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Chapter 5: Problem 126

A large lot of manufactured items contains \(10 \%\) with exactly one defect, \(5 \%\) with more than one defect, and the remainder with no defects. Ten items are randomly selected from this lot for sale. If \(Y_{1}\) denotes the number of items with one defect and \(Y_{2},\) the number with more than one defect, the repair costs are \(Y_{1}+3 Y_{2} .\) Find the mean and variance of the repair costs.

Short Answer

Expert verified
Mean: 2.5; Variance: 5.175

Step by step solution

01

Define the Random Variables

The random variables are defined as follows: \(Y_1\) represents the number of items with exactly one defect, and \(Y_2\) denotes the number of items with more than one defect. The total repair cost is given by the expression \(Y_1 + 3Y_2\).
02

Determine the Distribution of Y1 and Y2

Using the binomial distribution, \(Y_1\) is distributed as \(\text{Binomial}(10, 0.1)\) since there is a 10% chance of an item having exactly one defect. Similarly, \(Y_2\) follows \(\text{Binomial}(10, 0.05)\) due to the 5% probability of having more than one defect.
03

Calculate the Expectation of Y1 and Y2

The expected value for a binomial distribution \(\text{Binomial}(n, p)\) is \(np\). Thus, \(E(Y_1) = 10 \times 0.1 = 1\) and \(E(Y_2) = 10 \times 0.05 = 0.5\).
04

Compute the Expectation of the Repair Cost

The expectation of the total repair costs \(E(Y_1 + 3Y_2)\) is calculated as \(E(Y_1) + 3E(Y_2) = 1 + 3 \times 0.5 = 2.5\).
05

Calculate the Variance of Y1 and Y2

The variance for a binomial distribution \(\text{Binomial}(n, p)\) is \(np(1-p)\). Thus, \(Var(Y_1) = 10 \times 0.1 \times 0.9 = 0.9\) and \(Var(Y_2) = 10 \times 0.05 \times 0.95 = 0.475\).
06

Compute the Variance of the Repair Cost

The variance of the repair cost \(Y_1 + 3Y_2\) is calculated using the rule \(Var(aX + bY) = a^2 Var(X) + b^2 Var(Y)\), assuming independence. Thus, \(Var(Y_1 + 3Y_2) = Var(Y_1) + 3^2 Var(Y_2) = 0.9 + 9 \times 0.475 = 5.175\).

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Key Concepts

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

Random Variables
In probability theory, a random variable is a variable that can take different outcomes based on a random event. It's essentially a way to quantify randomness. Here, we have two key random variables:
  • \( Y_1 \), representing the count of items with exactly one defect.
  • \( Y_2 \), indicative of the count of items with more than one defect.
These variables help in predicting the results of random selections, like figuring out the repair costs in our exercise.Without random variables, structuring problems like these would be significantly harder, as they'd lack a foundation to base predictions on. It鈥檚 a crucial component in many statistical analyses.
Binomial Distribution
A binomial distribution is a probability distribution that summarizes the likelihood of a value taking one of two independent states under a given number of observations. It is defined by two parameters: the number of trials, \( n \), and the probability of success, \( p \).In the exercise:
  • For \( Y_1 \), which relates to items with one defect, the distribution is \( \text{Binomial}(10, 0.1) \). This indicates 10 trials, with a 10% chance of finding a defect.
  • \( Y_2 \), items with more than one defect, follows \( \text{Binomial}(10, 0.05) \). Here, we assess the same 10 items, but with a 5% defect probability.
Binomial distribution is essential because it helps calculate the probabilities of different outcomes, such as finding out the probability of a specific number of defects in selected items.
Expected Value
The expected value provides a measure of the "center" of a probability distribution and indicates the average or mean value to anticipate over lots of trials. In the binomial distribution world, this is calculated as \( np \).For our random variables:
  • \( E(Y_1) = 10 \times 0.1 = 1 \), suggesting on average, we'd find 1 item with exactly one defect.
  • \( E(Y_2) = 10 \times 0.05 = 0.5 \), implying typically we encounter half an item (on average) with more than one defect across multiple samples.
When evaluating the repair costs, the expected value helps determine overall cost expectation. Calculating the expected value for the total repair cost, \( E(Y_1 + 3Y_2) \), results in 2.5, giving a sense of the mean repair cost from these variables when scaling up the defect count of \( Y_2 \).
Variance
Variance measures how much the values of a random variable differ from the expected value. It reflects the "spread" or variability from the average. For a binomial distribution, the variance is computed as \( np(1-p) \).Let's break down our variance calculations:
  • \( Var(Y_1) = 10 \times 0.1 \times 0.9 = 0.9 \), showing the spread in the count of single defect items.
  • \( Var(Y_2) = 10 \times 0.05 \times 0.95 = 0.475 \), demonstrating variability in finding more than one defective item.
To calculate the variance of the repair costs, the independence of \( Y_1 \) and \( Y_2 \) is assumed, leading us to the formula \( Var(aX + bY) = a^2 \text{Var}(X) + b^2 \text{Var}(Y) \). For \( Y_1 + 3Y_2 \), this results in a variance of 5.175. Understanding variance is key to knowing how much fluctuation to expect around the expected cost, helping manage risks in decision-making.

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Most popular questions from this chapter

Let \(X_{1}, X_{2},\) and \(X_{3}\) be random variables, either continuous or discrete. The joint moment generating function of \(X_{1}, X_{2},\) and \(X_{3}\) is defined by $$m\left(t_{1}, t_{2}, t_{3}\right)=E\left(e^{t_{1} X_{1}+t_{2} X_{2}+t_{3} X_{3}}\right)$$ a. Show that \(m(t, t, t)\) gives the moment-generating function of \(X_{1}+X_{2}+X_{3}\) b. Show that \(m(t, t, 0)\) gives the moment-generating function of \(X_{1}+X_{2}\) c. Show that $$\left.\frac{\partial^{k_{1}+k_{2}+k_{3}} m\left(t_{1}, t_{2}, t_{3}\right)}{\partial t_{1}^{k_{1}} \partial t_{2}^{k_{2}} \partial t_{3}^{k_{3}}}\right]_{t_{1}=t_{2}=t_{3}=0}=E\left(X_{1}^{k_{1}} X_{2}^{k_{2}} X_{3}^{k_{3}}\right)$$

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