Question

The answers are in red. Please explain all parts, I really want to understand how to tell which quartile to use like part e!

A study discovered that Americans consumed an average of 13.5 pounds of chocolate per year. Assume that the annual chocolate consumption follows the normal distribution with a standard deviation of 3.9 pounds. Complete parts a through e below. a. What is the probability that an American will consume less than 9 pounds of chocolate next year? 0.1251 (Round to four decimal places as needed.) b. What is the probability that an American will consume more than 11 pounds of chocolate next year? 0.7389 (Round to four decimal places as needed.) c. What is the probability that an American will consume between 10 and 14 pounds of chocolate next year? 0.3676 (Round to four decimal places as needed.) d. What is the probability that an American will consume exactly 12 pounds of chocolate next year? 0 (Round to four decimal places as needed.) e. What is the annual consumption of chocolate that represents the 80th percentile? The 80th percentile is represented by an annual consumption of 16.8 pounds of chocolate (Type an integer or decimal rounded to one decimal place as needed.)

Answer 1

a)
P( X<9) = ?
  
I know that, z = (X-mean)/(sd)  
z1 = (9-13.5)/3.9) = -1.1538
hence,  
P( X<9)   = P(Z<-1.1538) = NORMSDIST(-1.1538) = 0.1243

b)
P( X>11) = 1 - P(X<11) = ?
  
I know that, z = (X-mean)/(sd)  
z1 = (11-13.5)/3.9) =   -0.6410
  
hence,  
P( X>11) = 1- P(Z<-0.641) = 1 - NORMSDIST(-0.641) =   0.7392

c)
P(10 < X < 14) = P(X<14) - P(X<10) = ?
  
I know that, z = (X-mean)/(sd)  
z1 = (10-13.5)/3.9) =   -0.8974
z2 = (14-13.5)/3.9) =   0.1282
  
hence,  
P(10 < X < 14) = P(Z<0.1282) - P(Z<-0.8974) = NORMSDIST(0.1282) - NORMSDIST(-0.8974) = 0.3663

d)
P(X=12) = 0.0000
probability at a point is zero as normal is continuous variable

e)
P(Z<z) = 80%
z = NORMSINV(0.8)
z = 0.841621234
(X-mean)/sd =   0.8416
X = 0.8416*3.9+13.5
X = 16.8