This problem deals with a binomial distribution, where each car spoiler part either passes inspection or is defective. The relevant parameters are:

• The probability of each spoiler part being defective is $p = 0.1$.
• The number of spoiler parts inspected is $n = 50$.
• We want to know the probability of returning the whole shipment if 20% or more of the parts are defective, meaning 10 or more defective parts (since 20% of 50 is 10).

The random variable $X$, representing the number of defective parts, follows a binomial distribution:

$X \sim \text{Binomial}(n = 50, p = 0.1)$

We need to calculate the probability that 10 or more parts are defective, i.e., $P(X \geq 10)$. This can be found using:

$P(X \geq 10) = 1 - P(X < 10) = 1 - P(X \leq 9)$

Using the binomial cumulative distribution function $P(X \leq 9)$, we can compute this probability.

Let me calculate that for you.The probability that the whole shipment will be returned, meaning 10 or more defective parts are found, is approximately 0.0245 or 2.45%.

Would you like more details on how this probability was calculated, or have any other questions?

Here are 5 related questions for further exploration:

1. How does the probability change if the defect rate increases to 15%?
2. What is the expected number of defective parts in a shipment of 50?
3. What would be the probability of finding exactly 5 defective parts?
4. How would the result differ if only 15% defective parts trigger a return?
5. How can you adjust the inspection procedure to reduce the risk of returning a good shipment?

Tip: In quality control, increasing the sample size or improving supplier quality can significantly reduce the chances of large-scale returns.