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Math  /  Data & Statistics

QuestionPatients arrive at an emergency department according to a poisson process with a mean of 6 per hour. What is the probability that more than 30 minutes is required for the third arrival?

Studdy Solution

STEP 1

What is this asking? If 6 patients arrive every hour, how likely is it that we have to wait more than half an hour for the third patient to arrive? Watch out! Don't forget to convert everything to the same units, like patients per half hour instead of per hour!

STEP 2

1. Define the random variable and distribution
2. Calculate the lambda parameter
3. Calculate the probability

STEP 3

Let XX be the **random variable** representing the *number of arrivals* in a 30-minute interval.
Since patients arrive according to a Poisson process, the number of arrivals in a given time interval follows a **Poisson distribution**.

STEP 4

A Poisson distribution is characterized by a single parameter, λ\lambda, which represents both the **mean** and the **variance** of the distribution.
The probability mass function (PMF) of a Poisson distribution is given by: P(X=k)=e−λ⋅λkk!P(X=k) = \frac{e^{-\lambda} \cdot \lambda^k}{k!} where kk is the number of arrivals, and ee is the base of the natural logarithm (e≈2.71828e \approx 2.71828).

STEP 5

We're given that the mean arrival rate is 6 patients per hour.
Since we're interested in the probability of waiting *more than* 30 minutes for the third arrival, we need to consider the arrival rate in a 30-minute interval.

STEP 6

Since there are two 30-minute intervals in an hour, the average number of arrivals in a 30-minute interval is: λ=6 patients2 30-min intervals=3 patients/30-min interval\lambda = \frac{6 \text{ patients}}{2 \text{ 30-min intervals}} = 3 \text{ patients/30-min interval} So, our λ\lambda is **3**!

STEP 7

We want to find the probability that more than 30 minutes is required for the third arrival.
This is equivalent to the probability that *at most* 2 patients arrive within 30 minutes.
Mathematically, we want to find P(X≤2)P(X \le 2).

STEP 8

Using the complement rule, we can express this as: P(X≤2)=1−P(X>2)=1−[P(X=3)+P(X=4)+P(X=5)+...]P(X \le 2) = 1 - P(X > 2) = 1 - [P(X=3) + P(X=4) + P(X=5) + ...] It's easier to calculate P(X≤2)P(X \le 2) directly: P(X≤2)=P(X=0)+P(X=1)+P(X=2)P(X \le 2) = P(X=0) + P(X=1) + P(X=2)

STEP 9

Using the Poisson PMF with λ=3\lambda = 3: P(X=0)=e−3⋅300!=e−3⋅11=e−3P(X=0) = \frac{e^{-3} \cdot 3^0}{0!} = \frac{e^{-3} \cdot 1}{1} = e^{-3} P(X=1)=e−3⋅311!=3e−3P(X=1) = \frac{e^{-3} \cdot 3^1}{1!} = 3e^{-3}P(X=2)=e−3⋅322!=9e−32P(X=2) = \frac{e^{-3} \cdot 3^2}{2!} = \frac{9e^{-3}}{2}

STEP 10

Adding these probabilities: P(X≤2)=e−3+3e−3+92e−3=(1+3+92)e−3=172e−3≈172⋅0.049787≈0.423P(X \le 2) = e^{-3} + 3e^{-3} + \frac{9}{2}e^{-3} = \left(1 + 3 + \frac{9}{2}\right)e^{-3} = \frac{17}{2}e^{-3} \approx \frac{17}{2} \cdot 0.049787 \approx 0.423

STEP 11

The probability that more than 30 minutes is required for the third arrival is approximately **0.423**.

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