Respuesta :
Using the normal distribution and the central limit theorem, it is found that there is a 0.1335 = 13.35% probability that 100 randomly selected students will have a mean SAT II Math score greater than 670.
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].
In this problem:
- The mean is of 660, hence [tex]\mu = 660[/tex].
- The standard deviation is of 90, hence [tex]\sigma = 90[/tex].
- A sample of 100 is taken, hence [tex]n = 100, s = \frac{90}{\sqrt{100}} = 9[/tex].
The probability that 100 randomly selected students will have a mean SAT II Math score greater than 670 is 1 subtracted by the p-value of Z when X = 670, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{670 - 660}{9}[/tex]
[tex]Z = 1.11[/tex]
[tex]Z = 1.11[/tex] has a p-value of 0.8665.
1 - 0.8665 = 0.1335.
0.1335 = 13.35% probability that 100 randomly selected students will have a mean SAT II Math score greater than 670.
To learn more about the normal distribution and the central limit theorem, you can take a look at https://brainly.com/question/24663213
