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- 3. Consider n independent observations X1,..., X, of a Poisson random variable with param- eter A. (a) Use Chebyshev's inequality to evaluate the minimum number of observations needed to achieve п 1 P(|Xn– A| 0.9, where Xn ΣΧ. п i=1 (b) Repeat the computations using the Central Limit Theorem. Note: Remember that the Poisson distribution is a discrete distribution with PMF fx(x;A) = х! where xe {0,1,2,3, .}, and 1 > 0 and hence E(X) = A , and Var (X) = A.A lot of 25 light bulbs consists of N₁ defectives and N₂ working bulbs. The buyer accepts the lot of N₁ + N₂ = 25 bulbs if the number of defectives, X among n = 5 items taken at random and without replacement from the lot is less than or equal to 1. The operating characteristic curve is defined as the probability of accepting the lot, namely OC (p) = P(X ≤ 1) = P(X = 0) + P(X = 1), where p = N₁/25 is the probability of a bulb being defective. (a) Show that the probability of the picking exactly X defectives follows a hypergeometric distribution. (b) Determine the operating characteristic curve for p = 0.04, p = 0.08, p = 0.12 and p = 0.16. (c) Plot, using R, the operating characteristic curve for N₁ = 1 to N₁ = 4.Could you solve (i), (ii), (iii)? Thank you.
- 14 Let X1 X, ...X, be 'n' random variable which are identically distributed such that with a prob. of I/2 Xx= 1 = 2 with a prob. of 1/3 =-1 with a prob. of 1/6. Find (a) E [X1+ X2 + .. .. X/ (b) E [X} + X3+ . .. X2IA.2) The time between successive customers coming to the market is assumed to have Exponential distribution with parameter lambda. a) If X1, X2. ..., Xp are the times, in minutes, between Successive customers selected randomly, estimate the parameter of the distribution. b) The randomly selected 15 times between successive customers are found as 1.8, 1.2, 0.8, 1.4, 1.2, 0.9, 0.6, 1.2, 1.2, 0.8, 1.5, 1.8, 0.9, 1.5 and 0.6 mins. Estimate the mean time between successive customers, and write down the distribution function. c) In order to estimate the distribution parameter with 0.4 error and 4% risk, find the minimum sample size.A continuous random variable X has the distribution function if X 3 Find K and f (x), P.d.f.
- Let {Xn}-1 be a sequence of random variables that converges in probability n=1 Prove that the limiting random variable is unique P-a.s., i.e., if Xn 4 X and Xn 4 Y, then P(X = Y) = 1.Theorem 27 (CENTRAL LIMIT THEOREM (CLT)) Let X1, X2,. .. be independent identically distributed random vari- ables with finite expectation and finite variance. Let S, => X4. 2=1 Then Sn – ESn 30), then the cdf of the rescaled random variable S, is approximately equal to the cdf of a standard-normal random variable Z ~ N(0,1), i.e. Sn – ES, P 3000) (i) with the Markov inequality, (ii) with the normal distribution.3. Let X1, X2,..., be a sequence of independent and identically distributed random variables. Let N be Poisson distributed with mean u and is independent of the X,'s. Define N W =EX;. i=1 We define W = 0 if N = 0. (a) Suppose each X; is normally distributed with mean 0 and variance 1. Work out the moment generating function for W given N. [4 marks] (b) Show that the moment generating function of W is given by Mw (t) = exp(ue"2 - ), t eR. [5 marks] (c) Calculate the mean and variance of W. [5 marks] (d) Now consider Z = NX1. Find the mean and variance of Z. [6 marks]
- need correct and urgentLet X and Y denote the weights (in grams) of apples which are imported from two different locations: Peshawar (Pakistan) and Zanzibar (Tanzania), respectively. Suppose that X has Normal distribution N(148, 3); and Y has Normal distribution N(151, 4). One apple is randomly selected from each city. We assume independence between X and Y. • [[a)] Find the probability that the weight of the apple from Peshawar exceeds the weight of the one from Zanzibar • [(b)]A random sample of 36 apples is selected from those imported from Peshawar. Compute the probability that their sample mean will fall between 150 and 152. Paragraph BI Path: p pe here to search II !!!