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GaryK [48]
3 years ago
11

During the past six months, 73.2 percent of US households purchased sugar. Assume that these expenditures are approximately norm

ally 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.
Mathematics
2 answers:
pashok25 [27]3 years ago
8 0

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.

lina2011 [118]3 years ago
4 0

Answer:polo g

Step-by-step explanation:

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