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Chapter 12: Problem 6

Find the expected value and variance for each random variable whose probability density function is given. When computing the variance, use formula (5). \(f(x)=\frac{3}{2} x-\frac{3}{4} x^{2}, 0 \leq x \leq 2\)

Short Answer

Expert verified
Expected value is 1 and the variance is 0.

Step by step solution

01

- Verify the Probability Density Function (PDF)

Ensure the given function is a valid probability density function. Integrate the PDF over the possible values of the random variable and ensure it equals 1: \[\int_{0}^{2} \left( \frac{3}{2} x - \frac{3}{4} x^2 \right) dx = 1\]
02

- Integrate the PDF

Compute the integral of the PDF to ensure it equals 1: \[\int_{0}^{2} \left( \frac{3}{2} x - \frac{3}{4} x^2 \right) dx = \left[ \frac{3}{4} x^2 - \frac{1}{4} x^3 \right]_{0}^{2} = \left( \frac{3}{4} \cdot 4 - \frac{1}{4} \cdot 8 \right) = 1\]
03

- Calculate the Expected Value

Determine the expected value (mean) by integrating \(x\) times the PDF over the possible values of the random variable: \[E(X) = \int_{0}^{2} x \left( \frac{3}{2} x - \frac{3}{4} x^2 \right) dx\]Compute the integral: \[E(X) = \int_{0}^{2} \left( \frac{3}{2} x^2 - \frac{3}{4} x^3 \right) dx = \left[ \frac{1}{2} x^3 - \frac{3}{16} x^4 \right]_{0}^{2} = \left( 4 - 3 = 1 \right)\]
04

- Calculate the Expected Value of \(X^2\)

Determine \(E(X^2)\) by integrating \(x^2\) times the PDF over the possible values of the random variable: \[E(X^2) = \int_{0}^{2} x^2 \left( \frac{3}{2} x - \frac{3}{4} x^2 \right) dx\]Compute the integral: \[E(X^2) = \int_{0}^{2} \left( \frac{3}{2} x^3 - \frac{3}{4} x^4 \right) dx = \left[ \frac{3}{8} x^4 - \frac{1}{4} x^5 \right]_{0}^{2} = 3 - 2 = 1\]
05

- Calculate the Variance

Use the formula for variance: \[Var(X) = E(X^2) - (E(X))^2\]Insert the previously calculated values: \[Var(X) = 1 - (1)^2 = 0\]

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Key Concepts

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

Probability Density Function
The probability density function (PDF) is a key concept in statistics and probability. It describes the likelihood of a continuous random variable taking on a particular value. For a function to qualify as a valid PDF, it must meet two criteria:
  • The function must be non-negative across its entire range.
  • The total area under the function, when integrated over all possible values, must equal 1.
This ensures that the probabilities sum to 1, as they should in any valid probability distribution. In our example, the given PDF is: \( f(x) = \frac{3}{2}x - \frac{3}{4}x^2 \) over the interval \(0 \leq x \leq 2 \). To verify its validity, we integrate this function over the interval from 0 to 2 and ensure the result is 1.
Expected Value
The expected value, or mean, of a random variable provides a measure of the 'central' value of a probability distribution. It's like the weighted average of all possible values that the random variable can take on, weighted by their probabilities. To calculate the expected value of a continuous random variable with a given PDF, we use the integral: \[ E(X) = \int_{a}^{b} x f(x) dx \] In the step-by-step solution, we determined the expected value by integrating the product of \(x\) and the PDF over the range from 0 to 2. The calculation yielded \( E(X) = 1 \), indicating that the mean value of the random variable is 1.
Variance
Variance measures the dispersion or spread of a set of values around their mean. It gives us an idea of how much the values deviate from the expected value. For a continuous random variable, the variance is calculated using the formula: \[ Var(X) = E(X^2) - (E(X))^2 \] Here, \( E(X^2) \) is the expected value of the square of the random variable. In our solution, we first computed \( E(X^2) \), then used the variance formula to find: \[ Var(X) = 1 - 1^2 = 0 \] This result tells us that there is no spread around the mean, indicating that all possible values of the random variable are concentrated at the mean value.
Integration
Integration is a fundamental concept in calculus, crucial for calculating quantities like area, volume, and in this case, probabilities and expected values. When calculating the integral of a function over a specific interval, you essentially sum up an infinite number of infinitesimally small quantities. In probability, we use integration to:
  • Verify that a PDF sums to 1 over its interval.
  • Calculate expected values and higher moments like \( E(X^2) \).
For our example, we computed multiple integrals to find the normalization of the PDF and to determine the expected value and variance of the random variable.
Random Variable
A random variable is a variable whose values depend on the outcomes of a random phenomenon. There are two main types of random variables:
  • Discrete Random Variables: These take on a countable number of values (e.g., the roll of a dice).
  • Continuous Random Variables: These take on an uncountable range of values, often over an interval (e.g., the temperature on a given day).
The function given in the exercise describes a continuous random variable. The probability density function (PDF) associated with this type of random variable tells us how likely different outcomes are within a particular range. By understanding the PDF, we can derive important characteristics of the random variable, such as its expected value and variance.

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Most popular questions from this chapter

Find the expected value and variance for each random variable whose probability density function is given. When computing the variance, use formula (5). \(f(x)=\frac{1}{4}, 1 \leq x \leq 5\)

Compute the cumulative distribution function corresponding to the density function \(f(x)=\frac{1}{2}(3-x)\), \(1 \leq x \leq 3 .\)

Probability Table, Expected Value The number of phone calls coming into a telephone switchboard during each minute was recorded during an entire hour. During 30 of the 1-minute intervals there were no calls, during 20 intervals there was one call, and during 10 intervals there were two calls. A 1-minute interval is to be selected at random and the number of calls noted. Let \(X\) be the outcome. Then, \(X\) is a random variable taking on the values 0,1 , and 2 . (a) Write out a probability table for \(X\). (b) Compute \(\mathrm{E}(X)\). (c) Interpret \(\mathrm{E}(X)\).

Decision Making Based on Expected Value A citrus grower anticipates a profit of \(\$ 100,000\) this year if the nightly temperatures remain mild. Unfortunately, the weather forecast indicates a \(25 \%\) chance that the temperatures will drop below freezing during the next week. Such freezing weather will destroy \(40 \%\) of the crop and reduce the profit to \(\$ 60,000 .\) However, the grower can protect the citrus fruit against the possible freezing (using smudge pots, electric fans, and so on) at a cost of \(\$ 5000\). Should the grower spend the \(\$ 5000\) and thereby reduce the profit to \(\$ 95,000 ?[\) Hint: Compute \(\mathrm{E}(X)\), where \(X\) is the profit the grower will get if he does nothing to protect the fruit.]

During a certain part of the day, the time between arrivals of automobiles at the tollgate on a turnpike is an exponential random variable with expected value 20 seconds. Find the probability that the time between successive arrivals is greater than 10 seconds and less than 30 seconds.
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