91Ó°ÊÓ

The manager of a construction job needs to figure prices carefully before submitting a bid. He also needs to account for uncertainty (variability) in the amounts of products he might need. To oversimplify the real situation, suppose that a project manager treats the amount of sand, in yards. needed for a construction project as a random variable \(Y_{1}\), which is normally distributed with mean 10 yards and standard deviation. 5 yard. The amount of cement mix needed, in hundreds of pounds, is a random variable \(Y_{2}\), which is normally distributed with mean 4 and standard deviation .2. The sand costs \(\$ 7\) per yard, and the cement mix costs \(\$ 3\) per hundred pounds. Adding \(\$ 100\) for other costs, he computes his total cost to be $$U=100+7 Y_{1}+3 Y_{2}$$. If \(Y_{1}\) and \(Y_{2}\) are independent, how much should the manager bid to ensure that the true costs will exceed the amount bid with a probability of only. \(01 ?\) Is the independence assumption reasonable here?

Step by step solution

01

Understand the Problem

We need to determine the bid price such that there is a 1% chance that the actual cost exceeds this bid. The manager computes total cost as \[ U = 100 + 7Y_1 + 3Y_2 \]where \( Y_1 \) and \( Y_2 \) are independent normal random variables representing the amount of sand and cement mix needed.

02

Calculate the Mean of Total Cost

To find the mean of \( U \), we use the linearity of expectation:\[ E(U) = 100 + 7E(Y_1) + 3E(Y_2) \]where \( E(Y_1) = 10 \) and \( E(Y_2) = 4 \), so:\[ E(U) = 100 + 7 \times 10 + 3 \times 4 = 100 + 70 + 12 = 182 \]

03

Calculate the Variance of Total Cost

Since \( Y_1 \) and \( Y_2 \) are independent, their covariance is 0. The variance of \( U \) is:\[ \text{Var}(U) = 7^2 \times \text{Var}(Y_1) + 3^2 \times \text{Var}(Y_2) \]where \( \text{Var}(Y_1) = 5^2 \) and \( \text{Var}(Y_2) = 0.2^2 \), so:\[ \text{Var}(U) = 49 \times 25 + 9 \times 0.04 = 1225 + 0.36 = 1225.36 \]

04

Determine the Standard Deviation of Total Cost

The standard deviation of \( U \) is the square root of the variance:\[ \text{SD}(U) = \sqrt{1225.36} \approx 35 \]

05

Find the 99th Percentile of the Cost Distribution

To find the bid amount where there's a 1% chance the cost exceeds this amount, we use the 99th percentile of the normal distribution. For a standard normal distribution, the 99th percentile corresponds to a z-score of approximately 2.33. Calculating the value:\[ U = E(U) + 2.33 \times \text{SD}(U) \]\[ U = 182 + 2.33 \times 35 \approx 182 + 81.55 = 263.55 \]

06

Conclusion

The manager should bid approximately \( \$ 263.55 \) to ensure the total costs exceed this value with only a 1% probability. The assumption of independence between the amounts of sand and cement mix is reasonable in this context, as they are different materials and their requirements are likely determined independently.

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Key Concepts

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

Random Variables

In statistics, a random variable is a numerical outcome of a random phenomenon. It's a way to quantify uncertainty. For instance, in a construction setting, the exact amount of materials required is uncertain but can be represented as a random variable. The random variables in our case are represented by \(Y_1\) for sand and \(Y_2\) for cement mix. Each of these variables is defined by a probability distribution, which describes how likely different outcomes are.

• They provide the foundational building block for many statistical models.
• They can be classified into discrete or continuous random variables. In our example, \(Y_1\) and \(Y_2\) are continuous random variables.

Understanding random variables helps in evaluating situations where outcomes are not deterministic, such as predicting the amount of materials needed for a project.

Normal Distribution

The normal distribution, also known as the Gaussian distribution, is one of the most important probability distributions in statistics. It is characterized by its bell-shaped curve which is symmetric around its mean.

• In our scenario, both \(Y_1\) (sand) and \(Y_2\) (cement mix) follow a normal distribution.
• It is defined by two parameters: the mean (average) and the standard deviation (measure of spread).
• Many real-world phenomena evolve in a manner that closely resembles a normal distribution, making it a suitable assumption for the manager's problem.

With a normal distribution, most outcomes are expected around the mean, with the probability tapering off equally on both sides.

Variance and Standard Deviation

Variance and standard deviation are measures of the spread in a set of numbers. In the context of random variables, they quantify the uncertainty associated with predictions.

• Variance measures the average squared deviation from the mean.
• For \(Y_1\), the variance is \(5^2 = 25\), and for \(Y_2\), it is \(0.2^2 = 0.04\).
• Standard deviation, being the square root of variance, provides an indication of how much the values deviate from the mean.

In our example, these figures help the manager determine how variable the actual material requirements could be, providing a basis for calculating potential cost ranges.

Linear Transformation of Random Variables

A linear transformation involves adjusting a random variable by applying a linear function, often for scaling and translating data.

• This technique is useful as it allows the transformation of simple variables like \(Y_1\) and \(Y_2\) into a more complex variable, \(U\), representing total cost.
• The transformation applied is \(U = 100 + 7Y_1 + 3Y_2\), which involves multiplying the random variables by constants (prices per unit) and adding a fixed cost.
• Through linear transformations, properties like the mean and variance of \(U\) can be calculated easily from \(Y_1\) and \(Y_2\).

Thus, linear transformations are crucial in simplifying computations and forecasting outcomes in real-world situations such as budgeting in construction projects.