91Ó°ÊÓ

The Cumulative Distribution Function (CDF) is essential for understanding probabilities related to continuous random variables. It reflects the cumulative probability that a random variable is less than or equal to some value. In our problem, the random variable represents repair costs.

To find the cumulative probability, we integrate the Probability Density Function (PDF). For the given PDF, we have:

• \( f(y) = 3(1-y)^2 \) for \( 0 < y < 1 \).

The CDF, denoted by \( F(y) \), is the integral of the PDF from the lower bound (here 0) up to the value \( y \):

• \[ F(y) = \int_0^y 3(1-t)^2 \, dt = (1-y)^3 - 1 \]

This function allows us to determine the probability of the machine repair cost being within specified limits, which is foundational for our budgeting task.

Cost budgeting involves estimating the amount of money needed to cover expenses, in this case, repair costs for a machine. The goal is to ensure the budget aligns with actual expenses, minimizing gaps where unexpected costs arise.

In our context, we want to find a budget amount \( b \) such that costs exceed this budget only 10% of the time. Thus, we aim for a 90% confidence that actual costs will be below \( b \). This is crucial for efficient financial management.

To achieve this, we use the CDF calculated earlier. Specifically, we set:

• \( F(b) = 0.9 \)

This condition ensures that 90% of cost instances fall at or below the budget \( b \). By solving this equation, we determine the appropriate budget, aligning expected expenditures with financial allocations.

Probability calculations are critical in making informed decisions under uncertainty. They utilize various mathematical tools to predict outcomes based on random variables' behavior.

For this exercise, we calculate the budget threshold using a probability condition. We want to find \( y \) (in hundreds of dollars), representing a cost from the PDF, which exceeds the budget only 10% of the time.

• Solve \((1-y)^3 = 0.9\) which ensures 90% of costs are less than or equal to \( y \)
• Use the cube root to find \( y = 1 - 0.9^{1/3} \)
• Calculate \( y \approx 0.046 \), meaning the budget should be \(0.046 \times 100 = 4.6\) dollars

This elegant combination of probability and integral calculus helps determine practical financial strategies, ensuring the budget is both reasonable and sufficient.