Using the <em>normal distribution and the central limit theorem</em>, it is found that there is a 0.5934 = 59.34% probability that there is an error of at most $1000.
<h3>Normal Probability Distribution</h3>
In a normal distribution with mean and standard deviation , the z-score of a measure X is given by:
- 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 .
Researching the problem on the internet, it is found that the population has mean and standard deviation given, respectively, by .
For samples of 64, the standard error is given by:
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 <u>p-value of Z when X = 58300 subtracted by the p-value of Z when X = 56300</u>, hence:
X = 58300:
By the Central Limit Theorem:
Z = 0.83
Z = 0.83 has a p-value of 0.7967.
X = 56300:
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 <em>normal distribution and the central limit theorem</em>, you can check brainly.com/question/24663213