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Chapter 7: Problem 289

Find the variance of a random variable \(\mathrm{X}\) that is uniformly distributed over the interval \([0,3]\).

Short Answer

Expert verified
The variance of a random variable X uniformly distributed over the interval \([0, 3]\) is found using the formula \(Variance(X) = E(X^2) - (E(X))^2\). By calculating, we get \(E(X) = \frac{3}{2}\) and \(E(X^2) = 3\). Finally, the variance of X is \(Variance(X) = 3 - (\frac{3}{2})^2 = \frac{3}{4}\).

Step by step solution

01

Find E(X)

To find E(X), let's compute its integral over the entire interval [0, 3]. E(X) = \(\int_{0}^{3} x \cdot f(x) dx = \int_{0}^{3} x \cdot \frac{1}{3} dx\) Now, integrate and evaluate the integral: E(X) = \(\frac{1}{3} \int_{0}^{3} x dx = \frac{1}{3} \left[\frac{1}{2}x^2\right]_0^3 = \frac{1}{3} \cdot \frac{1}{2} \cdot (3^2 - 0) = \frac{1}{2} \cdot 3 = \frac{3}{2}\) So, E(X) = 3/2.
02

Find E(X^2)

To find E(X^2), let's compute its integral over the entire interval [0, 3]. E(X^2) = \(\int_{0}^{3} x^2 \cdot f(x) dx = \int_{0}^{3} x^2 \cdot \frac{1}{3} dx\) Now, integrate and evaluate the integral: E(X^2) = \(\frac{1}{3} \int_{0}^{3} x^2 dx = \frac{1}{3} \left[\frac{1}{3}x^3\right]_0^3 = \frac{1}{3} \cdot \frac{1}{3} \cdot (3^3 - 0) = \frac{1}{3} \cdot 9 = 3\) So, E(X^2) = 3.
03

Calculate Variance(X)

Now, we can use the formula for variance to find the variance of X: Variance(X) = E(X^2) - (E(X))^2 = 3 - (3/2)^2 = 3 - 9/4 = (12 - 9)/4 = 3/4 Hence, the variance of the random variable X uniformly distributed over the interval [0, 3] is 3/4.

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Key Concepts

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

Uniform Distribution
Uniform distribution describes a scenario where all outcomes in a certain interval are equally likely. For example, if a variable is uniformly distributed over the interval \([0, 3]\), any value between 0 and 3 is equally probable.
  • The probability density function (PDF) for a uniform distribution in \([a, b]\) is given by \(f(x) = \frac{1}{b-a}\) for \(a \leq x \leq b\).
  • In our case, \(a = 0\) and \(b = 3\), so \(f(x) = \frac{1}{3}\).
The concept is simple yet powerful in probability, as it helps in modeling situations where each interval segment has the same chance of occurrence. This equal likelihood is what defines a uniform distribution.
Expected Value
The expected value or mean of a random variable gives an idea of where values are centered. It represents the average outcome if the experiment were repeated many times.
For a continuous variable, we compute this by integrating over the interval:
  • The expected value of a uniform distribution over \([a, b]\) is given by \(E(X) = \frac{a+b}{2}\).
  • For our variable X over \([0, 3]\), the expected value is \(E(X) = \frac{0 + 3}{2} = \frac{3}{2}\).
This formula helps in predicting outcomes in practical situations by providing the central value around which all measurements are expected to hover.
Random Variable
A random variable is a quantity whose outcome is determined by a random phenomenon. In simpler terms, it's a variable that takes on different values based on the outcome of a random event.
There are two main types:
  • Discrete: This type can take distinct, separate values, like rolling a die.
  • Continuous: This type can take any value within an interval, like the uniform distribution we explored.
Understanding random variables is essential for studying probability because they help in formalizing the way we talk about random processes and their respective outcomes.

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Most popular questions from this chapter

Use the properties of expectation to find the variance of the sum of two independent random variables.

The probability that a certain baseball player will get a hit on any given time at bat is \((3 / 10)\). If he is at bat one hundred times during the next month, find the theoretical mean and variance of \(\mathrm{x}\), the number of hits. Assume that the binomial distribution is applicable.

A retailer has the opportunity to sell a portion of slow-moving stock to a liquidator for \(\$ 1800\). Since the items in question are children's toys, he is also aware that he may do better financially by keeping them in stock through the approaching Christmas shopping season and selling them at a discount. The following table indicates the retailer's estimate of the consequences of his decisions given that a certain percentage of stock is sold in the Christmas season. \begin{tabular}{|c|c|c|} \hline Percentage Of Stock Sold & Revenue If Stock Is Kept & Revenue If Stock Is Sold Now \\ \hline \(70 \%\) & \(\$ 1400\) & \(\$ 1800\) \\ \hline \(80 \%\) & \(\$ 1600\) & \(\$ 1800\) \\ \hline \(90 \%\) & \(\$ 1800\) & \(\$ 1800\) \\ \hline \(100 \%\) & \(\$ 2000\) & \(\$ 1800\) \\ \hline \end{tabular} The retailer has kept good records of previous experience over 25 Christmas seasons and has found that: \(\operatorname{Pr}(70 \%\) of stock is sold \()=(4 / 25)\) \(\operatorname{Pr}(80 \%\) of stock is sold \()=(7 / 25)\) \(\operatorname{Pr}(90 \%\) of stock is sold \()=(12 / 25)\) \(\operatorname{Pr}(100 \%\) of stock is sold \()=(2 / 25)\) Evaluate the decisions using expected revenue as the criterion.

Suppose the earnings of a laborer, denoted by \(\mathrm{X}\), are given by the following probability function. \begin{tabular}{|c|c|c|c|c|} \hline \(\mathrm{X}\) & 0 & 8 & 12 & 16 \\ \hline \(\operatorname{Pr}=(\mathrm{X}=\mathrm{x})\) & \(0.3\) & \(0.2\) & \(0.3\) & \(0.2\) \\ \hline \end{tabular} Find the laborer's expected earnings.

Let the random variable \(X\) represent the number of defective radios in a shipment of four radios to a local appliance store. Assume that each radio is equally likely to be defective or non-defective, hence the probability that a radio is defective is \(\mathrm{p}=(1 / 2)\). Also assume that each radio is defective or non-defective independently of the other radios. Find the expected number of defective radios.
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