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Sagot :
Using the normal distribution and the central limit theorem, it is found that there is a 0.0029 = 0.29% probability that the sample average is LESS than 8.40.
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 [tex]\mu = 8.5[/tex].
- The standard deviation is of [tex]\sigma = 0.22[/tex].
- A sample of 37 is taken, hence [tex]n = 37, s = \frac{0.22}{\sqrt{37}} = 0.0362[/tex].
The probability that the sample average is LESS than 8.40 is the p-value of Z when X = 8.4, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{8.4 - 8.5}{0.0362}[/tex]
[tex]Z = -2.76[/tex]
[tex]Z = -2.76[/tex] has a p-value of 0.0029.
0.0029 = 0.29% probability that the sample average is LESS than 8.40.
To learn more about the normal distribution and the central limit theorem, you can take a look at https://brainly.com/question/24663213
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