Question

In​ 2008, the per capita consumption of soft drinks in Country A was reported to be 18.11 gallons. Assume that the per capita consumption of soft drinks in Country A is approximately normally​ distributed, with a mean of 18.11 gallons and a standard deviation of 4 gallons. Complete parts​ (a) through​ (d) below. a. What is the probability that someone in Country A consumed more than 13 gallons of soft drinks in​ 2008? The probability is nothing. ​(Round to four decimal places as​ needed.)

Answer 1

Answer:

There is a 89.97% probability that someone in Country A consumed more than 13 gallons of soft drinks in​ 2008.

Step-by-step explanation:

Problems of normally distributed samples can be 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

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

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. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

In this problem, we have that:

The per capita consumption of soft drinks in Country A is approximately normally​ distributed, with a mean of 18.11 gallons and a standard deviation of 4 gallons., so $\mu = 18.11, \sigma = 4$.

a. What is the probability that someone in Country A consumed more than 13 gallons of soft drinks in​ 2008?

The first step is finding the z score of $X = 13$.

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

$Z = \frac{13 - 18.11}{4}$

$Z = -1.28$

$Z = -1.28$ has a pvalue of 0.1003.

This means that there is a 1-0.1003 = 0.8997 = 89.97% probability that someone in Country A consumed more than 13 gallons of soft drinks in​ 2008.