Question #191081

A soft-drink machine is regulated so that it discharges an average of 200 mm/cup. Of the amount of drink is normally distributed with a standard deviation equal to 15 mm, how many cups will probably overflow if 230 mm cups are used for the next 1000 drinks?

Expert's answer

"z= \\frac{x-\\mu}{\\sigma} = \\frac{230-200}{15} = 2"

The probability of overflow if 230 cups is used

P(X>230) = P(Z>2)

= 1- P(Z<2)

= 1- 0.9772= 0.0228

by using binomial property we get

E(X)= "\\eta .p"

=1000 "\\times 0.0228" = 22.8

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