Z score with sample size problem

Question

A pizza restaurant’s order processing time is normally distributed. A random sample has mean 30 minutes and standard deviation five minutes.If the pizza restaurant promises their customers “35-minute delivery or free” and average order cost is $30, estimate the total cost of free food the restaurant has to give away with the current process variation

Answer 1

To answer this question without being provided additional information, we have to assume that delivery time is independent of order cost. It's perfectly possible that larger, more expensive orders are precisely the ones that take longer to deliver, but we need to assume that the (random) amount of time it takes the pizza shop to prepare and deliver an order doesn't depend on the dollar amount of the order. This will allow us to go from the facts we're given to the conclusion that each order that takes more than 35 minutes to deliver costs the company an average of $30. This is almost certainly what the problem wanted you to assume, but without assuming price is independent of delivery time, it's possible this isn't true. For example, if larger orders are exactly the ones that take more than 35 minutes to deliver, then each order that takes more than 35 minutes will cost the pizza shop more than $30.

Proceeding with our assumption above, we can find the average cost per order that the pizza shop incurs by running its "35 minutes or free" program. This average cost per order from the "35 or free" program is

P[the order takes more than 35 minutes] * (average cost of an order that takes more than 35 minutes)

where P denotes probability. Using the fact in bold type above, this means the average cost per order from the program is

= P[the order takes more than 35 minutes] * 30.

So all that's left to do is compute the probability that an order takes more than 35 minutes to deliver. Letting X denote the (random) amount of time it takes to deliver an order, we know that X has Normal(30,5) distribution: its mean is 30 minutes and its standard deviation is 5 minutes. Now recall that when we "standardize" X by subtracting its mean (30) and then dividing by its standard deviation (5), we get a Normal(0,1) random variable that I'll call Z. Therefore, to compute P[X > 35], we can just subtract 30 from both sides of the inequality and then divide both sides by 5 to find that

P[X > 35] = P[(X - 30)/5 > (35 - 30)/5] = P[Z > 1] = 0.16

I calculated P[Z > 1] using the so-called "68 95 99.7" rule, but in case you're not familiar with this, you can also use the fact that

P[Z > 1] = 1 - P[Z < 1] = 1 - 0.84 = 0.16.

Putting all of this back into the above, we have

Average cost per order from running "35 or free" program

= P[the order takes more than 35 minutes] * 30

= .16 * 30

=4.8,

so the company loses about $4.80 per order by choosing to run its "35 minute delivery or free" program.