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Weekly CPU time used by an accounting firm has probability density function (measured in hours) given by $$f(y)=\left\\{\begin{array}{ll} (3 / 64) y^{2}(4-y), & 0 \leq y \leq 4 \\ 0, & \text { elsewhere } \end{array}\right.$$ a. Find the expected value and variance of weekly CPU time. b. The CPU time costs the firm \(\$ 200\) per hour. Find the expected value and variance of the weekly cost for CPU time. c. Would you expect the weekly cost to exceed \(\$ 600\) very often? Why?

Short Answer

Expert verified
Expected cost: \$5,600. Cost exceeding \$600 is unlikely (low probability event).

Step by step solution

01

Understand the Problem and Given PDF

The probability density function (PDF) is given by \( f(y) = (3/64)y^2(4-y) \) for \( 0 \leq y \leq 4 \), and 0 elsewhere. We need to calculate the expected value (mean \( E(Y) \)) and variance (\( ext{Var}(Y) \)) of this distribution.
02

Calculate the Expected Value

The expected value \( E(Y) \) is given by the integral \( E(Y) = \int_{0}^{4} y f(y) \, dy \). Thus,\[ E(Y) = \int_{0}^{4} y \left( \frac{3}{64} y^2(4-y) \right) \, dy = \frac{3}{64} \int_{0}^{4} y^3(4-y) \, dy. \]Solve the integral to find \( E(Y) \).
03

Solve the Integral for Expected Value

Break down the integral:\[ \int_{0}^{4} y^3(4-y) \, dy = \int_{0}^{4} 4y^3 \, dy - \int_{0}^{4} y^4 \, dy. \]Calculate each part:1. \( \int_{0}^{4} 4y^3 \, dy = \left[ \frac{4y^4}{4} \right]_{0}^{4} = 4^4 = 256. \)2. \( \int_{0}^{4} y^4 \, dy = \left[ \frac{y^5}{5} \right]_{0}^{4} = \left( \frac{4^5}{5} \right) = \frac{1024}{5}. \)Thus, \( E(Y) = \frac{3}{64} (256 - \frac{1024}{5}). \)
04

Calculate the Variance

The variance \( \text{Var}(Y) \) is \( E(Y^2) - [E(Y)]^2 \). First find \( E(Y^2) \) using \( E(Y^2) = \int_{0}^{4} y^2 f(y) \, dy = \frac{3}{64} \int_{0}^{4} y^4(4-y) \, dy \).Carry out a similar integration to find \( E(Y^2) \), and use it to find \( \text{Var}(Y) = E(Y^2) - [E(Y)]^2 \).
05

Transition to Weekly Cost

The weekly cost is \( C = 200Y \), where \( Y \) is the CPU hours. \( E(C) = E(200Y) = 200E(Y) \) and \( ext{Var}(C) = 200^2 \times ext{Var}(Y) \). Calculate these values using the previously found \( E(Y) \) and \( ext{Var}(Y) \).
06

Evaluate Likelihood of Exceeding $600

The cost exceeds \(600 if \( Y > 3 \) because \( 600/200 = 3 \). Calculate \( P(Y > 3) = 1 - P(Y \leq 3) \) using the given PDF. Integrate to find \( P(Y \leq 3) \), then find \( P(Y > 3) \), and conclude based on the probability whether exceeding \)600 is likely.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Expected Value
Understanding expected value is like determining the average outcome if you repeat an experiment over and over. It involves using a probability density function, in this case, a formula representing how CPU time usage is spread over a week.
To find expected value, denoted as \( E(Y) \), we integrate the product of the variable \( Y \) and its PDF over its range. For this problem, the formula becomes:
  • \( E(Y) = \int_{0}^{4} y \left( \frac{3}{64} y^2(4-y) \right) \, dy \), which simplifies, upon solving, to finding the integral \( \int_{0}^{4} y^3(4-y) \, dy \).
Integration then requires solving two distinctive parts:
  • \( \int_{0}^{4} 4y^3 \, dy \)
  • \( \int_{0}^{4} y^4 \, dy \)
Each integral provides part of the whole picture, which when calculated, gives us the expected CPU hours used weekly. This mean reflects not just a guess but a calculable average which is vital in planning resources.
By understanding the expected value, firms can predict their average CPU usage, which can help in budget planning and forecasting potential technological resources needed.
Variance
Variance provides a measure of how much the CPU time usage fluctuates around the expected value. In other words, it tells us about the spread or variability of the weekly CPU time. If the variance is large, it means CPU time usage is highly unpredictable.
To calculate the variance of the CPU time, we must first find \( E(Y^2) \), an expectation involving each possible value squared and weighted by probability. The healthy formula is:
  • \( E(Y^2) = \int_{0}^{4} y^2 \left( \frac{3}{64} y^2(4-y) \right) \, dy \)
Solving this integral can be intricate, but once we have \( E(Y^2) \), variance \( \text{Var}(Y) \) is determined by subtracting the square of \( E(Y) \) from \( E(Y^2) \).
This difference shows us the extent of variation from the mean, or how spread out the CPU usage is from its expected value. Low variance would suggest that the CPU time stays close to the mean, indicating consistency, whereas high variance suggests more unpredictability.
Integration
Integration plays a central role in calculating both expected value and variance in probability density functions. It's a mathematical tool that helps sum up continuous variables over a given interval. In the context of CPU time, integration helps us determine both average values and their fluctuations over time.
As described in the context of solving for expected value and variance, integration takes the form of summing the area under a curve defined by the PDF. It's essential to understand that:
  • Integrals such as \( \int_{0}^{4} y^3(4-y) \, dy \) are solved by breaking them into simpler parts for easy calculation.
  • In the variance calculation, both \( \int_{0}^{4} 4y^3 \, dy \) and \( \int_{0}^{4} y^4 \, dy \) represent solving individual expressions that contribute to the overall spread of values.
This is part of why integration is powerful: by breaking down complex PDF expressions, it delivers precise insights into the behavior of data over time.
Efficient integration can provide critical reflections of a dataset's central tendencies and dispersions, helping companies make better data-driven decisions.

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