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Chapter 3: Problem 20

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

Short Answer

Expert verified
Therefore, the constant \(c\) that makes \(f(x) = c x(3-x)^{4}\) a valid pdf is \(c = \frac{2}{9}\).

Step by step solution

01

Setting Up the Integral

A pdf must satisfy the property that its integral over its full domain is 1. So we need to solve the equation \(\int_{0}^{3} f(x) dx = 1\). In this case, the integral to solve is \(\int_{0}^{3} c x(3-x)^{4} dx = 1\).
02

Substitution

Use substitution to simplify the integral. Let \(u = 3 - x\), then \(du = -dx\) and \(x = 3 - u\). The limits of the integral change to 3 for the lower limit and 0 for the upper limit. The integral to solve becomes \(-c \int_{3}^{0} u^{4}(3 - u)du\).
03

Switch the Limits of Integration and Simplify

Switching the limits of integration will get rid of the negative sign in front of the integral. This results in \(c \int_{0}^{3} u^{4}(3-u) du \). Expand the integrand \(u^{4}(3-u) = 3u^{4} - u^{5}\). So the integral to solve becomes \(c \int_{0}^{3} (3u^{4} - u^{5}) du\).
04

Evaluate the Integral

The integral can be solved using the power rule for integration: \(\int_{a}^{b} x^n dx = \frac{1}{n+1}x^{n+1} |_{a}^{b}\). Thus, the integral \(c \int_{0}^{3} (3u^{4} - u^{5}) du = c(\frac{3}{5}u^{5} - \frac{1}{6}u^{6}) |_{0}^{3} = c(45 - 81/2) = c(9/2)\). We set this equal to 1 and solve for \(c\).
05

Solve For \(c\)

\(\frac{9}{2}c = 1 \Rightarrow c = \frac{2}{9}\).

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Most popular questions from this chapter

Let \(X_{1}, X_{2}, X_{3}\) be iid random variables each having a standard normal distribution. Let the random variables \(Y_{1}, Y_{2}, Y_{3}\) be defined by $$ X_{1}=Y_{1} \cos Y_{2} \sin Y_{3}, \quad X_{2}=Y_{1} \sin Y_{2} \sin Y_{3}, \quad X_{3}=Y_{1} \cos Y_{3} $$ where \(0 \leq Y_{1}<\infty, 0 \leq Y_{2}<2 \pi, 0 \leq Y_{3} \leq \pi .\) Show that \(Y_{1}, Y_{2}, Y_{3}\) are mutually independent.

Suppose \(X\) is distributed \(N_{3}(\mathbf{0}, \mathbf{\Sigma})\), where $$ \boldsymbol{\Sigma}=\left[\begin{array}{lll} 3 & 2 & 1 \\ 2 & 2 & 1 \\ 1 & 1 & 3 \end{array}\right] $$ Find \(P\left(\left(X_{1}-2 X_{2}+X_{3}\right)^{2}>15.36\right)\).

. Let \(X_{1}, X_{2}, \ldots, X_{n}\) denote \(n\) mutually independent random variables with the moment-generating functions \(M_{1}(t), M_{2}(t), \ldots, M_{n}(t)\), respectively. (a) Show that \(Y=k_{1} X_{1}+k_{2} X_{2}+\cdots+k_{n} X_{n}\), where \(k_{1}, k_{2}, \ldots, k_{n}\) are real constants, has the mgf \(M(t)=\prod_{1}^{n} M_{i}\left(k_{i} t\right)\).

Let \(X\) be a random variable such that \(E\left(X^{2 m}\right)=(2 m) ! /\left(2^{m} m !\right), m=\) \(1,2,3, \ldots\) and \(E\left(X^{2 m-1}\right)=0, m=1,2,3, \ldots .\) Find the mgf and the pdf of \(X\).

. Let \(X\) and \(Y\) have a bivariate normal distribution with parameters \(\mu_{1}=\) \(5, \mu_{2}=10, \sigma_{1}^{2}=1, \sigma_{2}^{2}=25\), and \(\rho>0 .\) If \(P(4
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