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Rom4ik [11]
3 years ago
6

In 2013, a study found that U.S. consumers consumed soft drinks with a mean of 42.2 gallons per year and variance of 169. Ninety

-five percent of consumers in the U.S. consumed less than how many gallons?
Round to two decimals and include a leading zero if necessary.
Mathematics
1 answer:
melisa1 [442]3 years ago
4 0

Answer:

Ninety-five percent of consumers in the U.S. consumed less than 63.59 gallons.

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:

The standard deviation is the square root of the variance, so \sigma = \sqrt{169} = 13

Also, the mean is 42.2, so \mu = 42.2

Ninety-five percent of consumers in the U.S. consumed less than how many gallons?

The 95th percentile, which is the value of X when Z has a pvalue of 0.95. So X when Z = 1.645

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

1.645 = \frac{X - 42.2}{13}

X - 42.2 = 13*1.645

X = 63.59

Ninety-five percent of consumers in the U.S. consumed less than 63.59 gallons.

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