We are given that the total number of car spoiler parts is $n = 50$ and the probability of each part being defective is $p = 0.1$. This is a binomial distribution problem, where each part either is defective or not.

The binomial distribution formula is:

$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$

where:

• $n$ is the total number of trials (here, 50 parts),
• $k$ is the number of successes (defective parts),
• $p$ is the probability of success on a single trial (probability of a defective part),
• $\binom{n}{k}$ is the binomial coefficient, representing the number of ways to choose $k$ defective parts from $n$ parts.

a) What is the probability that the whole shipment gets returned?

The shipment will be returned if 20% or more of the parts are defective. Since 20% of 50 parts is:

$0.20 \times 50 = 10 \text{ parts}$

So, the shipment will be returned if 10 or more parts are defective. We need to find $P(X \geq 10)$, where $X$ is the number of defective parts.

This is:

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

We can compute $P(X \leq 9)$ using the binomial distribution cumulative probabilities.

b) What is the probability of finding at most 5 defects per shipment?

This is the probability that $X \leq 5$, where $X$ is the number of defective parts. We will calculate the cumulative binomial probability for $X \leq 5$.

$P(X \leq 5) = \sum_{k=0}^{5} \binom{50}{k} (0.1)^k (0.9)^{50-k}$

c) Determine the average number of defects found in a shipment.

The average (or expected) number of defects in a binomial distribution is given by:

$E(X) = n \times p$

Substituting the values $n = 50$ and $p = 0.1$:

$E(X) = 50 \times 0.1 = 5$

So, the average number of defective parts per shipment is 5.

I'll now calculate the probabilities for parts a and b using the binomial distribution.### Results:

a) The probability that the whole shipment gets returned (i.e., 10 or more defective parts) is approximately 0.0245 (or 2.45%).

b) The probability of finding at most 5 defective parts per shipment is approximately 0.6161 (or 61.61%).

c) The average number of defects found in a shipment is 5.

Would you like more details on any of the calculations?

Here are 5 related questions to explore further:

1. How would the probabilities change if the shipment size increased to 100 parts?
2. What if the probability of a part being defective was 0.05 instead of 0.1?
3. How does the variance of the binomial distribution affect the quality control process?
4. What is the probability of finding exactly 7 defective parts in a shipment?
5. Can we approximate this problem using a normal distribution for large shipment sizes?

Tip: For large values of $n$ and small $p$, the binomial distribution can often be approximated by a Poisson distribution, which simplifies calculations.