Respuesta :
Answer:
a: 0.6578
b: 0.6578
c: 0.6578
Step-by-step explanation:
For a: See attached photo 1 for the calculation of the probability
There is no more work needed to calculate the other probabilities. The mean is the only thing that's different. We still want a range of being within 10 points of the population mean, so the z-scores won't change, even though the mean has. We would still wind up with the same numerator in the test statistic, and since the sample size didn't change, we'd get the same z-value!
Using the normal distribution and the central limit theorem, it is found that there is a:
- 0.6578 = 65.78% probability a sample of 90 test takers will provide a sample mean test score within 10 points of the population mean of 502 on the Critical Reading part of the test.
- 0.6578 = 65.78% probability a sample of 90 test takers will provide a sample mean test score within 10 points of the population mean of 515 on the Math part of the test.
- 0.6826 = 68.26% probability a sample of 90 test takers will provide a sample mean test score within 10 points of the population mean of 494 on the writing part of the test.
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].
The population standard deviation is 100, thus [tex]\sigma = 100[/tex].
On Critical Reading:
- Mean of 502, thus [tex]\mu = 502[/tex].
- Sample of 90, thus [tex]n = 90[/tex].
We want the probability of scores between 492 and 512, thus, it is the p-value of Z when X = 512 subtracted by the p-value of Z when X = 492.
X = 512:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem:
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{512 - 502}{\frac{100}{\sqrt{90}}}[/tex]
[tex]Z = 0.95[/tex]
[tex]Z = 0.95[/tex] has a p-value of 0.8289.
X = 492:
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{492 - 502}{\frac{100}{\sqrt{90}}}[/tex]
[tex]Z = -0.95[/tex]
[tex]Z = -0.95[/tex] has a p-value of 0.1711.
0.8289 - 0.1711 = 0.6578.
0.6578 = 65.78% probability a sample of 90 test takers will provide a sample mean test score within 10 points of the population mean of 502 on the Critical Reading part of the test.
On Math:
- Mean of 515, thus [tex]\mu = 502[/tex].
- Sample of 90, thus [tex]n = 90[/tex].
We want the probability of scores between 505 and 525, thus, it is the p-value of Z when X = 525 subtracted by the p-value of Z when X = 515.
X = 525:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem:
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{525 - 515}{\frac{100}{\sqrt{90}}}[/tex]
[tex]Z = 0.95[/tex]
[tex]Z = 0.95[/tex] has a p-value of 0.8289.
X = 505:
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{505 - 515}{\frac{100}{\sqrt{90}}}[/tex]
[tex]Z = -0.95[/tex]
[tex]Z = -0.95[/tex] has a p-value of 0.1711.
0.8289 - 0.1711 = 0.6578.
0.6578 = 65.78% probability a sample of 90 test takers will provide a sample mean test score within 10 points of the population mean of 515 on the Math part of the test.
On Writing:
- Mean of 494, thus [tex]\mu = 494[/tex].
- Sample of 100, thus [tex]n = 100[/tex].
We want the probability of scores between 484 and 504, thus, it is the p-value of Z when X = 504 subtracted by the p-value of Z when X = 484.
X = 504:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem:
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{504 - 494}{\frac{100}{\sqrt{100}}}[/tex]
[tex]Z = 1[/tex]
[tex]Z = 1[/tex] has a p-value of 0.8413.
X = 484:
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{484 - 494}{\frac{100}{\sqrt{100}}}[/tex]
[tex]Z = -1[/tex]
[tex]Z = -1[/tex] has a p-value of 0.1587.
0.8413 - 0.1587 = 0.6826
0.6826 = 68.26% probability a sample of 90 test takers will provide a sample mean test score within 10 points of the population mean of 494 on the writing part of the test.
A similar problem is given at https://brainly.com/question/24663213
