Final answer:
To calculate the probability that more than 12% of the sampled orders would have been returned, we used the normal approximation to the binomial distribution and calculated the standard error and z-score. Based on our z-score calculation and typical z-table values, the probability is approximately 2%, indicating option b is most likely correct.
Step-by-step explanation:
To find the approximate probability that more than 12% of the sampled orders would have been returned, we need to understand that we are dealing with a sampling distribution of the sample proportion. Since we are estimating a proportion with a sample size of 350 and an assumed population proportion of 9% (0.09), we can use the normal approximation to the binomial distribution because the sample size is large enough (np > 10 and n(1-p) > 10).
First, find the standard error (SE) of the sampling distribution of the sample proportion:
SE = sqrt(p(1-p)/n)
Using the population proportion (p = 0.09) and sample size (n = 350), we calculate SE.
SE = sqrt(0.09(0.91)/350) = sqrt(0.0819/350) = sqrt(0.000234) ≈ 0.0153
Next, calculate the z-score for the sample proportion of 0.12:
Z = (sample proportion - population proportion) / SE
Z = (0.12 - 0.09) / 0.0153 ≈ 1.96
We use the standard normal distribution to find P(Z > 1.96), which is approximately 0.025 (or 2.5%). However, as this is not an option provided, we would need the exact probabilities corresponding to the choices given (0, 0.02, 0.04, 0.06). Using the Z-score table or a standard normal distribution calculator would give us the closest option.
Assuming we've done this, the correct answer is likely to be closest to option b, indicating that the probability is 0.02 or 2%.