Question

A pizza chain monitors the total weight of pepperoni that goes on its deluxe pepperoni pizzas to make sure customers are satisfied and product isn't being wasted. Suppose that for pizzas in this population, these weights are strongly skewed to the left with a mean of 250 g and a standard deviation of 8 g. Management takes a random sample of 64 of these pizzas and calculates the mean weight of the pepperoni on the pizzas. Assume that the pizzas in the sample are independent.

What is the probability that the mean weight of the pepperoni from the sample of 64 pizzas LaTeX: \bar x is within 1 g of the true mean?

Answer 1

68.26% probability that the mean weight of the pepperoni from the sample of 64 pizzas is within 1 g of the true mean.

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 = 250, \sigma = 8, n = 64, s = (8)/(√(64)) = 1`

What is the probability that the mean weight of the pepperoni from the sample of 64 pizzas is within 1 g of the true mean?

This is the pvalue of Z when X = 64 + 1 = 65 subtracted by the pvalue of Z when X = 64 - 1 = 63. So

X = 65

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

By the Central Limit Theorem

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

`Z = (65 - 64)/(1)`

`Z = 1`

`Z = 1` has a pvalue of 0.8413.

X = 63

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

`Z = (63 - 64)/(1)`

`Z = -1`

`Z = -1` has a pvalue of 0.1587

0.8413 - 0.1587 = 0.6826

68.26% probability that the mean weight of the pepperoni from the sample of 64 pizzas is within 1 g of the true mean.