1 The amounts a soft drink machine is designed to dispense for each drink are no
ID: 3220533 • Letter: 1
Question
1 The amounts a soft drink machine is designed to dispense for each drink are normally distributed, with a mean of 123 fuidounces and a standard deviation of 02 fluid ounce. A drink is randomly selected. (a) Find the probability that the drink is less than 12.1 fuid ounces. (b) Find the probability that the drink is between 11.8 and 12.1 fluid ounces (c) Find the probability that the drink is more than 12.8 fluid ounces. Can this be considered an unusual event? Explain your reasoning. (a) The probability that the drink is less than 12.1 fluid ounces is (Round to four decimal places as needed.) (b) The probability that the drink is between 11.8 and 12.1 fluid ounces is (Round to four decimal places as needed.) (c) The probability that the drink is more than 12.8 fluid ounces is (Round to four decimal places as needed.) ls a drink containing more than 12,8 fluid ounces an unusual event? Choose the correct answer below. O A. Nn the nrnhabilitv that a drink onntains more than 12 fluid nunnAs is inss than o o5 this event is not unusual. because Click to select your answer(s). Algebra Probability and statistics s SOLUTI The amount of a. into 12 ounce can soda follows a normal distribution with a mean of 12.03 ounces and a Then mark off units right and a left by adding the standard deviation 0.2 the to the right and subtracting 02 from the mean to get to points on the left side.Explanation / Answer
Mean ( u ) =12.3
Standard Deviation ( sd )=0.2
Normal Distribution = Z= X- u / sd ~ N(0,1)
a.
P(X < 12.1) = (12.1-12.3)/0.2
= -0.2/0.2= -1
= P ( Z <-1) From Standard Normal Table
= 0.1587
b.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 11.8) = (11.8-12.3)/0.2
= -0.5/0.2 = -2.5
= P ( Z <-2.5) From Standard Normal Table
= 0.00621
P(X < 12.1) = (12.1-12.3)/0.2
= -0.2/0.2 = -1
= P ( Z <-1) From Standard Normal Table
= 0.15866
P(11.8 < X < 12.1) = 0.15866-0.00621 = 0.1524
c.
P(X > 12.8) = (12.8-12.3)/0.2
= 0.5/0.2 = 2.5
= P ( Z >2.5) From Standard Normal Table
= 0.0062
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