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Chapter 4: Problem 25

Determine the constant \(c\) so that \(f(x)=c x(3-x)^{4}, 0

Short Answer

Expert verified
The value of the constant c should be \( \frac{5}{3} \)

Step by step solution

01

Write down the integral for the PDF

The integral of a PDF over its entire possible space is given by \(\int_{-\infty}^{\infty} f(x) dx\). In given problem, x ranges from 0 to 3, hence the equivalent is \(\int_{0}^{3} f(x) dx = 1 \)
02

Substitute f(x) into the integral

Substitute f(x) = \(c x(3-x)^{4}\) into the integral: \(\int_{0}^{3} c x(3-x)^{4} dx = 1 \)
03

Solve the integral

By solving the integral we get: \(\frac{3}{5}c = 1\)
04

Solve for the constant c

Solving for c from the equation: c = \(\frac{5}{3} \)

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Key Concepts

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

Integral of a PDF
To understand the concept of the integral of a Probability Density Function (PDF), it's essential to know that a PDF represents the likelihood of a random variable falling within a specific range of values. The total probability over all possible values must equal 1. In mathematical terms, this means the integral of the PDF over its range must be equal to 1.

You start by setting up the integral for the given range of a PDF. With the provided function, \[ f(x)=c x(3-x)^{4} \] we need to integrate over the interval from 0 to 3. The expression becomes:\[ \int_{0}^{3} f(x) \, dx = 1 \] This equation ensures that the total probability is normalized to 1 over its defined range. This process is critical for ensuring that the function qualifies as a probability density function.
Solving Integrals
Once the PDF is substituted into the integral, the next step is solving it. The expression:\[ \int_{0}^{3} c x(3-x)^{4} \, dx = 1 \] is tackled using integration techniques.

Let's break it down:
  • The term \( x(3-x)^{4} \) is a polynomial, a type of algebraic expression, which can be integrated by first multiplying out or through substitution if required.
  • Refining this involves calculating the antiderivative or the indefinite integral of the equation within the defined limits of 0 to 3.
By computing this integral, you arrive at \[ \frac{3}{5}c \] which equals 1. Understanding this part is crucial because solving integrals often requires different techniques depending on the form of the function.
Determining Constants
After solving the integral, the next step is determining the constant, often a crucial part of finding probabilities. In the equation \[ \frac{3}{5}c = 1 \] which resulted from the integral, we simply solve for the constant \(c\).

Start by isolating \(c\) on one side of the equation:
  • Multiply both sides by the reciprocal of the fraction \( \frac{3}{5} \).
  • This operation yields \( c = \frac{5}{3} \).
The constant \(c\) ensures that the integral of the PDF across its domain equals 1, completing the normalization process. Determining constants in this manner is vital in statistics to transform mathematical expressions into valid probability statements.

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Most popular questions from this chapter

Let \(X_{1}, X_{2}, X_{3}\) be a random sample from a distribution of the continuous type having p.d.f. \(f(x)=2 x, 0

Let \(X_{1}\) and \(X_{2}\) be stochastically independent random variables of the discrete type with joint p.d.f. \(f_{1}\left(x_{1}\right) f_{2}\left(x_{2}\right),\left(x_{1}, x_{2}\right) \in \mathscr{A} .\) Let \(y_{1}=u_{1}\left(x_{1}\right)\) and \(y_{2}=u_{2}\left(x_{2}\right)\) denote a one-to-one transformation that maps \(\mathscr{A}\) onto \(\mathscr{B}\). Show that \(Y_{1}=u_{1}\left(X_{1}\right)\) and \(Y_{2}=u_{2}\left(X_{2}\right)\) are stochastically independent.

Let \(X_{1}\) and \(X_{2}\) be two stochastically independent random variables so that the variances of \(X_{1}\) and \(X_{2}\) are \(\sigma_{1}^{2}=k\) and \(\sigma_{2}^{2}=2\), respectively. Given that the variance of \(Y=3 X_{2}-X_{1}\) is 25, find \(k\).

Let the stochastically independent random variables \(X_{1}\) and \(X_{2}\) be \(b\left(n_{1}, p\right)\) and \(b\left(n_{2}, p\right)\), respectively. Find the joint p.d.f. of \(Y_{1}=X_{1}+X_{2}\) and \(Y_{2}=X_{2}\), and then find the marginal p.d.f. of \(Y_{1} .\) Hint. Use the fact that $$ \sum_{w=0}^{k}\left(\begin{array}{c} n_{1} \\ w \end{array}\right)\left(\begin{array}{c}. n_{2} \\ k-w \end{array}\right)=\left(\begin{array}{c} n_{1}+n_{2} \\ k \end{array}\right) $$ This can be proved by comparing the coefficients of \(x^{k}\) in each member of the identity \((1+x)^{n_{1}}(1+x)^{n_{2}} \equiv(1+x)^{n_{1}+n_{2}}\)

Let \(F\) have an \(F\) distribution with parameters \(r_{1}\) and \(r_{2}\). Prove that \(1 / F\) has an \(F\) distribution with parameters \(r_{2}\) and \(r_{1}\).
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