Question

A coffee vending machine is set to fill a cup with an average of 6.5 ounces of soft drink. The amount of fill varies. Sometimes the machine overfills the cup until it overflows and sometimes it fills it under the legal minimum. Suppose the amount of fill, X, follows the Normal distribution with an average of 6.5 ounces and a standard deviation of 0.3 ounces.

What is the probability that a 7-ounce cup overflows, that is, find P(X>7)?

Given the information above, what is the probability that the average amount of the 8 cups of coffee exceeds 6.7 ounces, that is, find P(x¯>6.7)P(x¯>6.7).

Answer 1

Solution:

What is the probability that a 7-ounce cup overflows, that is, find P(X>7)?

P(X>7) = 1 – P(X<7)

Z = (X - µ)/σ

µ = 6.5

σ = 0.3

Z = (7 – 6.5)/0.3

Z = 1.666667

P(Z< 1.666667) = P(X<7) = 0.95221

(by using z-table or excel)

P(X>7) = 1 – P(X<7)

P(X>7) = 1 – 0.95221

P(X>7) = 0.04779

Required probability = 0.04779

Given the information above, what is the probability that the average amount of the 8 cups of coffee exceeds 6.7 ounces, that is, find P(x̄>6.7)

P(x̄>6.7) = 1 - P(x̄<6.7)

Z = (X - µ)/(σ/sqrt(n))

µ = 6.5

σ = 0.3

n = 8

Z = (6.7 - 6.5)/(0.3/sqrt(8))

Z = 1.885618

P(Z<1.885618) = P(x̄<6.7) = 0.970327

(by using z-table)

P(x̄>6.7) = 1 - P(x̄<6.7)

P(x̄>6.7) = 1 - 0.970327

P(x̄>6.7) = 0.029673

Required probability = 0.029673