TABLE 13-6-10.13-21. Let X and Y be two random variables eachYtaking three values -1, 0 and 1, and having the joint probability-11.distribution as given in Table 13-6.2Obtain the marginal probability distributions of X and Yand hence their expected values.

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Answered: TABLE 13-6 13-21. Let X and Y be two…

A First Course in Probability (10th Edition)

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Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

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Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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The random variable X takes values -1, 0, 1 with probabilities 1/8, 3/8, 4/8 respectively. a) Write the CDF of X. b) Write the PMF of Y = X² + 2. %3D c) Compute E(Y).
17. Conditional Distributions Consider two random variables X and Y for which we know Pr(X = x) = Pr(X = x|Y = y)for all possible values of x and y. In this case we would say that X and Y are.... Pick ONE option Independent Perfectly correlated Equal to each other More than one of these choices None of these choices
B5. Let X₁, X₂, ..., Xn be IID random variable with common expectation µ and common variance o², and let X = (X₁ + + X₂)/n be the mean of these random variables. We will be considering the random variable S² given by (a) By writing or otherwise, show that S² (b) Hence or otherwise, show that n S² = (x₁ - x)². = Ĺ(X₂ i=1 X₁ X = (X₁-μ) - (x-μ) = Σ(X; -μ)² - n(X - μ)². i=1 ES² = (n-1)0². You may use facts about X from the notes provided you state them clearly. (You may find it helpful to recognise some expectations as definitional formulas for variances, where appropriate.) (c) At the beginning of this module, we defined the sample variance of the values x₁, x2,...,xn to be S = 1 n-1 n i=1 ((x₁ - x)². Explain one reason why we might consider it appropriate to use 1/(n-1) as the factor at the beginning of this expression, rather than simply 1/n. B6. (New) Roughly how many times should I toss a coin for there to be a 95% chance that between 49% and 510/ of my nain toon land Honda?
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- Let x be a discrete random variable with probability distribution function f(x)=k( x2 +20) and x= −1,1,2,3. Find the value of k. Find the Variance of X. B- Let x denote a discrete random variable which can take the values −2,0, and 5. Given that the expectation of X is 8/100 and P(X=−2)=8/20 , find P(X=5).
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The probability generating function for a random variable X is given by: 9x(z) = 1 1 10. 1 I 20 8 -25 + 4 P(X > 10 |X < 22) equals Select one: a. 0.286 b. 0 c. 0.8 d. 1 е. 0.2
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TABLE 13-6 13-21. Let X and Y be two random variables each Y taking three values -1, 0 and 1, and having the joint probability -1 distribution as given in Table 13-6. 2 Obtain the marginal probability distributions of X and Y and hence their expected values. O20