Question

Chocolate consumption A study conducted by Hershey’s discovered that Americans consumed an average of 11.3 pounds of chocolate pre year. Let’s assume that the annual chocolate consumption follows a normal distribution with a standard deviation of 3.7 pounds. What is the probability of that an American will consume:

(a) Less than 7 pounds of chocolate next year

(b) More than 9 pounds of chocolate next year

(c) Between 8 and 12 pounds of chocolate next year

(d) Exactly 10 pounds of chocolate next year

(e) What is the annual consumption of pounds of chocolate that represents the 80th ?percentile?

Answer 1

Mean = = 11.3

Standard deviation = = 3.7

a) We have to find P(X < 7)

For finding this probability we have to find z score.

That is we have to find P( Z < - 1.16)

P( Z < - 1.16) = 0.1226 ( Using z table)

b) We have to find P(X > 9)

For finding this probability we have to find z score.

That is we have to find P( Z < - 0.62)

P( Z < - 0.62) = 0.2671 ( Using z table)

c) In this part we have to find P( 8 < X < 12)

For finding this probability we have to find z score.

That is we have to find P( - 0.89 < Z < 0.19)

P( - 0.89 < Z < 0.19) = P(Z < 0.19) - P( Z < - 0.89)

P( Z < - 0.89) = 0.1862 ( Using z table)

P(Z < 0.19) = 0.5750 ( Using z table)

P( - 0.89 < Z < 0.19) = 0.5750 - 0.1862 = 0.3888

d) We have to find P(X = 10)

X follows normal distribution and it is continuous.

P(X = 10) = 0 ( Because any exact probability in coutinuous distribution is zero)