2. Inspecting incoming car parts from suppliers is a crucial step in automobile industry quality
control procedures. Let's say they received a shipment of 50 car spoiler parts. Each part has
a probability of 0.1 being defective. After thoroughly examining them, they'll return the
whole shipment if 20% or more defectives were found.
a. What is the probability of a whole shipment to get returned?
b. What is the probability of finding at most 5 defects per shipment?
c. Determine the average number of defects found in a shipment.

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.

Inspecting incoming car parts from suppliers is a crucial step in automobile industry quality control procedures. Let's say they received a shipment of 50 car spoiler parts. Each part has a probability of 0.1 being defective. After thoroughly examining them, they'll return the whole shipment if 20% or more defectives were found.
a. What is the probability of a whole shipment to get returned?
b. What is the probability of finding at most 5 defects per shipment?
c. Determine the average number of defects found in a shipment.

Analyze a quality control problem in the automobile industry by determining the probability of a shipment being returned, finding at most 5 defects, and calculating the average defects in a batch of 50 car parts using binomial distribution. Suitable for students in Grades 11-12.

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