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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
\(c = \frac{5}{9}\)

Step by step solution

01

Setup the Integral

The first step is to set up the integral from 0 to 3 for the given function. This is written as \(\int_{0}^3 c x(3-x)^4 \, dx = 1\). This equation represents the total probability equals to 1.
02

Solve the Integral

Next, we need to solve the integral part separately without considering the constant \(c\) because \(c\) is a constant that will not affect the operation of the integral. The integral of \(x(3-x)^4\) dx between 0 and 3 is \(-\frac{1}{5}x(3-x)^5 |_{0}^3 + \frac{3}{10}(3-x)^5 |_{0}^3\). After evaluating the integrals at the limits 0 and 3, we get \(\frac{9}{5}\).
03

Solve for the Constant

After obtaining the result of the integral, insert it back into the equation and solve for \(c\). Therefore, \(c \times \frac{9}{5} = 1\). So, solving for \(c\), we get \(c = \frac{5}{9}\).

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Most popular questions from this chapter

Consider a standard deck of 52 cards. Let \(X\) equal the number of aces in a sample of size 2 . (a) If the sampling is with replacement, obtain the pmf of \(X\). (b) If the sampling is without replacement, obtain the pmf of \(X\).

Let \(X_{1}\) and \(X_{2}\) be two independent random variables. Suppose that \(X_{1}\) and \(Y=X_{1}+X_{2}\) have Poisson distributions with means \(\mu_{1}\) and \(\mu>\mu_{1}\), respectively. Find the distribution of \(X_{2}\).

Three fair dice are cast. In 10 independent casts, let \(X\) be the number of times all three faces are alike and let \(Y\) be the number of times only two faces are alike. Find the joint pmf of \(X\) and \(Y\) and compute \(E(6 X Y)\).

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Let \(X\) be \(N(0,1)\). Use the moment generating function technique to show that \(Y=X^{2}\) is \(\chi^{2}(1)\)
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