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The mean per capita consumption of milk per year is 126 liters with a standard deviation of 24 liters. If a sample of 100 people is randomly selected, what is the probability that the sample mean would be less than 122.27 liters? Round your answer to four decimal places.

User P Burke
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Answer:

0.0606 = 6.06% probability that the sample mean would be less than 122.27 liters

Explanation:

To solve this question, we need to understand the normal probabiliy distribution and the central limit theorem.

Normal probability distribution

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 = (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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean
\mu and standard deviation
\sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
\mu and standard deviation
s = (\sigma)/(√(n)).

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:


\mu = 126, \sigma = 24, n = 100, s = (24)/(√(100)) = 2.4

What is the probability that the sample mean would be less than 122.27 liters?

This is the pvalue of Z when X = 122.27. So


Z = (X - \mu)/(\sigma)

By the Central Limit Theorem


Z = (X - \mu)/(s)


Z = (122.27 - 26)/(2.4)


Z = -1.55


Z = -1.55 has a pvalue of 0.0606

0.0606 = 6.06% probability that the sample mean would be less than 122.27 liters

User Aknew
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