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Please answer asap 1(b). The integral of the p.d.f. as displayed over the support is 2; the x-axis should be scaled down by 1/2 (so the vertex should (1,3/4) instead of (1,3/2) as displayed). Thank You

Please answer asap 1(b). The integral of the p.d.f. as displayed over the support is 2; the x-axis should be scaled down by 1/2 (so the vertex should (1,3/4) instead of (1,3/2) as displayed). Thank You

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter7: Analytic Trigonometry
Section7.6: The Inverse Trigonometric Functions
Problem 91E
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Please answer asap

1(b). The integral of the p.d.f. as displayed over the support is 2; the x-axis should be scaled down by 1/2 (so the vertex should (1,3/4) instead of (1,3/2) as displayed).
Thank You

Problem 1. Continuous Distributions. For the following, provide a random variable from the formula sheet and (possibly) a transformation of that variable that produces a distribution satisfying the given constraints. For example, your answer could look something like the random variable X = g(Z) where Z~ {distribution from sheet} and g(x) = {some function}. a) Find a distribution that satisfies the following: Supp(X) = {x:x>0} E[X] = 3 V[X] = 5. b) The distribution of X has a p.d.f. given by the following parabola:
Transcribed Image Text:Problem 1. Continuous Distributions. For the following, provide a random variable from the formula sheet and (possibly) a transformation of that variable that produces a distribution satisfying the given constraints. For example, your answer could look something like the random variable X = g(Z) where Z~ {distribution from sheet} and g(x) = {some function}. a) Find a distribution that satisfies the following: Supp(X) = {x:x>0} E[X] = 3 V[X] = 5. b) The distribution of X has a p.d.f. given by the following parabola:
on support p.d.f. 3/2- 1- 1/2- 0- 1/2 Supp(X) = {x : x = [0, 2]}. c) The distribution of X has the following p.d.f.: Supp(X) = {x : x > 0} 1 fx(x) = 2πT 3/2 xe 2V27. 2V2T -~ 2
Transcribed Image Text:on support p.d.f. 3/2- 1- 1/2- 0- 1/2 Supp(X) = {x : x = [0, 2]}. c) The distribution of X has the following p.d.f.: Supp(X) = {x : x > 0} 1 fx(x) = 2πT 3/2 xe 2V27. 2V2T -~ 2
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