Respuesta :
Answer:
40.13% probability that 9 randomly selected packages will have an average weight in excess of 16.025 ounces
Step-by-step explanation:
To solve this question, we have to understand the normal probability distribution and the central limit theorem.
Normal probability distribution:
Problems of normally distributed samples can be 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 random normally distributed variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], 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]
In this problem, we have that:
[tex]\mu = 16, \sigma = 0.3, n = 9, s = \frac{0.3}{\sqrt{9}} = 0.1[/tex]
What is the probability that 9 randomly selected packages will have an average weight in excess of 16.025 ounces
This is 1 subtracted by the pvalue of Z when X = 16.025. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{16.025 - 16}{0.1}[/tex]
[tex]Z = 0.25[/tex]
[tex]Z = 0.25[/tex] has a pvalue of 0.5987
1 - 0.5987 = 0.4013
40.13% probability that 9 randomly selected packages will have an average weight in excess of 16.025 ounces
