Answer to Question #247032 in Statistics and Probability for jonats

Question

A certain fast-food chain sells cheeseburger. On a typical weekday, the demand for these burgers can be approximated by a normal distribution with the mean of 313 burgers and a standard deviation of 57 burgers. If the fast-food chain has a stock of 400 burgers, what is the probability that it will run out of burgers on that day

Answer 1

Let the random variable Y represent the cheeseburgers sold. Then, $Y\sim N(\mu,\sigma^2)$ where,

$\mu=313$

$\sigma^2=57^2$

For the stock of 400 burgers to run out, the orders for these burgers must exceed this amount, therefore, we determine the probability that the burgers exceed 400 as expressed below.

$p(Y\gt400)$ . This can be found as follows.

$p((Y-\mu)/\sigma\gt(400-\mu)/\sigma)=p(Z\gt(400-313)/57)=p(Z\gt1.5263)$

This can also be written as,

$1-p(Z\lt1.5263)=1-0.9370=0.0630$

Thus, the probability that a stock of 400 burgers run out is 0.0630.