In 25 polcapitas contemptions of sofiat drinks in Country A was reported to be 18.26 gallons. Assume that per capita consumption of soft drinks in Country A is approximately normally distributed, with a mean of 18.26 gallons and a standard deviation of 4 gallons. What is the probibility that Someone in Country A consumed more than 13 gallons of soft drinks in 2008?
The Correct Answer and Explanation is :
The probability that someone in Country A consumed more than 13 gallons of soft drinks in 2008 is approximately 90.57%.
Explanation:
To solve this problem, we assume that the per capita consumption of soft drinks follows a normal distribution with a given mean (μ) = 18.26 gallons and standard deviation (σ) = 4 gallons.
We need to determine the probability that an individual consumed more than 13 gallons. This is done by calculating the Z-score, which standardizes the given consumption value:
[
Z = \frac{X – \mu}{\sigma}
]
Substituting the values:
[
Z = \frac{13 – 18.26}{4} = \frac{-5.26}{4} = -1.315
]
The Z-score tells us how many standard deviations the value is from the mean. Next, we use the cumulative distribution function (CDF) of the standard normal distribution to find the probability that someone consumed less than 13 gallons:
[
P(X < 13) = P(Z < -1.315) \approx 0.0943
]
Since we need the probability of consuming more than 13 gallons, we subtract this value from 1:
[
P(X > 13) = 1 – 0.0943 = 0.9057
]
Thus, the probability that a randomly selected person in Country A consumed more than 13 gallons of soft drinks in 2008 is 90.57%.
Interpretation:
This result suggests that most people (about 90.57%) in Country A consume more than 13 gallons of soft drinks annually. Since 13 gallons is below the mean (18.26 gallons), it makes sense that a large proportion of the population would exceed this amount. The normal distribution allows us to estimate how consumption is spread among individuals.