612 In 2011 the per capita consumption of coffee in Swe den

6.12 In 2011, the per capita consumption of coffee in Swe den was reported to be 7.27 kg, or 15.994 pounds. Data extracted from www.ico.org.) Assume that the per capita consumption of coffee in Sweden in 2011 is approximately normally distributed with a mean of 15.994 pounds and a standard deviation of 5 pounds. a. What is the probability that someone in Sweden con- sumed more than 10 pounds of coffee in 2011? b. What is the probability that someone in Sweden con- sumed between 3 and 5 pounds of coffee in 20 c. What is the probability that someone in Sweden con- sumed less than 5 pounds of coffee in 2011? d. 99% of the people in Sweden consumed less than how many pounds of coffee in 2011?

Solution

Mean ( u ) =15.994
Standard Deviation ( sd )=5
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
P(X > 10) = (10-15.994)/5
= -5.994/5 = -1.1988
= P ( Z >-1.199) From Standard Normal Table
= 0.8847
b)                  
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 3) = (3-15.994)/5
= -12.994/5 = -2.5988
= P ( Z <-2.5988) From Standard Normal Table
= 0.00468
P(X < 5) = (5-15.994)/5
= -10.994/5 = -2.1988
= P ( Z <-2.1988) From Standard Normal Table
= 0.01395
P(3 < X < 5) = 0.01395-0.00468 = 0.0093                  
c)
P(X < 5) = (5-15.994)/5
= -10.994/5= -2.1988
= P ( Z <-2.1988) From Standard Normal Table
= 0.0139                  
d)
P ( Z < x ) = 0.99
Value of z to the cumulative probability of 0.99 from normal table is 2.326
P( x-u/s.d < x - 15.994/5 ) = 0.99
That is, ( x - 15.994/5 ) = 2.33
--> x = 2.33 * 5 + 15.994 = 27.624