The scenario given by the problem is as follows: Suppose a restaurant has 4 possible entrées, chicken, beef, pork, and tofu, and the manager believes that orders for each entrée arrive independently in a Poisson manner at rates 15, 20, 10, and 5 per hour, respectively.
Q1: If there are twenty pork entrées ordered in the next hour, consider the number of chicken entrées ordered in that same hour. Find the manager’s distribution, expected value, and variance.
My ans:
Since it is given that orders for each entree arrives independently, the number of pork entrees ordered has no bearing on the number of chicken entrees ordered.
Hence, we can use a poisson distribution with parameter 15 to model this; i.e. Pois(15)
E[X] = 15
Var[x] = 15
Q2: What is the manager’s approximate probability that the next one hundred and fifty orders will include at least fifty-five but less than seventy beef entrée orders? Please answer with a simple expression and a number rather than a sum over many values.
My ans:
Since the orders for beef entrees arrive independently in a Poisson manner, we can calculate the probability using the cumulative distribution function for Poisson distribution.
Inputting lambda = 20, and the summation from i = 55,...,69, we arrive at the probability 9.05 x 10^-11
Would really appreciate it if someone can verify whether my reasoning and answers are correct. Thank you for your help in advance.
Your answer to the first question is correct, but the second answer is incorrect. The reason is because the stopping criterion for the counting process is a function of the number of events observed, and not a function of time. In other words, the value you calculated is the probability that, in the next hour, there will be between $55$ and $70$ orders for beef. But the question is actually asking, among the next 150 orders, what is the probability that between $55$ and $70$ of these will be for beef.
Because of this, the conditional distribution of the random number of beef orders $B$, given a fixed number of observed total orders $n = 150$, is binomial with parameter $$p = \frac{20}{15+20+10+5} = 0.4.$$ This value represents the probability that any given order among the $150$ that were received was a beef entree.
Then, the desired probability is exactly $$\Pr[55 \le B \le 70] = \sum_{b=55}^{70} \binom{150}{b} (0.4)^b (0.6)^{150-b}.$$
Since the question is asking for an "approximate" probability, this suggests that you may use a normal approximation with continuity correction; i.e., consider $$X \sim \operatorname{Normal}(\mu = np, \sigma^2 = np(1-p)),$$ where $n = 150$ and $p = 0.4$ as we found for $B$. What we are doing here is approximating the exact binomial distribution with a normal distribution with the same mean and variance. Then $$\Pr[55 \le B \le 70] \approx \Pr[54.5 \le X \le 70.5]$$ where we have applied continuity correction to capture the full probability mass at the endpoints. Then we obtain by standardizing $X$ $$\Pr[54.5 \le X \le 70.5] = \Pr\left[\frac{54.5 - 60}{6} \le \frac{X - \mu}{\sigma} \le \frac{70.5 - 60}{6}\right] \approx \Pr[-0.91667 \le Z \le 1.75].$$
I leave it to you to compute the exact and approximate probabilities above and compare them. How well does the approximation work? What would be your approximation if you did not use continuity correction? Which is better?