Example 5-1. A random variable X has the following probability function :Values of X, x :13457p(x) :k2k2k3kk22k²7k? + k> find1(i) Find k, () Evaluate Р(X < 6), Р(Х26), and P(0 2'the minimum value of a, and (iv) Determine the distribution function of X.CS Scanned with CamScanner

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Answered: Example 5-1. A random variable X has…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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A random variable X has a following probability function, X 0 1 3 4 5 6 P(X) 0k 2k 2k | 3k k 2k 7k2 + k Find the value of, (i) k (ii) P(X < 6).
According to Bayes' Theorem, the probability of event A, given that event B has occurred, is as follows. P(A) • P(B| A) P(A) • P(B| A) + P (A') •P(B| A') P(A| B) = %3D Use Bayes' Theorem to find P(A B) using the probabilities shown below. P(A) =1. P(A') =, P(B| A) =5, 1 and P(B| A') = %3D %3D 10 %3D 4 The probability of event A, given that event B has occurred, is P(A B) =|. (Round to the nearest thousandth as needed.)
Suppose that X, Y, and Z are jointly distributed random variables, that is, they are defined on the same sample space. Suppose that we also have the following. E(X)=-1 E(Y) = 4 E (Z)=0 Var(X)=20 Var(Y) = 27 Var (Z) = 38 Compute the values of the expressions below. E (5Z-2) = 0 2Z+X -5 Var(2X-3) = 0 E = E (32²) = 0 Explanation Check X O hp Ⓒ2022 McGraw Hill LLC. All Rights Reserved. Term
1. The random variables X and Y describe the outcome of two independent tosses with a dice. Let Z = X+Y be the sum of the results. Determine (a) The probability distributions for X, Y and Z. (b) H(X), H(Y) and H(Z). (c) H(YX), H(Z|X) and H(X|Z). (d) I(X;Y) and I(X; Z).
The cumulative distribution function F(x) of a random variable X is given in the following table: 1 2 3 F(x) 0.5 b 1 a Let a and b are positive constants. If P(1 3) = 0.3. Then a and b equal to: a=0.1, b=0.3 None of these a=0.1, b=0.7 a=0.5, b=0.8
2. Some properties of Expected value and variance of a random variable. a) Assume that X is an arbitrary discrete random variable, and a and b are constant. Using the definitic Show: and E(aX + b) = a · E(X) + b V(aX + b) = a² · V(X ) Stat 3128 Ali Mahzarnia P STAT 3128 Ali Mahzarnia b) Justify the computational formula of Variance of a random variable which is to justify : V(X) = E[(X – µ°] = Ex – µ)P • ptx) = | 2: Here needs justification By Cauchy Schwarz inequality it can be shown that the right hand side is always positive. Analogs expression in Mechanic : Parallel axis theorem Iem = I– md² Moment of Moment of inetria about Inertia of an an axis shifted object about the center of a mass by d from center of mass (a parallel shift) Icm is dispersion around the mean and is like second central moment (variance) I is like second moment if d is mean m is like sum total all the weight of each of the x which all add up to 1 d squared is like squared of mean since we,
A discrete random variable Y has the probability function as given below. Y -2 2 4 P(Y) 0.25 0.50 0.25 If F(Y) denotes the cumulative distribution function for Y, select the correct statement below Group of answer choices F(-3) = 0, F(1)=0.25 and F(3)=0.50 F(-3) = 0, F(1)=0.25 and F(5)=0 F(-3) = 0, F(0) = 0.25 and F(5) = 1 F(-3) = 0, F(1)=0.25 and F(6)=0 F(0)= 0, F(1)= 0 and F(3)=0
Let (X1, X2, X3) be the random outcome of tossing a fair six-faced die three times, and let (Y(1), Y(2), Y(3) be the order statistics from (X1, X2, X3). We have Y(1) = min (X1, X2, X3), Y(3) = max (X1, X2, X3), and Y(1) < Y(2) < Y(3). Find Pr (Y(1) = 3).
Business Calculus 2Section 11.2: Continuous Random Variables
Possible values of X, the number of components in a system submitted for repair that must be replaced, are 1, 2, 3, and 4 with corresponding probabilities 0.35, 0.15, 0.35, and 0.15, respectively. (a) Calculate E(X) and then E(5 - X). E(X) E(5-X) = (b) Would the repair facility be better off charging a flat fee of $70 or else the amount $ 150 (5-X) The repair facility ---Select--- be better off charging a flat fee of $70 because E ]= 150 (5-X)] ? Note: It is not generally true that : E ( =) = E(Y)

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Example 5-1. A random variable X has the following probability function: Values of X, x : 1 3 4 7 p(x) : k 2k 2k 3k 2k2 7k2 + k (i) Find k, (ii) Evaluate P(X < 6), P(X 2 6), and P(0 < X < 5), (iii) If P(X sa) >5, find the minimum value of a, and (iv) Determine the distribution function of X.