Saturday, October 12, 2013

Commication

DMD Assignment 1. (3.7) (a) firstly, suppose it is Normally distribution, we know the mean of the fill out apples weight is 4.2(µ=4.2), the banal deviation is 1.0(?=1.0). suck that the weight of tomato is dependent. The tomatoes ar sold in box of three. ?~N(nµ, ?n?). E(?)=3*µ=12.6 ?(?)= ?3*?=1.7321 (b) P(11.0???13.0)=P{(11-12.6)/1.7321?(?-E(?))/?(?) ?(13.0-12.6)/1.7321} =0.5910-0.1788=0.4122=41.2% 2. (3.9) Let X denote the region increase in the Dow Jones Index. Let Y denote the parting increase in S& angstrom unit;P five hundred index. view X and Y obey a conjugation Normal distribution. The mean of X is 11% (µx=0.11), the regular deviation of X is 13% (?x=0.13). the mean of Y is 10%(µy=0.10), the standard deviation of Y is 12% (?y=0.12). Suppose CORR(X,Y)=0.43. (a)P(X?0.11)=1-P{(X-µx)/ ??(0.11-0.11)/0.13} =1-F (0) =1/2=50% (b)P(X?-0.11)=P{(X-µx)/?x?(-0.11-0.11)/0.13} =F(-1.69)=0.0455=45.5% (c)P(0?Y?0.15)=P{(X-µy)/ ?y?(0.15-0.10)/0.12}-P{(X-µy)/?y?(0-0.10)/0.12} =0.6628-0.20=46.3% (d) Suppose A is the portfolio of Dow Jones index and S&P index. So, E(A)=0.3*0.11+0.7*0.10=0.103 Var(A)=0.32*0.132+0.72*0.122+2*0.3*0.7*0.43*0.13*0.12=0.01138 ?(A)=0.1067 (e) (X-Y) in any case obeys Normal distribution. µ(x-y) =0.01, ?(x-y) =0.13. (X-Y)~N (0.01, 0.
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13) P{ (X-Y) ?0}=1-P{ (X-Y)- µ(x-y)/ ?(x-y)?(0-0.01)/0.13}=1-F(-0.0415)=0.532=53.2% 3. (3.15) (a) We quite a little pick up from the content of the problem, assume and infer that the quantity of X ~ binominal (n, p). So, assume S=X1+X2++Xn, n=2500, p=0.1. E(S) = n*p=250, ?s= 15 (b ) Revenue: assume Y be the returned revenue ! item. Y~ Binomial (n, p). E( Y) =500*250-Sn*500=125,000-500*Sn =1125000; ?y=335.4 (c ) assume the Z be the item, P(Z?1,300,000) =1-P(Z?1,300,000-1125000/335.4) =1-P(521.77)=0 4.(3.18) (a) The assumptions are that the probability of hazardous problem happened is in depended from each other, and also obey the identically distribution. The assumptions appear to be...If you want to get a full essay, put up it on our website: BestEssayCheap.com

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