Respuesta :
Using the normal distribution and the central limit theorem, 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 [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:
- The mean is of 430, hence [tex]\mu = 430[/tex].
- The standard deviation is of 120, hence [tex]\sigma = 120[/tex].
Item a:
The probability is the p-value of Z when X = 500, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{500 - 430}{120}[/tex]
[tex]Z = 0.58[/tex]
[tex]Z = 0.58[/tex] 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 [tex]n = 35, s = \frac{120}{\sqrt{35}}[/tex]
Then:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{500 - 430}{\frac{120}{\sqrt{35}}}[/tex]
[tex]Z = 3.45[/tex]
[tex]Z = 3.45[/tex] 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 normal distribution and the central limit theorem, you can take a look at https://brainly.com/question/24663213
