Question

The amounts a soft drink machine is designed to dispense for each drink are normally distributed, with the mean of 11.0 fluid ounces and a standard deviation of 0.3 fluid ounce. A drink is randomly selected.

(a) Find the probability that the drink is than 11.7 fluid ounces.

(b) Find the probability that the drink is between 11.5 and 11.7 ounces

(c) Find the probability that the drink is more than 12.4 fluid ounces. Can this be considered an unusual event? Explain your reasoning.

Answer 1

P(X < A) = P(Z < (A - mean)/standard deviation)

Mean = 11 fluid ounces

Standard deviation = 0.3 fluid ounce

a) P(the drink is less than 11.7 fluid ounces) = P(X < 11.7)

= P(Z < (11.7 - 11)/0.3)

= P(Z < 2.33)

= 0.9901

P(the drink is greater than 11.7 fluid ounces) = 1 - 0.9901)

= 0.0099

b) P(the drink is between 11.5 and 11.7 ounces) = P(11.5 < X < 11.7)

= P(X < 11.7) - P(X < 11.5)

= 0.9901 - P(Z < (11.5 - 11)/0.3)

= 0.9901 - P(Z < 1.67)

= 0.9901 - 0.9525

= 0.0376

c) P(the drink is more than 12.4 fluid ounces) = P(X > 12.4)

= 1 - P(X < 12.4)

= 1 - P(Z < (12.4 - 11)/0.3)

= 1 - P(Z < 4.67)

= 1 - 1

= 0

The probability of this event is less than 0.05. (Z score corresponding to the event is not between -2 and 2). Therefore, the event can be considered unusual.