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Sagot :
Using the normal distribution and the central limit theorem, it is found that there is a 0.1635 = 16.35% probability of a sample result with 68% or fewer returns prior to the third day.
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, for a proportion p in a sample of size n, the sampling distribution of sample proportions has mean [tex]\mu = p[/tex] and standard error [tex]s = \sqrt{\frac{p(1 - p)}{n}}[/tex]
In this problem:
- Sample of 500 customers, hence [tex]n = 500[/tex].
- Amazon believes that the proportion is of 70%, hence [tex]p = 0.7[/tex]
The mean and the standard error are given by:
[tex]\mu = p = 0.7[/tex]
[tex]s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.7(0.3)}{500}} = 0.0205[/tex]
The probability is the p-value of Z when X = 0.68, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.68 - 0.7}{0.0205}[/tex]
[tex]Z = -0.98[/tex]
[tex]Z = -0.98[/tex] has a p-value of 0.1635.
0.1635 = 16.35% probability of a sample result with 68% or fewer returns prior to the third day.
A similar problem is given at https://brainly.com/question/25735688
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