Answer to Question #247034 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

The probability of run out of burgers on single day is:

P(x > 400) = P(z > 400)

= P[z > (400 - 313) / 57]

= P[z > (87 / 57)]

= P(z > 1.53)

From normal standard table values, we find that:

P = 0.063 or 6.63%

This means there is 6.63% probability that fast food chain will run out of burgers on a single day when stick is 400.