In 2012, the per capita consumption of soft drinks in the United States was repo
ID: 3158741 • Letter: I
Question
In 2012, the per capita consumption of soft drinks in the United States was reported to be 44 gallons. Assume that the per capita consumption of soft drinks in the USA is approximately normally distributed with a mean of 44 gallons and a standard deviation of 14 gallons. What is the probability that someone in the United States consumed more than 60 gallons of soft drinks in 2012? Refer back to the soft drink consumption data. What is the probability that someone in the USA consumed between 15 and 30 gallons of soft drinks in 2012? Refer back to the soft drink consumption data. What is the probability that someone in the USA consumed less than 18 gallons of soft drinks in 2012? Refer back to the soft drink consumption data. You must quantify the soft drink consumption of an individual who was among the top 2% of soft-drink consumers in the USA in 2012. How many gallons of soft drinks did such an individual consume?
Explanation / Answer
Normal Distribution
Mean ( u ) =44
Standard Deviation ( sd )=14
Normal Distribution = Z= X- u / sd ~ N(0,1)
a.
P(X > 60) = (60-44)/14
= 16/14 = 1.1429
= P ( Z >1.143) From Standard Normal Table
= 0.1265
b.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 15) = (15-44)/14
= -29/14 = -2.0714
= P ( Z <-2.0714) From Standard Normal Table
= 0.01916
P(X < 30) = (30-44)/14
= -14/14 = -1
= P ( Z <-1) From Standard Normal Table
= 0.15866
P(15 < X < 30) = 0.15866-0.01916 = 0.1395
c.
P(X < 18) = (18-44)/14
= -26/14= -1.8571
= P ( Z <-1.8571) From Standard Normal Table
= 0.0316
d.
P ( Z > x ) = 0.02
Value of z to the cumulative probability of 0.02 from normal table is 2.05
P( x-u/ (s.d) > x - 44/14) = 0.02
That is, ( x - 44/14) = 2.05
--> x = 2.05 * 14+44 = 72.756
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