Question

The mean per capita consumption of milk per year is 105105 liters with a standard deviation of 2626 liters. If a sample of 220220 people is randomly selected, what is the probability that the sample mean would be less than 107.81107.81 liters? Round your answer to four decimal places.

Answer 1

0.9452 = 94.52% probability that the sample mean would be less than 107.81 liters.

Explanation:

To solve this question, we need to understand the normal probability 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 random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sample means with size n of at least 30 can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`

In this problem, we have that:

`\mu = 105, \sigma = 26, n = 220, s = (26)/(√(220)) = 1.7529`

Probability that the sample mean would be less than 107.81 liters?

pvalue of Z when X = 107.81. So

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

By the central limit theorem

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

`Z = (107.81 - 105)/(1.7529)`

`Z = 1.60`

`Z = 1.60` has a pvalue of 0.9452

0.9452 = 94.52% probability that the sample mean would be less than 107.81 liters.