A company sells eggs whose individual weights are normally distributed with a mean of 70\,\text{g}70g70, start text, g, end text and a standard deviation of 2\,\text{g}2g2, start text, g, end text. Suppose that these eggs are sold in packages that each contain 444 eggs that represent an SRS from the population. What is the probability that the mean weight of 444 eggs in a package \bar x x ˉ x, with, \bar, on top is less than 68.5\,\text{g}68.5g68, point, 5, start text, g, end text?

6.68% probability that the mean weight is below 68.5g.

Step-by-step 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 and standard deviation , the zscore of a measure X is given by:

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 and standard deviation , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean and standard deviation .

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

In this question:

Probability that the mean weight is below 68.5g:

This is 1 subtracted by the pvalue of Z when X = 68.5. So

By the Central Limit Theorem

has a pvalue of 0.0668

6.68% probability that the mean weight is below 68.5g.