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.

0% probability that a household spent more than $16.00 on sugar.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

Find the probability that a household spent more than $16.00 on sugar.

This is 1 subtracted by the pvalue of Z when X = 16. So

has a pvalue of 1.

So there is a 1-1 = 0% probability that a household spent more than $16.00 on sugar.

120 seems like the most sensible answer.

step-by-step explanation:

if there is only one topping on all caramel apples and there's 120 of them. then all apples should/will be covered in sprinkles. make sense?