Question

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

Answer 1

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 $\mu$ 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 mean and the standard error 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 p-value of Z when X = 0.68, 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