In: Statistics and Probability
A study discovered that Americans consumed an average of 11.1 pounds of chocolate per year. Assume that the annual chocolate consumption follows the normal distribution with a standard deviation of 3.4lbs. Complete parts a through e below.
a. What is the probability that an American will consume less than 7 pounds of chocolate next year? (Round to four decimal places as needed.)
b. What is the probability that an American will consume more than 9 pounds of chocolate next year? (Round to four decimal places as needed.)
c. What is the probability that an American will consume between 8 and 12 pounds of chocolate next year? (Round to four decimal places as needed.)
d. What is the probability that an American will consume exactly 10 pounds of chocolate next year? (Round to four decimal places as needed.)
e. What is the annual consumption of chocolate that represents the 60th percentile?
The 60th percentile is represented by an annual consumption of ________________ pounds of chocolate.
(Type an integer or decimal rounded to one decimal place as needed.)
Solution:-
a) The probability that an American will consume less than 7 pounds of chocolate next year is 0.113.
Mean = 11.1, S.D = 3.40
x = 7
By applying normal distribution:-
z = - 1.21
P(z < -1.21) = 0.113
b) The probability that an American will consume more than 9 pounds of chocolate next year is 0.732.
Mean = 11.1, S.D = 3.40
x = 9
By applying normal distribution:-
z = - 0.62
P(z > -0.62) = 0.732
c) The probability that an American will consume between 8 and 12 pounds of chocolate next year is 0.423.
x1 = 8
x2 = 12
By applying normal distruibution:-
z1 = - 0.912
z2 = 0.265
P( -0.912 < z < 0.265) = P(z > - 0.912) - P(z > 0.265)
P( -0.912 < z < 0.265) = 0.819 - 0.396
P( -0.912 < z < 0.265) = 0.423
d) the probability that an American will consume exactly 10 pounds of chocolate next year is 0.1114.
x = 10
By applying normal distribution:-
z = - 0.324
P(z = -0.324) = 0.1114
e) The 60th percentile is represented by an annual consumption of 13.6 pounds of chocolate.
p-value for the bottom 60% = 0.60
z-score for the p-value = 0.726
By applying normal distruibution:-
x = 13.568