly selected office hour in the 1030 am time slot exactly 5 students will arrive.
Poisson distribution with a mean of m:
P(x) = e^(-m) m^x / x!
P(5) = e^(-3) 3^5 / 5!
I'll let you finish the computation.
Thank you.. It is 0.1008..
Correct!
To solve this problem using the Poisson distribution, we need to know the average number of students that arrive during the 1030 am office hour. In this case, the average is given as 3 students.
The Poisson distribution formula is: P(x; λ) = (e^(-λ) * λ^x) / x!
Where:
- P(x; λ) represents the probability of x events occurring given an average rate of λ events.
- e is the base of the natural logarithm (approximately equal to 2.71828).
- x is the number of events we're interested in (in this case, 5).
- λ is the average rate of events (in this case, 3).
Now, let's calculate the probability that exactly 5 students will arrive during a randomly selected office hour at the 1030 am time slot:
P(5; 3) = (e^(-3) * 3^5) / 5!
= (e^(-3) * 3^5) / (5 * 4 * 3 * 2 * 1)
= (e^(-3) * 243) / 120
≈ 0.10082 (rounded to 5 decimal places)
Therefore, the probability that exactly 5 students will arrive during a randomly selected office hour in the 1030 am time slot is approximately 0.10082.