Answer:
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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:
[tex]\mu = 8.22, \sigma = 1.10[/tex]
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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{16 - 8.22}{1.10}[/tex]
[tex]Z = 7.07[/tex]
[tex]Z = 7.07[/tex] has a pvalue of 1.
So there is a 1-1 = 0% probability that a household spent more than $16.00 on sugar.
