Per capita coffee consumption last year was reported to be 4

Per capita coffee consumption last year was reported to be 4.2k or 9.24 pounds. Assume the per capita consumption in 2014 is approximately distributed as a normal, random variable, with a mean of 9.24 lbs and standard deviation of 3 lbs.

a. What is the probability that someone consumed more than 10 lbs of coffee in 2014?

b. What is the probability that someone consumed between 3-5 lbs if coffee in 2014?

c. What is the probability that someone consumed less than 5 lbs of coffee in 2014?

d. 99% of the people consumed less than how many pounds of coffee?

Solution

                  
Normal Distribution
Mean ( u ) =9.24
Standard Deviation ( sd )=3
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
P(X > 10) = (10-9.24)/3
= 0.76/3 = 0.2533
= P ( Z >0.253) From Standard Normal Table
= 0.4                  
b)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 3) = (3-9.24)/3
= -6.24/3 = -2.08
= P ( Z <-2.08) From Standard Normal Table
= 0.01876
P(X < 5) = (5-9.24)/3
= -4.24/3 = -1.4133
= P ( Z <-1.4133) From Standard Normal Table
= 0.07878
P(3 < X < 5) = 0.07878-0.01876 = 0.06                  
c)
P(X < 5) = (5-9.24)/3
= -4.24/3= -1.4133
= P ( Z <-1.4133) From Standard Normal Table
= 0.0788                  
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 - 9.24/3 ) = 0.99
That is, ( x - 9.24/3 ) = 2.33
--> x = 2.33 * 3 + 9.24 = 16.218