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Lilit [14]
2 years ago
5

4. Amazon executives believe that at least 70% of customers would return a product 2 days after it arrives at their home. A samp

le of 500 customers found 68% returned the product they purchased prior to the third day. Given the executives' estimate, what would be the probability of a sample result with 68% or fewer returns prior to the third day
Mathematics
1 answer:
inysia [295]2 years ago
7 0

Using the <u>normal distribution and the central limit theorem</u>, 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 \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

  • 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 \mu = p and standard error s = \sqrt{\frac{p(1 - p)}{n}}

In this problem:

  • Sample of 500 customers, hence n = 500.
  • Amazon believes that the proportion is of 70%, hence p = 0.7

The <u>mean and the standard error</u> are given by:

\mu = p = 0.7

s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.7(0.3)}{500}} = 0.0205

The probability is the <u>p-value of Z when X = 0.68</u>, hence:

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{0.68 - 0.7}{0.0205}

Z = -0.98

Z = -0.98 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 brainly.com/question/25735688

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