Math Problem Statement
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.
Solution
Let's analyze the given problem.
Problem Overview:
We are given the probability density function (PDF) of the cost of production (in thousands of dollars) as follows:
\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).Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Probability Density Function
Conditional Probability
Expected Value
Sampling Probability
Formulas
Conditional Probability: P(A | B) = P(A ∩ B) / P(B)
Expected Value of C: E[C] = 1.1E[X] + 0.25
Sampling Probability Formula
Theorems
Fundamental Theorem of Probability
Expectation Theorem
Suitable Grade Level
University (Statistics/Probability)
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