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Suppose a state allows individuals filing tax returns to itemize deductions only if the total of all itemized deductions is at least $\$ 5,000 .$ Let $X$ (in 1000 s of dollars) be the total of itemized deductions on a randomly chosen form. Assume that $X$ has the pdf$$f(x ; \alpha)=\left\{\begin{array}{cl}{k / x^{\alpha}} & {x \geq 5} \\ {0} & {\text { otherwise }}\end{array}\right.$$(a) Find the value of $k .$ What restriction on $\alpha$ is necessary?(b) What is the cdf of $X$ ?(c) What is the expected total deduction on a randomly chosen form? What restriction on $\alpha$ is necessary for $E(X)$ to be finite?(d) Show that $\ln (X / 5)$ has an exponential distribution with parameter $\alpha-1$ .

(a) k = (? 1)5$^{?1}$, ? > 1 (b) F(x) = 1 (5/x)$^{?-1}$ for x ? 5 (c) 5(? 1)/(? 2), ? > 2

Intro Stats / AP Statistics

Chapter 3

Continuous Random Variables and Probability Distributions

Section 9

Supplementary Exercises

Continuous Random Variables

Temple University

Piedmont College

Oregon State University

University of St. Thomas

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were given the probability density function shown here for part A were asked to solve for K to do this, we know that if we integrate over the entire distribution, we should get one. So this distribution is only non zero for X is at least five. So we can integrate from X equals five to infinity. And so we get K is equal to Alfa minus one times five to the exponents, Alfa minus one. And this only works if Alfa is greater than one. That's because this integral Onley converges. If the total value in the exponents here is negative, if X approaches infinity, we must have a negative exponents for this. Integral to converge for Part B were asked to find the c d f of X. So that's given by the integral from five. Two x. And this comes out to one minus X over five to the exponents. Alfa minus one. That's for X is at least five, and it is equal to zero for exes. Less than five Next for part C were asked to find the expected value. So this is the integral from five to infinity of X times A FedEx so we have K times X to the exponents to minus alpha over two minus Alfa evaluated on five to infinity. Now, this integral is only going to converge If the argument and the exponents here is negative, which means that we need Alfa to be bigger than two. And this comes out to Alfa minus one over. Alfa minus two times five and last for part B were given a transformation. Why is equal to the natural law algorithm of x over five and were asked to show that this is an exponential distribution with parameter alfa minus one. Now we can show this using the CDF. So the CDF for why secret of the probability that natural algorithm of X over five is that most why now This is the CDF for X, evaluated at five times E to the exponents. Why now We've solved for X o CDF in Part B. I believe so. We simply have to put this value in for X in the CDF and we get the following. This is equal to one minus e to the negative alfa minus one times why Also the domain for X was X is greater than five greater than or equal to five. That means that why is greater than or equal to zero. Because why is given by this equation? So if we put five in place of X here, natural lug of one is zero. And as we can see, this is the CDF for the exponential distribution with parameter gamma equals Alfa minus one. Sorry, that's parameter Lambda equals Alfa minus one.

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