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Suppose that the random variable \(X\) represents the length of a punched part in centimeters. Let \(Y\) be the length of the part in millimeters. If \(E(X)=5\) and \(V(X)=0.25,\) what are the mean and variance of \(Y ?\)

Short Answer

Expert verified
The mean of \( Y \) is 50 mm, and the variance of \( Y \) is 25 \( \text{mm}^2 \).

Step by step solution

01

Understand the Conversion

We know that 1 centimeter equals 10 millimeters. Therefore, the relationship between the random variables can be described as \( Y = 10X \). We will use this relationship to find the mean and variance of \( Y \).
02

Calculate the Mean of Y

The expected value of \( Y \), denoted \( E(Y) \), is calculated by the formula \( E(Y) = E(10X) = 10E(X) \). Given \( E(X) = 5 \), we find \( E(Y) = 10 \times 5 = 50 \).
03

Calculate the Variance of Y

The variance of \( Y \) is calculated using the formula \( V(Y) = V(10X) = 10^2V(X) \). Given \( V(X) = 0.25 \), we have \( V(Y) = 100 \times 0.25 = 25 \).

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Key Concepts

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

Expected Value
Expected Value, often symbolized as \(E(X)\), represents the average or mean outcome of a random variable \(X\). It is a fundamental concept in probability theory, providing insight into the average behavior that can be anticipated over numerous repetitions of a random event. In the context of our exercise, if \(X\) is a random variable representing the length of a part in centimeters, the expected value \(E(X) = 5\) conveys that the average length of the part over many instances is 5 centimeters. This concept helps in predicting the outcome without performing numerous actual measurements.To compute the expected value for a function of a random variable, say \(Y = 10X\), we apply the linear property:
  • \(E(Y) = E(10X) = 10 \times E(X)\)
This direct scaling reflects how transformations of variables, like converting units, impact their expected values in a predictable manner.
Variance
Variance is a measure that describes the spread of a set of data points around their mean, providing insight into the level of uncertainty or risk associated with the random variable. It is denoted as \(V(X)\) for a random variable \(X\).In our exercise, \(V(X) = 0.25\) signifies that the lengths of the parts in centimeters have a small variance, suggesting that individual measurements are close to the expected value of 5 centimeters.When it comes to transforming random variables, variance behaves differently than expected value. For a linear transformation \(Y = aX\), the variance is given by:
  • \(V(Y) = a^2V(X)\)
This formula incorporates an additional squared term \(a^2\), allowing us to determine the variance of \(Y\) when \(Y = 10X\). Consequently, the variance \(V(Y)\) becomes \(100 \times 0.25 = 25\), illustrating increased variability due to the scaling factor.
Random Variables
Random Variables are a core concept in probability theory that allow us to quantitatively describe random phenomena. A random variable can be thought of as a numerical outcome applied to a chance process or event.In our scenario, we have two random variables:
  • \(X\): Represents the length of a part in centimeters.
  • \(Y\): Represents the length of the same part in millimeters, where \(Y = 10X\).
By understanding how random variables work, we can model real-life situations, transforming measurements and applying operations such as calculating mean and variance.The study of random variables is integral to probability theory because it lays the foundation for deeper analysis, enabling us to:
  • Predict outcomes with expected values.
  • Gauge uncertainty with variance.
  • Conduct further stochastic analysis.
By approaching problems with these tools, students can effectively translate theoretical knowledge into practical solutions.

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Most popular questions from this chapter

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