Question

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.

Answer 1

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.