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Chapter 10: Problem 3

Verify the two properties of a probability density function over the given interval. $$ f(x)=\frac{3}{26} x^{2},[1,3] $$

Short Answer

Expert verified
The function \( f(x) = \frac{3}{26} x^2 \) is non-negative and integrates to 1 over [1, 3].

Step by step solution

01

Check non-negativity

The probability density function (pdf) must be non-negative over the given interval [1, 3]. \( f(x) = \frac{3}{26} x^{2} \). Since the square of a number is always non-negative and the constant \(\frac{3}{26}\) is positive, \( f(x)\geq 0\) for all \( x\) in [1, 3].
02

Integrate the pdf over the interval

To verify the second property, integrate \( f(x)\) over the interval [1, 3]: \[ \int_{1}^{3} f(x) \, dx = \int_{1}^{3} \frac{3}{26} x^2 \, dx \].
03

Compute the integral

Find the antiderivative of \( \frac{3}{26} x^2 \): \[ \int \frac{3}{26} x^2 \, dx = \frac{3}{26} \cdot \frac{x^3}{3} = \frac{1}{26} x^3 \]. \[ \text{Evaluate it from 1 to 3}: \left. \frac{1}{26} x^3 \right|_{1}^{3} = \frac{1}{26} (3^3 - 1^3) = \frac{1}{26} (27 - 1) = \frac{26}{26} = 1 \]. Thus, \( \int_{1}^{3} \frac{3}{26} x^2 \, dx = 1 \).
04

Conclusion

Since the function \(f(x) = \frac{3}{26} x^{2}\) is non-negative over [1, 3] and the integral over this interval equals 1, it satisfies both properties of a probability density function.

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Key Concepts

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

Non-Negativity
Non-negativity is a crucial property of a probability density function (pdf). This means that the function must be greater than or equal to zero for all values within the given interval. In the example with the function \(f(x) = \frac{3}{26} x^2\) on the interval [1, 3], it is important to note the following points:
  • The term \(x^2\) is always non-negative because the square of any real number is never negative.
  • The constant \(\frac{3}{26}\) is positive, meaning it does not change the non-negativity of \(x^2\).
Hence, \(f(x) \geq 0\) for all \(x\) in [1, 3], satisfying the non-negativity criterion for a pdf.
Integral of a Function
The integral of a function over a specific interval is essential in verifying the second property of a probability density function. To satisfy this property, the integral of the pdf over the entire interval must equal 1.
For the given function, \(\int_{1}^{3} \frac{3}{26} x^2 \, dx\), we follow these steps:
  • Find the antiderivative of \(\frac{3}{26} x^2\), which is \(\frac{1}{26} x^3\).
  • Evaluate this antiderivative at the upper and lower limits of the interval, 3 and 1 respectively.
  • Subtract the value at the lower limit from the value at the upper limit to get \(\left. \frac{1}{26} x^3 \right|_{1}^{3} = \frac{1}{26} (3^3 - 1^3) = \frac{26}{26} = 1\).
Thus, the integral of \(\frac{3}{26} x^2\) from 1 to 3 equals 1, confirming the property.
Probability Theory
Probability theory provides the foundation for understanding how probability density functions work. Here are some key points:
  • A probability density function describes the likelihood of a continuous random variable taking on a specific value within a given range.
  • Unlike discrete probabilities, which sum to 1, the area under the curve of a pdf over its interval must equal 1.
  • PDFs allow us to compute probabilities for continuous variables by integrating the function over the desired range.
This theoretical framework is central to using and understanding pdfs in probability.
Calculus for Life Sciences
Calculus is a powerful tool in life sciences, particularly in modeling and understanding complex systems. Probability density functions often appear in various fields such as biology, epidemiology, and environmental sciences.
  • In biological research, PDFs can model the distribution of traits or behaviors within a population.
  • Epidemiologists use PDFs to predict the spread of diseases by analyzing infection rates over time.
  • Environmental scientists apply PDFs in studying the distribution of pollutants or animal species in a given area.
Understanding how to integrate and differentiate these functions allows scientists to make informed decisions based on their data and models.

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Most popular questions from this chapter

The probability density function of a random variable and a significance level \(\alpha\) are given. Find the critical value. $$ f(x)=0.32 e^{-x}+1.36 e^{-2 x} \text { over }[0, \infty) ; \alpha=0.01 $$

Find \(k\) such that the function is a probability density function over the given interval. Then write the probability density function. If there is no \(k\) that makes the function a probability density function, state why. $$ f(x)=k\left(x^{2}-x\right),[0,2] $$

A random variable \(X\) is assumed to have a standard normal distribution. Find the observed significance level if the random variable is equal to the given value in an experiment. Observed value of \(-1.24\)

In a stable population, on average, three mutations occur every minute. a) Find the probability that exactly one mutation occurs in \(2 \mathrm{~min}\). b) Find the probability that exactly five mutations occur in \(3 \mathrm{~min}\). c) Find the probability that no mutation occurs in a minute.

Suppose that an individual is randomly selected from a population that satisfies the Hardy-Weinberg Law. Let \(p\) and \(q\) denote the proportions of \(A\) and \(a\) alleles in the population. a) The allele inherited by the offspring may be visualized with the following multiplication tree. Fill in the probabilities of the unmarked arrows. b) In the multiplication tree, there are three ways corresponding to the transmission of the \(A\) allele. Show that the probability that the \(A\) allele is transmitted is \(p\). c) Show that the probability that the \(a\) allele is Iransmitted is \(q .\) d) Use the result of Example 9 to show that, if one generation satisfies the Hardy-Weinberg Law and survives to mate, then the offspring generation should also satisfy it.
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