Respuesta :
Using the normal distribution and the central limit theorem, it is found that:
a) The 90th percentile for an individual teacher's salary is $50,216.
b) The 90th percentile for the average teacher's salary is $43,914.
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 given by [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
In this problem:
- Mean of $41,000, thus [tex]\mu = 41000[/tex].
- Standard deviation of $7,200, thus [tex]\sigma = 7200[/tex].
Item a:
The 90th percentile is X when Z has a p-value of 0.9, so X when Z = 1.28. Thus:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.28 = \frac{X - 41000}{7200}[/tex]
[tex]X - 41000 = 1.28(7200)[/tex]
[tex]X = 50216[/tex]
The 90th percentile for an individual teacher's salary is $50,216.
Item b:
Average for a sample of 10, thus [tex]n = 10, s = \frac{7200}{\sqrt{10}}[/tex].
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]1.28 = \frac{X - 41000}{\frac{7200}{\sqrt{10}}}[/tex]
[tex]X - 41000 = 1.28\frac{7200}{\sqrt{10}}[/tex]
[tex]X = 43914[/tex]
The 90th percentile for the average teacher's salary is $43,914.
A similar problem is given at https://brainly.com/question/24663213
