EE131A hw8sol W15
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EE 131A Homework #8 Solution Winter 2015 K. Yao 1. The relative frequency f A (n) has mean 2/10 and variance (1/n)p(1 - p)=0.16/n . P [|f Z (n) - 0.2| < 0.02] = P [0.18 <f A (n) < 0.22] = P 0.18 - 0.20 0.16/n < f A (n) - 0.1 0.16/n < 0.22 - 0.20 0.16/n =1 - 2Q 0.02 0.16/n =0.95 ⇒ Q √ n 20 =0.025 ⇒ √ n 20 =1.95 ⇒ n = 1521 . 2. We have n = 80 and λ =1/4 . Then E[S n ]= nλ = 20 and V AR(S n ) = 20 . A comparison of the exact values of P (S n = k) and the above approximations is given below. Tabulate k = 1 : 1 : 30, P (S k )=(λ k /k!)exp(-λ), where λ = 20, and P approx (k) = (1/ √ 40π)exp(-(k - 20) 2 /40). 1.00000000000000 0.00000004122307 0.00001073710917 2.00000000000000 0.00000041223072 0.00002707757482 3.00000000000000 0.00000274820483 0.00006495572526 4.00000000000000 0.00001374102415 0.00014822122007 5.00000000000000 0.00005496409660 0.00032172781337 6.00000000000000 0.00018321365533 0.00066428142199 7.00000000000000 0.00052346758665 0.00130467061811 8.00000000000000 0.00130866896663 0.00243744560806 9.00000000000000 0.00290815325917 0.00433165899778 10.00000000000000 0.00581630651835 0.00732249128096 11.00000000000000 0.01057510276063 0.01177466994075 12.00000000000000 0.01762517126771 0.01801042233608 13.00000000000000 0.02711564810417 0.02620500987264 14.00000000000000 0.03873664014882 0.03626853674196 15.00000000000000 0.05164885353176 0.04774864115336 16.00000000000000 0.06456106691470 0.05979670798364 17.00000000000000 0.07595419637023 0.07123260215139 18.00000000000000 0.08439355152248 0.08071711293577 19.00000000000000 0.08883531739209 0.08700369673863 20.00000000000000 0.08883531739208 0.08920620580764 21.00000000000000 0.08460506418294 0.08700369673863 22.00000000000000 0.07691369471176 0.08071711293577 23.00000000000000 0.06688147366240 0.07123260215139 24.00000000000000 0.05573456138533 0.05979670798364 25.00000000000000 0.04458764910827 0.04774864115336 26.00000000000000 0.03429819162174 0.03626853674196 27.00000000000000 0.02540606786796 0.02620500987264 28.00000000000000 0.01814719133426 0.01801042233608 29.00000000000000 0.01251530436845 0.01177466994075 30.00000000000000 0.00834353624564 0.00732249128096 3. a. f X (x 1 ,x 2 )= 1 2π |Λ| 1/2 e -(x-μ) T Λ -1 (x-μ)/2 . |Λ| =6 - 1=5. Λ -1 = 1 5 3 1 1 2 = 0.6 0.2 0.2 0.4 . 1
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EE131A hw8sol W15
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EE 131A Homework #8 Solution Winter 2015K. Yao
1. The relative frequency fA(n) has mean 2/10 and variance (1/n)p(1 p) = 0.16/n .
P [|fZ(n) 0.2| < 0.02] = P [0.18 < fA(n) < 0.22]
= P
[0.18 0.20
0.16/n
