Question

Please solve step by step. Let X be the amount released by a soft-drink dispensing machine into a 6-ounce cup. Assume that X is normally distributed with standard deviation 0.25 ounces and that the mean can be set by the vendor. At what quantity should the mean be set so that the probability of a release overflowing the cup is 0.004?

Answer 1

Let the mean be set by the vendor.

Then X ~ N(, 0.25)

Probability of a release overflowing the cup is 0.004

=> P(X > 6) = 0.004

=> P[Z > (6 - ) / 0.25] = 0.004

=> (6 - ) / 0.25 = 2.65

=> 6 - = 0.25 * 2.65

=> = 6 - 0.25 * 2.65 = 5.3375 ounces