In 2011, the per capita consumption of chocolate in Switzerland was reported to

Question

In 2011, the per capita consumption of chocolate in Switzerland was reported to be 20.63 pounds. Assume that the per capita consumption of chocolate in Switzerland is normally distributed with a mean of 20.63 pounds and a standard deviation of 6 pounds. a. What is the probability that someone in Switzerland consumed more than 12 pounds of chocolate in 2011? b. What is the probability that someone in Switzerland consumed between 5 and 17 pounds of chocolate in 2011? c. What is the proability that someone in Switzerland consumed less than 17 pounds of chocolate in 2011? d. 99% of the people in Switzerland consumed less than how many pounds of chocolate? e. 95% of the people in Sweden consumed more than how many pounds of chocolate? f. The probability is 50% that a person in Sweden consumed between these two amounts of coffee (symmetrically distributed around the mean)? In 2011, the per capita consumption of chocolate in Switzerland was reported to be 20.63 pounds. Assume that the per capita consumption of chocolate in Switzerland is normally distributed with a mean of 20.63 pounds and a standard deviation of 6 pounds. a. What is the probability that someone in Switzerland consumed more than 12 pounds of chocolate in 2011? b. What is the probability that someone in Switzerland consumed between 5 and 17 pounds of chocolate in 2011? c. What is the proability that someone in Switzerland consumed less than 17 pounds of chocolate in 2011? d. 99% of the people in Switzerland consumed less than how many pounds of chocolate? e. 95% of the people in Sweden consumed more than how many pounds of chocolate? f. The probability is 50% that a person in Sweden consumed between these two amounts of coffee (symmetrically distributed around the mean)?

Explanation / Answer

Normal Distribution
Mean ( u ) =20.63
Standard Deviation ( sd )=6
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
P(X > 12) = (12-20.63)/6
= -8.63/6 = -1.4383
= P ( Z >-1.438) From Standard Normal Table
= 0.9248
b)              
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 5) = (5-20.63)/6
= -15.63/6 = -2.605
= P ( Z <-2.605) From Standard Normal Table
= 0.00459
P(X < 17) = (17-20.63)/6
= -3.63/6 = -0.605
= P ( Z <-0.605) From Standard Normal Table
= 0.27259
P(5 < X < 17) = 0.27259-0.00459 = 0.268                  
c)
P(X < 17) = (17-20.63)/6
= -3.63/6= -0.605
= P ( Z <-0.605) From Standard Normal Table
= 0.2726                  
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 - 20.63/6 ) = 0.99
That is, ( x - 20.63/6 ) = 2.33
--> x = 2.33 * 6 + 20.63 = 34.586                  
e)
P ( Z > x ) = 0.95
Value of z to the cumulative probability of 0.95 from normal table is -1.64
P( x-u/ (s.d) > x - 20.63/6) = 0.95
That is, ( x - 20.63/6) = -1.64
--> x = -1.64 * 6+20.63 = 10.76                  

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