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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
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:
[tex]\mu = 7.9, \sigma = 0.16, n = 3, s = \frac{0.16}{\sqrt{3}} = 0.0924[/tex]
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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{8.09 - 7.9}{0.0924}[/tex]
[tex]Z = 2.06[/tex]
[tex]Z = 2.06[/tex] has a pvalue of 0.9803
X = 7.76
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{7.76 - 7.9}{0.0924}[/tex]
[tex]Z = -1.52[/tex]
[tex]Z = -1.52[/tex] 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
