Joe owns a pizza place and is trying to figure out how much supply to buy. He st

Question

Joe owns a pizza place and is trying to figure out how much supply to buy. He starts keeping track of the number of pizzas sold each week and starts to notice that it looks like a Poisson distribution. a. Joe finds after taking some data that the average number of pizzas per week (N) is 900, but to be safe, he wants to buy enough supplies for one standard deviation (sigma) extra pizzas. How many more should he order? b. As described in Activity F, what is the relative uncertainty for this number of pizzas? c. Joe changes to a new supplier who only lets him order once even 4 weeks. How many pizzas should he expect (N) to sell in 4 weeks? d. What is the new relative uncertainty for this new number of pizzas?

Explanation / Answer

(a) Suppose X = number of Pizzas to be ordered per week.

Then, X follows Poission Distribution with mean as N = 900

Now, 1 standard deviation = square root of variance of X

But, Var(X) = E(X) = N = 900

So, 1 Standard Deviation = sqrt(900) = 30

Hence, total pizzas to keep in supply = N + 30 = 900 + 30 = 930

(b) Absolute Uncertainty as per definition is:

A = (930-E(X)) = 30

So, Relative Uncertainty as per definition = A/E(X) = 30/900 = 1/30

(c) Now, for 4 weeks, each of the weeks order are independent of each other. So,

if Y = number of pizzas in 4 weeks, then

Y follows Poission with mean as n = 4*N = 3600

So, he should expect to sell 3600 pizzas in 4 weeks.

To be on safe side though, he can keep 3600 + sqrt(3600) = 3660 Pizzas

(d) Relative Uncertainty here is:

60/3660 = 1/61

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