Using the <u>normal distribution and the central limit theorem</u>, it is found that there is a:
a) 0.281 = 28.1% probability that the sample mean is above 500.
b) 0.0003 = 0.03% probability that the sample mean is above 500.
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
.
In this problem:
- The mean is of 430, hence
.
- The standard deviation is of 120, hence
.
Item a:
The probability is the <u>p-value of Z when X = 500</u>, hence:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{500 - 430}{120}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B500%20-%20430%7D%7B120%7D)
![Z = 0.58](https://tex.z-dn.net/?f=Z%20%3D%200.58)
has a p-value of 0.719.
1 - 0.719 = 0.281
0.281 = 28.1% probability that the sample mean is above 500.
Item b:
Sample of 35, hence ![n = 35, s = \frac{120}{\sqrt{35}}](https://tex.z-dn.net/?f=n%20%3D%2035%2C%20s%20%3D%20%5Cfrac%7B120%7D%7B%5Csqrt%7B35%7D%7D)
Then:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
By the Central Limit Theorem
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
![Z = \frac{500 - 430}{\frac{120}{\sqrt{35}}}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B500%20-%20430%7D%7B%5Cfrac%7B120%7D%7B%5Csqrt%7B35%7D%7D%7D)
![Z = 3.45](https://tex.z-dn.net/?f=Z%20%3D%203.45)
has a p-value of 0.9997.
1 - 0.9997 = 0.0003
0.0003 = 0.03% probability that the sample mean is above 500.
To learn more about the <u>normal distribution and the central limit theorem</u>, you can take a look at brainly.com/question/24663213