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![(18) Let X = (X₁, X2, X3, X₁)' have probability distribution function
20!
f(T1, T2, T3, T4)
#₁!₂!3!₁!
Find
(0.3) ¹ (0.1) 2 (0.4) ³ (0.2) x₁ + x₂ + x3 + ₁ = 20
(i) Pr(X₁ = 6, X₂ = 2, X3 = 8, X₁ = 4)
(ii) E [X2, X₁ X1 = 4, X3 = 6]
(iii) the distribution of Z = X₁ + 4X₂ - 2X3 +21](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0dc7fdfc-2d05-4936-8b9d-5ee1ad16be99%2F1c9ea65a-421a-48f6-afad-2701c0a3847b%2F2cgt8bn_processed.jpeg&w=3840&q=75)
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- Consider the three functions I. f(x)= {1-x if 0 < x < 2 {0 otherwise II. f(x)= {1+x if 0 < x < 2 {0 otherwise III. f(x)= {(3/27)x^2) if 0< x < 3) {0 otherwise Which of these above functions can be probability distributions? And why?(1) Let X = b(16,- find E(4-3x) and distribution function.The lung cancer hazard rate λ(t) of a t-year-old male smoker is such that λ(t) = 0.027 + 0.00025(t-40)2, and t >= 40. Assuming that a 40-year-old male smoker survives all other hazards, (a) what is the probability that he survives to age 50 without contracting lung cancer?(b) what is the probability that he survives to age 60 without contracting lung cancer?
- 2. Let fx.y (x, y) be the joint probability function for the random vector (X, Y) (discrete or continuous). Then fx (1) = 0 implies fx.x(1, y) = 0 for all y E R. True FalseLet X be a positive random variable (i.e. P(X 1/E(X) (b) E(-log(X)) 2 -log(E(X)) (c) E(log(1/X)) > log(1/E(X)) (d) E(X³) > (E(X))³3. Show that the following are probability density functions: 1 (a) f(x) = -2*, x = 1, 2, ..., N, and zero elsewhere 2N+1-2 (b) f(x)=p(1-p)*, x=0, 1, 2, ..., and zero elsewhere; 0Let X₁, ..., X₂ be a random sample from an exponential distribution with the probability density function ƒx (2) = e-*/², for x > 0, and zero elsewhere. Which statement is correct. (1) X² is a consistent estimator of 3². (II) X is a consistent estimator of 3². (III) X(1 + X) is a consistent estimator of 32. (IV) All of the above. O (1) (IV) O (III) ☆Q 4. (a) (i) Show that if U and V are independent random variables having chi-square distributions with vị and v2 degrees of freedom respectively, then X = is a random variable having an F distribution with vị and v2 degrees of freedom , i.e. a random variable whose probability density function is given by f (x) = I (4)r (4) (v½. 1+-x (ii) Using (i), or otherwise, show that if S? and S are respective variances of independent random samples of sizes nį and n2 from normal populations with variances a? and ož, then Y %3D is a random variable having an F distribution with n1 – 1 and n2 – 1 degrees of freedom. (b) Let X be a random variable modeling the heights of all adults in the population where o² is known. (i) For a random sample X1, X2, ..., Xn, state the sampling distribution of Š when n > 30. (ii) A random sample of 50 adults gives a mcan of 174.5 cm. Given that o = 6.9 construct a 98% confidence interval for µ, the mean height of all adults. Based on the confidence interval…5. Suppose that X is a discrete random variable with probability density function p(x) = cx', (a) Find the value of c. (c) Find Var(X). x = 1, 2, 3, 4. (b) Find E(X) and E(X(X – 1)).Use the probability generating functions to derive the formula for the average small components size (s) when GC is present: u² g (1) (s) = 1 + go(u) [1-g(u)]' where u is the solution of u = 9₁ (u). (Essentially to derive the same result we obtained not using generating function method earlier.1. Let f be the probability density function of a continuous random variable X whose range is (-0, 00), where if a sxsb f(x) ={32 0 otherwise. (a) Explain why a, hence b, must be nonnegative. Graph the function f. (b) Find the value of b'-a. 7 (c) If it is known that P(X 8).Q6. (a) The continuous random variable X has a probability density function given by 2 1 1SEE MORE QUESTIONSRecommended textbooks for youMATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons Inc
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