Mathematics, 05.05.2020 02:09 jjiopppotdd5638

The amount of coffee that a filling machine puts into an 8-ounce jar is normally distributed with a mean of 8.2 ounces and a standard deviation of 0.18 ounce. What is the probability that the sampling error made in estimating the mean amount of coffee for all 8-ounce jars by the mean of a random sample of 100 jars will be at most 0.02 ounce?

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73.30% probability that the sampling error made in estimating the mean amount of coffee for all 8-ounce jars by the mean of a random sample of 100 jars will be at most 0.02 ounce

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:

$\mu = 8.2, \sigma = 0.18, n = 100, s = \frac{0.18}{\sqrt{100}} = 0.018$

What is the probability that the sampling error made in estimating the mean amount of coffee for all 8-ounce jars by the mean of a random sample of 100 jars will be at most 0.02 ounce?

This is the pvalue of Z when X = 8.2 + 0.02 = 8.22 subtracted by the pvalue of Z when X = 8.2 - 0.02 = 8.18. So

X = 8.22

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

By the Central Limit Theorem

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

$Z = \frac{8.22 - 8.2}{0.018}$

Z = 1.11

Z = 1.11 has a pvalue of 0.8665

X = 8.18

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

$Z = \frac{8.18 - 8.2}{0.018}$

Z = -1.11

Z = -1.11 has a pvalue of 0.1335

0.8665 - 0.1335 = 0.7330

73.30% probability that the sampling error made in estimating the mean amount of coffee for all 8-ounce jars by the mean of a random sample of 100 jars will be at most 0.02 ounce

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