# Chapter 5

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See, ~:10'1\ 'S,So 52 I. Random Sample Y\ population - N 'I. l :. ~\.-.(, \J~ \ l,\ e Of \ ~O..Yr\f sample·- "observed values of a sequence of random variables" - n I ~A.Def- If each of the samples has an equal probability of being selected, the sampling is said to be random and the result is said to be a random sample (RS). 1. X 1 , ••• , X 0 r.v.'s constitute a RS of size n if the Xi's are independent and have the same prob. dist. That is, if they are independent and identically dist. (iid). ---- - ~ -- -, 2. IfX 1 , ••. ,~is a RS, then for Xi~(μ, cr 2 ), Ll e1V\ 'i--~ Cul --..; , xi ~ (μ, cr 2 ), i = 1, ... , n ·> ..h-' -:. t et' 1 t V'iD~ \1'1\ta.-r\~ ,-- .) 1 '- l 'b~CAM~e • Cov(Xi, Xj) = 0, i * j. \JJ V\t '<'t E;, -::. ,·+"' -('tJ.. 't\ tiVV\ '(C.\O., 'f u,. \ I d ,. . . -- X · X- , ev,o.. -t\\l \fl ~'<"t>'M JV. W\'r ½ t_ QI/\~ - ' Q'f~ \V\<}._~~~eJ...eV\~· e" \Vl,t\9 . l I. Linear om 1nation \ ( 0 , /J ) '1 t~ ~A. Def- 1. Conditions: X 1 , ••• , X 0 are n r.v.'s and a 1 , ••• , a 0 are n constants. n 2. + a X = " a. X. is a linear combination of nn 11 i=l the X's. B. Expected Value: n n n E[Y] = E[L aiXJ = L aiE[XJ = L i=l i=l i=l
n n C. Variance 1. V(Y) a.a . Cov(x.,x.) LL I J I J i= I j = I 2. IfX 1, • •• , Xn are indep. r.v.'s, then D. \ Sample Mean: 1 n x j X = L n i=l N E. Sample Total: To = L xi + ... + xn i=l n L Note: X is a L.C. with ai = 1/n and T 0 is a L.C. with ai = 1. F. Let Xi ~ (μ,cr 2 ), i = 1, .. . , n. Then - L E[xJ = = μ B ln ni =l n V(x) = 1 n L VAR(x) = n 2 i=l n = L E[xJ = i=l V(T 0 ) = L Var(x) = no 2 _1_ no2 n2 Example: A transformer core is made up of 50 layers of sheet metal (s .m .) and 49 layers of insulating paper (i.p.). Let Z 1 , ..• , Z 50 and W 1 , . . . W 49 be the r. v.'s (independent) for s.m. and i.p. thicknesses, respective ly . Let μ 2 ; = μ 2 = 0.5 mm and a 2 ;2 = a 2 2 = 0. 01 mm, i = 1, .. . , 50; and μ wj = μw = 0. 05 mm and a w/= aw 2 = 0.005 mm, j = 1, .. . , 49. What is the expected thickness and standard de vi a ti on of thickness? 53
54 Solution: Let T = thickness of core => 50 49 T = L zi + L w. Thus, . 1 . J I= J=l 50 49 μT = E[T] = L μ 2 + L μw = 50 (0.5)' + 49 (0.05) i=l i=l 50 49 Oi = L ai + L a~ = 50 (0.01) + 49 (0.005) i=l i=l OT = (o~) 112 = [50 (0.01) + 49 (0.005)] 112 III. Linear Combination: Normal R. '7r. 's. ( S0c:\-°I o V\ CS. S lo) A. Conditions: X 1 , ••• , Xn are n independent r.v .. ' :s ~fr ~om ~ n~o~rm ~ a g_ l ____ ~:- - ~ I ,.. distributions. That is, Xi ~ N(μi,o/). L, "'-e °'-"' (J)'v-1\ '-o \ t>-.. \ '> ' 'fl B. al, ... ,an are n constants. c\ '-r'l01£''1V)O , ,\ ~"' ~\) 'M\ 2 \}Q..\r \O,. \o \ (~ \-S Y\Of""'IY'<J .... \ y = alx + ... + asxn => y ~ N{μy,Oy) n n 2 2 2 where μY = aiμi and ay = ai ai i=l i=l Proof: Beyond the scope of this course. Previous Example Cont'd. Assume Zi - N(μz,cr/), i = 1, .. ., 50; and Wj - N (μw,crw 2 ) j = 1, ... , 49. What is the dist. of T? What is the prob. that the thickness is within 2 std of its mean? Solutions: 1. T ~ N(μy,cr/)