Answer:
73.3% probability that the sampling error made in estimating the mean amount of coffee for all 8 dash ounce 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 theore.
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:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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, we have that:
![\mu = 8.2, \sigma = 0.18, n = 100, s = \frac{0.18}{\sqrt{100}} = 0.018](https://tex.z-dn.net/?f=%5Cmu%20%3D%208.2%2C%20%5Csigma%20%3D%200.18%2C%20n%20%3D%20100%2C%20s%20%3D%20%5Cfrac%7B0.18%7D%7B%5Csqrt%7B100%7D%7D%20%3D%200.018)
What is the probability that the sampling error made in estimating the mean amount of coffee for all 8 dash ounce 8-ounce jars by the mean of a random sample of 100 jars, will be at most 0.02 ounce
That is, probability of the sample mean between 8.2-0.02 = 8.18 and 8.2 + 0.02 = 8.22, which is the pvalue of Z when X = 8.22 subtracted by the pvalue of Z when X = 8.18.
X = 8.22
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
By the Central Limit Theorem
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
![Z = \frac{8.22 - 8.2}{0.018}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B8.22%20-%208.2%7D%7B0.018%7D)
![Z = 1.11](https://tex.z-dn.net/?f=Z%20%3D%201.11)
has a pvalue of 0.8665.
X = 8.18
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
![Z = \frac{8.18 - 8.2}{0.018}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B8.18%20-%208.2%7D%7B0.018%7D)
![Z = -1.11](https://tex.z-dn.net/?f=Z%20%3D%20-1.11)
has a pvalue of 0.1335.
0.8665 - 0.1335 = 0.7330
73.3% probability that the sampling error made in estimating the mean amount of coffee for all 8 dash ounce 8-ounce jars by the mean of a random sample of 100 jars, will be at most 0.02 ounce