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Using the normal distribution and the central limit theorem, it is found that there is a 0.5934 = 59.34% probability that there is an error of at most $1000.
Normal Probability Distribution
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].
Researching the problem on the internet, it is found that the population has mean and standard deviation given, respectively, by [tex]\mu = 57300, \sigma = 9600[/tex].
For samples of 64, the standard error is given by:
[tex]s = \frac{9600}{\sqrt{64}} = 1200[/tex]
The probability of an error of at most $1000 is the probability of a sample mean between $56,300 and $58,300, which is the p-value of Z when X = 58300 subtracted by the p-value of Z when X = 56300, hence:
X = 58300:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem:
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{58300 - 57300}{1200}[/tex]
Z = 0.83
Z = 0.83 has a p-value of 0.7967.
X = 56300:
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{56300 - 57300}{1200}[/tex]
Z = -0.83
Z = -0.83 has a p-value of 0.2033.
0.7967 - 0.2033 = 0.5934.
0.5934 = 59.34% probability that there is an error of at most $1000.
To learn more about the normal distribution and the central limit theorem, you can check https://brainly.com/question/24663213
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