Question

The amount of potato chips an 18-ounce bag contains follows a normal distribution with a mean of 18.5 ounces and a standard deviation of 0.2 ounces. 100 bags of chips were randomly selected. What is the probability that the sample mean weight of these 100 bags is less than 18.6 ounces

Answer 1

100% probability that the sample mean weight of these 100 bags is less than 18.6 ounces

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 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 question, we have that:

`\mu = 18.5, \sigma = 0.2, n = 100, s = (0.2)/(√(100)) = 0.02`

What is the probability that the sample mean weight of these 100 bags is less than 18.6 ounces

This is the pvalue of Z when X = 18.6. So

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

By the Central Limit Theorem

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

`Z = (18.6 - 18.5)/(0.02)`

`Z = 5` has a pvalue of 1

100% probability that the sample mean weight of these 100 bags is less than 18.6 ounces