Question

The distribution of actual weights of 8-oz chocolate bars produced by a certain machine is normal with mean 7.9 ounces and standard deviation 0.16 ounces.
1. What is the probability that the average weight of a bar in a simple random sample of three of these chocolate bars is between 7.76 and 8.09 ounces?

Answer 1

Answer:

91.60% probability that the average weight of a bar in a simple random sample of three of these chocolate bars is between 7.76 and 8.09 ounces

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 $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 $s = \frac{\sigma}{\sqrt{n}}$.

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:

$\mu = 7.9, \sigma = 0.16, n = 3, s = \frac{0.16}{\sqrt{3}} = 0.0924$

What is the probability that the average weight of a bar in a simple random sample of three of these chocolate bars is between 7.76 and 8.09 ounces?

This is the pvalue of Z when X = 8.09 subtracted by the pvalue of Z when X = 7.76. So

X = 8.09

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{8.09 - 7.9}{0.0924}$

$Z = 2.06$

$Z = 2.06$ has a pvalue of 0.9803

X = 7.76

$Z = \frac{X - \mu}{s}$

$Z = \frac{7.76 - 7.9}{0.0924}$

$Z = -1.52$

$Z = -1.52$ has a pvalue of 0.0643

0.9803 - 0.0643 = 0.9160

91.60% probability that the average weight of a bar in a simple random sample of three of these chocolate bars is between 7.76 and 8.09 ounces