Question

The amounts a soft drink machine is designed to dispense for each drink are normally? distributed, with a mean of 12.3fluid ounces and a standard deviation of 0.3 fluid ounce. A drink is randomly selected. ?(a) Find the probability that the drink is less than 12 fluid ounces. ?(b) Find the probability that the drink is between 11.9 and 12 fluid ounces. ?(c) Find the probability that the drink is more than 12.9 fluid ounces. Can this be considered an unusual? event? Explain your reasoning.

Answer 1

(a) P(X 12) Normal Distribution, μ-12.3, σ = 0.3, P(X < 12)-Area to the left of 12 we convert this to standard normal using z = 12-(12.3) z = --100 0.1587 P(X < 12) = P(Z <-1.00) = 0.1587 (from z-table) P(Z<-1): in a z-table having area to the left of z, locate -1 in the left most column. Move across the row to the right under column 0.00 and get value 0.1587 P(X < 12)= 0.1587 15.87% b) P(11.9< X < 12)

Normal Distribution-1-12.3, ơ-0.3, we convert this to standard normal using 11.9 (12.3) 0.3 1.333333 -1.33 12 - (12.3) 100 22- P(-1.33< Z<-1.00)Area in between -1.33 and-1.00 0.1587 0.09180.0669 0.1587 0.0918 1.00 P(11.9 < X < 12) = P(-1.33 < Z <-1.00) = P(Z <-1.00)-P(Z <-1.33) 0.1587-0.0918-0.0669 P(Z<1): in a z-table having area to the left of z, locate -1 in the left most column. Move across the row to the right under column 0.00 and get value 0.1587 P(Z<-1.33): in a z-table having area to the left of z, locate -1.3 in the left most column. Move across the row to the right under column 0.03 and get value 0.0918 P(11.9 < X < 12)-0.0669-6.69%

(c) P(X 12.9) Normal Distribution, 1-123, ơ-0.3, we convert this to standard normal using 12.9 (12.3)_200 0.3 P(Z > 2.00) Area to the right of 2.00 0.0228 2.00 P(Xx> 12.9)- P(Z > 2.00) 1- P(Z<2.00) 1-0.9772 0.0228 PZ <2): in a z-table having area to the left of z, locate 2 in the left most column. Move across the row to the right under column 0.00 and get value 0.9772 > 12.9-0.0228-2.28% This is an unusual event because its probability is less than 0.05