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 can approach this problem by modeling it using a **binomial distribution**, since each car spoiler part can either be defective or not. The binomial distribution is used when we have a fixed number of independent trials, each trial resulting in a success or failure, and the probability of success is constant.

### Given:
- \( n = 50 \) (number of car spoiler parts),
- \( p = 0.1 \) (probability of a part being defective),
- \( q = 1 - p = 0.9 \) (probability of a part not being defective),
- Shipment will be returned if 20% or more parts are defective, which corresponds to at least \( 0.2 \times 50 = 10 \) defective parts.

### a. What is the probability of a whole shipment getting returned?

We need to find the probability of having 10 or more defective parts in the shipment, i.e., \( P(X \geq 10) \), where \( X \) is the number of defective parts.

The probability mass function of the binomial distribution is:

\[
P(X = k) = \binom{n}{k} p^k q^{n-k}
\]

The probability of returning the shipment is:

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

We'll calculate \( P(X \leq 9) \), which is the cumulative probability of having 9 or fewer defects.

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

Here we need to find \( P(X \leq 5) \), i.e., the cumulative probability of having 5 or fewer defective parts.

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

The expected value (mean) of a binomial distribution is given by:

\[
E(X) = n \cdot p
\]

Let me calculate each part now.### Results:

a. The probability of the whole shipment being returned (i.e., having 10 or more defective parts) is approximately **0.0245** or **2.45%**.

b. The probability of finding at most 5 defective parts in the shipment is approximately **0.6161** or **61.61%**.

c. The average number of defects expected in a shipment is **5**.

Would you like more details on any of these calculations or any related questions?

### Follow-up questions:
1. What is the probability of finding exactly 5 defects in a shipment?
2. What is the probability of finding between 5 and 10 defects (inclusive) in a shipment?
3. How would the result change if the probability of a defect increases to 0.2?
4. How does increasing the sample size (e.g., to 100 car parts) affect the probability of return?
5. What is the variance of the number of defective parts per shipment?

### Tip:
In quality control, small probabilities of return like 2.45% can suggest a well-functioning process, but continuous monitoring is crucial to prevent sudden shifts in defect rates.

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.

Explore how to calculate the probability of defective car parts in a shipment using binomial distribution. Understand key concepts such as calculating the probability of returning a shipment, finding at most 5 defects, and determining the average number of defects. This solution is suitable for college-level or advanced high school students.

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