During the past six months,73.2 percent of US households purchased sugar.Assume that these expenditures are approximately normally distributed with a mean of $8.22 and a standard deviation of $1.10.Find the probability that a household spent more than $16.00 on sugar.
A)1.00
B)0.00
C)0.50
D)0.98

Question

Quizplus · Accepted Answer

Since the mean expenditure on sugar is $8.22 and the standard deviation is $1.10, it is highly unlikely for a household to spend more than $16.00 on sugar. We can calculate the z-score as follows:

z-score = (x - μ) / σ
z-score = (16.00 - 8.22) / 1.10
z-score = 7.09

Using a z-table or calculator, we can see that the probability of a z-score being greater than 7.09 is essentially 0. Therefore, the probability of a household spending more than $16.00 on sugar is essentially 0 or 0.00.

During the Past Six Months,73