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Binomial and geometric distributions 
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#1
Dec2907, 02:08 PM

P: 773

i was doing some exercises nut i'm not sure if my answers are correct
1) X~B(5,0.25) i have to find: a) E(x^2) and my answer was 2.5, is this correct? b) P(x(>or=to)4) and my answer was 0.0889, is this correct? 2) X~Geom(1/3) i have to find: a) E(x) my answer is 1/3 b) E(x^2) c) var(x) d) P(X=4) my answer is 0.988 e) P(X>2) my answer is..... (1/18) i dont know how to compute b) and c) and i think d) is wrong and i know e) is wrong. can someone help please. Thank you very much 


#2
Dec2907, 04:16 PM

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P: 39,559

Would "yes" or "no" answers help you? Show what you did to get those answers and we can point out exactly where, if at all, you went wrong.



#3
Dec3007, 05:56 AM

P: 773

my working out is:
1) a) E(x^2)= var(x)+E(x)^2 = np(1p+np)= 5(1/4)(11/45/4)=2.5 b) P(x(>or = to)4)= 1P(x=3)P(x=2)P(x=1)P(x=0) using the formula P(x=m)=(nCm)P^m(1p)^(nm) =1(15/1024)(135/512)(405/1024)(243/1024)=91/1024=0.0889 2) a)E(x)=1/p=1/(1/3)=3 b) and c) i have no idea d) using the formula P(x=m)=P(1P)^(m1) P(x=4)=1/3(11/3)^3=8/81=0.988 e) P(x>2)=1P(x(<or=to)2)=1P(x=2)P(x=1)P(x=0) =1(1/3)(2/3)(1/3)(2/3)^0(1/3)(2/3)^1=1/18 thank u very much 


#4
Jan208, 04:03 PM

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P: 2,481

Binomial and geometric distributions
1a is correct except you meant +5/4) not 5/4).
For 1b can you write out the nCm terms? 2a is correct. For 2b use the same formula as in 1a. 2c: Var(Geometric(p)) = (1p)/p^2; see http://en.wikipedia.org/wiki/Geometric_distribution 2d: 1/3(11/3)^3 = 8/81 < 0.1 so it cannot be > 0.9 (i.e., you have made a fraction computation error). 2e: if the exponent is k1 then k = 1, 2, .... OTOH, if the exponent is k, then k = 0, 1, 2, ... In either case, exponent term > 0 so it cannot be 1. See http://en.wikipedia.org/wiki/Geometric_distribution 


#5
Jan908, 01:15 PM

P: 773

for 1b:
my working including the ncm terms: P(x(>or = to)4)= 1P(x=3)P(x=2)P(x=1)P(x=0) using the formula P(x=m)=(nCm)P^m(1p)^(nm) =1[(5C3)(1/4)^3(11/4)^2][(5C2)(1/4)^2(11/4)^3][(5C1)(1/4)(11/4)^4][(5C0)(1/4)^0(1/1/4)^5] =1(15/1024)(135/512)(405/1024)(243/1024) =91/1024=0.0889 for 2e: i looked at the link but i still havent figured out what i had done wrong with my working out. thank u very much for everything else. 


#6
Jan908, 01:52 PM

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P: 2,481

My guess is you should have 1  (1/3)(2/3)^2  (1/3)(2/3)^1  (1/3)(2/3)^0, but you should verify that. For 1b, you have: 1  Binomial[5, 3] (1/4)^3 (1  1/4)^2  Binomial[5, 2] (1/4)^2 (1  1/4)^3  Binomial[5, 1] (1/4) (1  1/4)^4  Binomial[5, 0] (1/4)^0 (1/1/4)^5, where Binomial[n,m] stands for nCm. Binomial[5, 3] = Binomial[5, 2] = 10 Binomial[5, 1] = 5 Binomial[5, 0] = 1 so the expression is: 1  10 (1/4)^3 (1  1/4)^2  10 (1/4)^2 (1  1/4)^3  5 (1/4) (1  1/4)^4  1 (1/4)^0 (1  1/4)^5 =1  10 (1/4)^3 (3/4)^2  10 (1/4)^2 (3/4)^3  5 (1/4) (3/4)^4  1 (1/4)^0 (3/4)^5 = 1  10 (9/1024)  10 (27/1024)  5 (81/1024)  1 (243/1024) = 1 (90 + 270 + 405 + 243)/1024 = 1  1008/1024 = 1  504/512 = 8/512 = 1/64 = 0.015625 


#7
Jan908, 02:42 PM

P: 773

oh okay i see so the formula i used will only work if k=1,2,3... (as u said before)
so if i used the formula P(x=m)=P(1P)^(m1) and did 1P(x=2)P(x=1) =1(1/3)(2/3)(1/3)(2/3)^0 =4/9 would that work? 


#9
Jan1208, 03:09 PM

P: 773

no it should only be for x=2 and x=1 as these are the ones strictly less than three



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