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Sagot :
Using the normal distribution and the central limit theorem, it is found that there is a 0.6328 = 63.28% probability that a random sample of 100 accounting graduates will provide an average that is within $902 of the population mean.
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]
Hence, the probability of sample having mean within M of the population mean is the p-value of [tex]Z = \frac{M}{\frac{\sigma}{\sqrt{n}}}[/tex] subtracted by the p-value of [tex]Z = -\frac{M}{\frac{\sigma}{\sqrt{n}}}[/tex].
In this problem, we suppose [tex]\sigma = 10000[/tex], and thus, with a sample of 100, we have that [tex]s = \frac{10000}{\sqrt{100}} = 1000[/tex].
- Within $902, hence [tex]M = 902[/tex].
[tex]Z = \frac{M}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]Z = \frac{902}{1000}[/tex]
[tex]Z = 0.902[/tex]
[tex]Z = 0.902[/tex] has a p-value of 0.8164.
[tex]Z = \frac{M}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]Z = -\frac{902}{1000}[/tex]
[tex]Z = -0.902[/tex]
[tex]Z = -0.902[/tex] has a p-value of 0.1836.
0.8164 - 0.1836 = 0.6328.
0.6328 = 63.28% probability that a random sample of 100 accounting graduates will provide an average that is within $902 of the population mean.
A similar problem is given at https://brainly.com/question/24663213
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