Ans-Q4.pdf - IME602 Probability and Statistics Quiz-4 An A...

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Unformatted text preview: IME602: Probability and Statistics Quiz-4, October 27, 2016 An A c0 t‘< Duration: 1 hour Maximum Marks: 10 Name: ....................................... Roll No: This is a closed book/note examination. Useful information, whenever necessary, is provided along with the question. Attempt all questions. 1. Let X1 and X 2 denote your quiz-1 and 2 marks. Let us assume X1 and X 2 to be independent and uniformly distributed in [0,10]. Let Y = X1 + X 2 denote the total marks in quiz-1 and 2. Derive the density function of Y. Represent fy (y) pictorially. [Marks: 3] Info: If X1 and X2 are independent, fx1+x2 (y) = L: fx1 (x) sz (y — x)dx. 2. Let C denote the melting point of a newly developed material. Measurements of c, due to error, are of the form: Mi = c + 61-, where ei’s are the errors. Let us take ei’s to be iid random variables with mean 0 and variance 3. Then Mi’s are iid random variables too. We consider IV! = Z?=1Mi/n as the measurement for c. Using Chebyshev’s inequality, in quiz-3, we got 11 2 120 to be 90% sure that [V] is in c i 0.5. Using the central limit theorem, find the new minimum of n to ensure the same accuracy. [Marks: 2] Note: According to the central limit theorem, X = Z?=1Xi/n, where Xi’s are iid random variables with mean ,u and variance 02, approximately follows N (u, 02 / n) for large n. 3. We want to estimate the distribution function value of an arbitrary random variable at an arbitrary point, i.e., we want to estimate 9 = Fx(a) for any X and a. We have a random sample of X, denoted by (X1,X2, , Xn). Then it makes sense to estimate 0 by the proportion of Xi’s that do not exceed a, i.e., 0 = (221:1 1x,-5a)/na where 1Xi5a = 1 if X,- S a, otherwise 1X,-sa = 0. Study the properties of g (as an estimator of 0). [Marks: 2.5] Hint: 1xi5a’5 are iid random variables. Identify their mean and variance first. 4. Estimate k and p for N e gB in(k, p) using the method of moments. [Marksz 2.5] Hint: Geometric(p) has mean 1 / p and variance (1 — p) / p2. I am talking about geometric distribution when the question is about negative-binomial distribution! i~ Km u(o)1o) (W3 X2” 0 (0,10) m imfldmAeW. “:2 w x1. (mama, Y Wm mm 'm wan. HM, #chzo A 4 ¢Y0'“°i‘-““@ Pom % € [0,?»011 by 41M) comramkizm {lommflm N063 «Mm =r{f OS) 2 "1% {k 1 e ‘0’”) Q owotmi’it HI 1,; 1 2:; m x A L_C \M‘m (gum) — WWW (0’3"!) ‘50 W. = VA“; M awm W (30%me 16% 3600.10] Q3? H gem; One (AM (imam ‘HAK 5W1 WWW {Mica Fa“ Ward/3b 8M“ \02 <1 Tka in? 04m 710%»,qu tvm 647 \o'na (xx/3 AM OWWMDUt V3 COMQ} \ x ‘ swam} 54%“ \pe mwfiw in mm empty. WW § 0 ID m» “‘“k¢N}—-~“" 2. MMJz; am; m (Mg): 76» TAU“ AV FILL C(MJTM’I Aim/f afifificximfida y‘allcN/s 3 ”V4“ 411 2'. _~ ‘ EMK M... :2. ‘ Ilmdrtw, 7%" Qfiufl’é n“ h ”<413/4)‘ ""“""‘””—‘ x [email protected] wflmm‘P(C—0§§R<c+05)>/09 __________________ @ E ‘P *0‘6’< F4~c£ o-s >>0 < 3A 4?; I22 , 9 W 1660.1) :: 0'5 "05‘ > 0.6) ‘‘‘‘‘‘‘‘‘‘ KZ‘ " “mywm ’ C? r 0; N : l~2<f>(~ M3205 By W a r - ¢(——J:/;> 5°'05%45<-16453(‘§“o~n~(¢m¢~“ @ .a ()‘S‘ ‘ : - W7: $ ‘1‘64S ms 69' [,3 Mum-H3 MUM/32m 1" K. \4 i J 1'6“ W3 E W :> “E 324% X; has :5me Al‘m‘m'LrA/ydh M ”MAM/$410. Av», PM“ ~ 2? EMMA]: 5%” 6M We Mag {0 Mug; ZSHOygfi‘ZH-fi»; 772 on «we 4471M flzafiarl—I'w. x.“ W RA mmgm“ Vomubw), ._ in“ )/H wrmqw c9: ting, (-1 ’0 >=I~fi<w {g V[&(1X£5a)2 ECfiO-gl—[iQ/D} #5 M m &¢;mfl;»rc #8. Wt 64/»? demM/JL 2,5 414 ”SE (:0 film” 012 Inhale. C) do‘nbim‘mca .‘ 537162 m ® => E‘*1:'bk,EYY|1: V10. » —» ' , ~ \mcx): “$53100: ko-bfi/bg‘; , , — -- ‘ * “ @ To mmm Laws, ’9 04; ngLEvaMB ‘03 4m wamfl fi wowmél {mm MW‘. 3;: 1/; - ~~»@ Since 4kg «bow WVJHO'M Mm Lo Ml- M‘WUW} XE MHz for! '2 mg I; we efwr-fka w {2 - (Ts/“1+ EC‘h Wok: LB ane (McuW’m Mum th Mtr‘tuncz if NC; 3??? (Ida) Qc‘wdlgg, {MI Mm) La. 1 MK/M be fiivw. ...
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