Question

I run a pizza restaurant that is only open on Thursday, Friday and Saturday evenings from 5:30 to 9:30. Over that time, orders for pizza come into the restaurant randomly with a distribution that is approximately normal with mean 20 and standard deviation of 5 pizzas per hour. A pizza requires 1 kg of dough, 250 g of cheese and a mix of other ingredients depending on the order.On Thursday Morning I have 100 kg of dough on hand . How likely is it that I will run out of dough before the end of day Thursday?

Answer 1

given that the order comes approximately in form of normal distribution with mean 20 and standard deviation 5 per hour, so for 4 hours the mean would be 80 and the standard deviation would be 10 this is shown below

Let X be the number of orders from 5.30 to 6.30

Let Y be the number of orders from 6.30 to 7.30

Let Z be the number of orders from 7.30 to 8.30

Let W be the number of orders from 8.30 to 9.30

As all X,Y,Z,W follows normal distribution with mean 20 and variance 25

X+Y+Z+W will follow normal with mean 80 and variance 100.

therefore the standard deviation would be 10

Let T=X+Y+Z+W

So T follows normal with 80 mean and 10 standard deviation

To find

given 100 kg of dough and 1 kg of it is required for 1 pizza then atmost 100 pizza can be made out of it, and if orders exceed 100 pizza then dough will be over that is he will run out of dough.

T represents the number of orders in a given day as specificed above

So if T>100 the dough will be over.

To find

$P(T>100)=P((T-80)/10>(100-80)/10)$

$=P(Z>2)=0.028$

where Z is the standard normal variate, so using the table for standard normal variate the probability is computed.

So the probability that the owner will run out of dough is 0.028