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Sagot :
Using the normal distribution and the central limit theorem, it is found that his risk is of 0.0456.
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- It 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, which is the percentile of X.
- By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
In this problem:
- Mean of 20 ounces and standard deviation of 2 ounces, hence [tex]\mu = 20, \sigma = 2[/tex].
- Sample of 4 packages, hence [tex]n = 4, s = \frac{2}{\sqrt{4}} = 1[/tex]
The risk is the probability of being more than 2 ounces from the mean, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
Applying the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{2}{1}[/tex]
[tex]Z = 2[/tex]
The risk is P(|Z| > 2), which is 2 multiplied by the p-value of Z = -2.
- Looking at the z-table, Z = -2 has a p-value of 0.0228.
0.0228 x 2 = 0.0456
The risk is of 0.0456.
For more on the normal distribution and the central limit theorem, you can check https://brainly.com/question/24663213
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