91Ó°ÊÓ

When dealing with a continuous uniform distribution, you are looking at a scenario where a variable is equally likely to fall anywhere within a specified interval. In this exercise, the variable in focus is the waiting time for delivery, represented by \( Y \), which can occur from 1 to 4 days.

A continuous uniform distribution is characterized by a constant probability density function across its interval. Specifically for this problem, the density function is given by \( f(y) = \frac{1}{3} \) when \( 1 \leq y \leq 4 \). This means each day within this time frame is equally probable, which simplifies many calculations related to probability and expected outcomes.

In practice, understanding this type of distribution helps us anticipate potential outcomes (in this case, delivery times) and helps in calculating costs or benefits associated with different time intervals.

The cost function in this exercise outlines how much the builder has to pay based on the time it takes for the supplies to arrive. Initially, for any delay within the first two days, the cost remains flat at \(\\( 100\). This means no additional cost is incurred if the wait is within this period.

However, once waiting time exceeds 2 days, additional costs accrue at \(\\) 20\) for each day (or part thereof) beyond the 2-day mark. This creates a piecewise cost function where the first part covers \( Y \leq 2 \) with a fixed cost, and the second part adds a daily penalty for \( Y > 2 \). Therefore, the comprehensive cost function is:

• \(100\) for \(1 \leq Y \leq 2\)
• \(100 + 20(y - 2)\) for \(Y > 2\)

Clearly distinguishing these sections of the cost function is crucial, as it directly affects the overall cost associated with waiting times.

Integral calculus plays a pivotal role in this problem as it is used to determine the expected cost due to delays. Given that we have a continuous distribution of waiting times, we employ integration to accumulate all possible costs over the specified interval.

To find the expected value, we compute the integral of the cost function times the probability density function over the interval of interest.

• The first integral from 1 to 2 represents the cost within the initial 2 days: \[\int_{1}^{2} 100 \cdot \frac{1}{3} \, dy = \frac{100}{3}.\]
• The second integral from 2 to 4 considers both the flat fee and added daily costs: \[\int_{2}^{4} 100 \cdot \frac{1}{3} \, dy + \int_{2}^{4} 20(y - 2) \cdot \frac{1}{3} \, dy.\]

By computing these integrals, we can sum them to find the total expected cost, representing an average scenario over time.

The expected cost is a concept that provides an average monetary expense one can expect over a given time period or scenario, taking into account all possible outcomes weighed by their probabilities. For the builder waiting on supplies, it boils down to calculating how much, on average, the delay will cost.

Using integral calculus, we computed the expected cost by evaluating how the continuous distribution spreads over time, and how the cost function assigns values to these times. After integrating and summing the costs across the possible waiting times, the expected cost was calculated to be \(\frac{340}{3}\), or approximately \(\$113.33\).

Expected cost is valuable as it offers insight into operational budgeting and financial planning. It helps the builder predict financial impacts with better precision across different scenarios.