Question

A study discovered that Americans consumed an average of 12.3 pounds of chocolate per year. Assume that the annual chocolate consumption follows the normal distribution with a standard deviation of 3.2 pounds.

a. What is the probability that an American will consume less than 9 pounds of chocolate next? year?

?(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?

c. What is the probability that an American will consume between 10 and 13 pounds of chocolate next? year?

d. What is the probability that an American will consume exactly 12 pounds of chocolate next? year?

e. What is the annual consumption of chocolate that represents the 80th percentile?

Answer 1

a)

for normal distribution z score =(X-)/x
here mean=       =	12.3
std deviation   ==	3.200

  probability that an American will consume less than 9 pounds of chocolate next? year:

probability =

P(X<9)

=

P(Z<-1.03)=

0.1515

b)

probability that an American will consume more than 11 pounds of chocolate next? year:

probability =

P(X>11)

=

P(Z>-0.41)=

1-P(Z<-0.41)=

1-0.3409=

0.6591

c)

  probability that an American will consume between 10 and 13 pounds of chocolate next? year:

probability =

P(10<X<13)

=

P(-0.72<Z<0.22)=

0.5871-0.2358=

0.3513

d)

probability that an American will consume exactly 12 pounds of chocolate next? year =0 (as being continuous distribution ; point probability is 0)

e)

for 80th percentile ; crtiical value z =0.84

hence annual consumption of chocolate that represents the 80th percentile =mean+z*std deviaton =12.3+0.84*3.2

=14.99