Respuesta :
The z-score for 36,000 is (36,000 - 40,000) / (15,000 / √25) = -1.333
The z-score for 42,000 is (42,000 - 40,000) / (15,000 / √25) = 0.6667
P(36,000 < x < 42,000) = P(-1.3333 < z < 0.6667)
P(z < 0.6667) - P(z < -1.3333)
0.7475 - 0.0913
0.6562
Using the normal probability distribution and the central limit theorem, it is found that there is a 0.6568 = 65.68% probability that the average for this sample is between $36,000 and $42,000.
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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, working with samples of size n, the standard deviation is [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
In this problem:
- Mean of $40,000, thus [tex]\mu = 40000[/tex].
- Standard deviation of $15,000, thus [tex]\sigma = 15000[/tex].
- Sample of 25, thus [tex]n = 25, s = \frac{15000}{\sqrt{25}} = 3000[/tex].
The probability is the p-value of Z when X = 42000 subtracted by the p-value of Z when X = 36000, thus:
X = 42000
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{42000 - 40000}{3000}[/tex]
[tex]Z = 0.67[/tex]
[tex]Z = 0.67[/tex] has a p-value of 0.7486.
X = 36000
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{36000 - 40000}{3000}[/tex]
[tex]Z = -1.33[/tex]
[tex]Z = -1.33[/tex] has a p-value of 0.0918.
0.7486 - 0.0918 = 0.6568.
0.6568 = 65.68% probability that the average for this sample is between $36,000 and $42,000.
A similar problem is given at https://brainly.com/question/22934264
