Question 1204803: During my office hours (12pm to 6pm), an average of 2 students per hour comes in for help. Assuming that the probability of a student coming in is uniform throughout my office hours, what are the odds that on a particular day I would have 6 students come in for help? What is the mean number of students that will come see me on a particular day? Variance? Standard deviation?
Answer by ElectricPavlov(122)   (Show Source):
You can put this solution on YOUR website! **1. Define the Random Variable**
* Let X be the number of students who come in for help during your office hours.
**2. Determine the Distribution**
* Since the students arrive at a constant average rate within a given interval (your office hours) and the arrivals are independent, X follows a **Poisson distribution**.
**3. Calculate the Probability of 6 Students**
* The Poisson distribution has a single parameter, λ, which represents the average number of occurrences within the given interval.
* In this case, λ = 2 students/hour * 6 hours = 12 students/day
* The probability mass function (PMF) of a Poisson distribution is:
P(X = k) = (λ^k * e^(-λ)) / k!
where:
* k is the number of occurrences (in this case, the number of students)
* λ is the average number of occurrences
* e is the base of the natural logarithm (approximately 2.71828)
* To find the probability of 6 students:
P(X = 6) = (12^6 * e^(-12)) / 6!
P(X = 6) ≈ 0.0113
Therefore, the odds of having 6 students come in for help on a particular day are approximately 0.0113.
**4. Calculate Mean, Variance, and Standard Deviation**
* For a Poisson distribution:
* Mean (μ) = λ = 12 students/day
* Variance (σ²) = λ = 12 students²/day
* Standard Deviation (σ) = √λ = √12 ≈ 3.46 students/day
**In summary:**
* The probability of having 6 students come in for help is approximately 0.0113.
* The mean number of students is 12 per day.
* The variance is 12 students²/day.
* The standard deviation is approximately 3.46 students/day.
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