Two-thirds of U.S. households purchased ground coffee. Consider the annual ground coffee expenditures for households purchasing ground coffee, assuming that these expenditures are approximately distributed as a normal random variable with a mean of $65.16 and a standard deviation of $10.00.
A. Find the probability that a household spent less than $35.00
P(X<35) = P(x-65.16/10 < 35-65.16/10) = P(Z < -3.016) = 0.001281 13% OF HOUSEHOLDS SPENT LESS THAN $35.00
B. Find the probability that a household spend more than $60.00
P(X > 60) = 1- P(X<60) = 1-P(Z<60-65.16/10) = 1 -P(Z< -0.516) = 1-0.3029=0.6971 70% OF HOUSEHOLDS SPENT MORE THAN $60.00
C. What proportion of the households spent between $40.00 and $50.00?
P(40
D. 99% of the households spent less than what amount?
P(Z < K-65.16/10) = 0.99 = P(Z <2.326) K=65.16+10*2.326=88.42 99% OF HOUSEHOLDS SPENT LESS THAN $88.00