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Let's analyze the given problem.

### Problem Overview:
We are given the probability density function (PDF) of the cost of production $ X $ (in thousands of dollars) as follows:

\[
f(x) = \begin{cases} 
\frac{1}{9}x^2 & \text{if } 0 < x \leq 3 \\
0 & \text{otherwise}
\end{cases}
\]

This function represents a continuous random variable with support on $ [0, 3] $.

### 3.1 - Finding the Probability:

The question asks: **If the production cost exceeds $1,600, what is the probability that the cost is less than $2,000?**

We need to convert the dollar values into thousands. So:
- $1,600 = 1.6 (thousands)
- $2,000 = 2.0 (thousands)

We are looking for:
\[
P(X < 2.0 \mid X > 1.6)
\]

Using conditional probability:
\[
P(A \mid B) = \frac{P(A \cap B)}{P(B)}
\]
Where:
- $ A = \{X < 2.0\} $
- $ B = \{X > 1.6\} $

So:
\[
P(X < 2.0 \mid X > 1.6) = \frac{P(1.6 < X < 2.0)}{P(X > 1.6)}
\]

To calculate this, we first find the individual probabilities by integrating the PDF.

### 3.2 - Expected New Cost:

The new cost $ C $ is defined as $ C = 1.1X + 0.25 $.

We need to find the expected value of $ C $, $ \mathbb{E}[C] $, which can be calculated as:
\[
\mathbb{E}[C] = 1.1 \mathbb{E}[X] + 0.25
\]

First, we calculate $ \mathbb{E}[X] $ by integrating the original PDF $ f(x) $:
\[
\mathbb{E}[X] = \int_0^3 x f(x) \, dx = \int_0^3 x \cdot \frac{1}{9}x^2 \, dx
\]

Once $ \mathbb{E}[X] $ is determined, plug into the equation for $ \mathbb{E}[C] $.

### 3.3 - Probability of a Specific Order in Sampling:

We are asked to find the probability that only the last selected item has a cost below $1,600 (1.6 \text{ in thousands})$ in a random sample of 5 articles.

This scenario involves finding the probability that in a sequence of 5 independent samples:
- The first 4 articles have a cost greater than 1.6.
- The 5th article has a cost less than 1.6.

This can be written as:
\[
P(\text{first 4} > 1.6, \text{last} < 1.6) = \left( P(X > 1.6) \right)^4 \cdot P(X < 1.6)
\]

Where $ P(X > 1.6) $ and $ P(X < 1.6) $ are calculated from the CDF of $ X $.

---

I'll now proceed with the necessary calculations.Here are the results for the given problems:

### 3.1 - Conditional Probability:
We are tasked with finding $ P(X < 2.0 \mid X > 1.6) $.

- $ P(1.6 < X < 2.0) = 0.1446 $
- $ P(X > 1.6) = 0.8483 $

So the conditional probability is:
\[
P(X < 2.0 \mid X > 1.6) = \frac{0.1446}{0.8483} \approx 0.1704
\]

### 3.2 - Expected New Cost:
We calculated the expected value of $ X $ as $ \mathbb{E}[X] = 2.25 $ (in thousands of dollars).

The expected new cost $ \mathbb{E}[C] = 1.1 \times 2.25 + 0.25 = 2.725 $ (in thousands of dollars), or $2,725.

### 3.3 - Probability in Random Sample:
The probability that the first 4 articles have a cost greater than $1,600 and the 5th article has a cost less than $1,600 is:
\[
P(\text{first 4} > 1.6, \text{last} < 1.6) = (P(X > 1.6))^4 \times P(X < 1.6)
\]
Using the values:
\[
= (0.8483)^4 \times 0.1517 \approx 0.0701
\]

### Let me know if you'd like more details or have any questions!

---

Here are 5 related questions:
1. How would the results change if the cost distribution were different?
2. What is the significance of the expected value $ \mathbb{E}[X] $ in real-world cost prediction?
3. How can the conditional probability concept be used in decision-making scenarios?
4. How would you compute the variance of the new cost $ C $?
5. What would happen if the sampling size in problem 3.3 increased?

**Tip**: When working with continuous probability distributions, always remember to normalize your results (ensure the total probability is 1).

The cost of production X is a random variable with a given probability density function. Three questions involve conditional probability, expected value, and probability in a random sample.

Explore a probability density function for production cost and solve related problems, including calculating conditional probabilities, expected values, and probabilities in random sampling. This problem is suitable for university-level probability and statistics courses.

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