Question

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.

Answer 1

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 $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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:

$\mu = 8.22, \sigma = 1.10$

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

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{16 - 8.22}{1.10}$

$Z = 7.07$

$Z = 7.07$ has a pvalue of 1.

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

Answer 2

Answer:polo g

Step-by-step explanation: