Binomial and geometric distributions


by sara_87
Tags: binomial, distributions, geometric
sara_87
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#1
Dec29-07, 02:08 PM
P: 774
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
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HallsofIvy
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#2
Dec29-07, 04:16 PM
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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.
sara_87
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#3
Dec30-07, 05:56 AM
P: 774
my working out is:
1) a) E(x^2)= var(x)+E(x)^2 = np(1-p+np)= 5(1/4)(1-1/4-5/4)=2.5
b) P(x(>or = to)4)= 1-P(x=3)-P(x=2)-P(x=1)-P(x=0)
using the formula P(x=m)=(nCm)P^m(1-p)^(n-m)
=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(1-P)^(m-1)
P(x=4)=1/3(1-1/3)^3=8/81=0.988
e) P(x>2)=1-P(x(<or=to)2)=1-P(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

EnumaElish
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#4
Jan2-08, 04:03 PM
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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)) = (1-p)/p^2; see http://en.wikipedia.org/wiki/Geometric_distribution

2d: 1/3(1-1/3)^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 k-1 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
sara_87
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#5
Jan9-08, 01:15 PM
P: 774
for 1b:
my working including the ncm terms:
P(x(>or = to)4)= 1-P(x=3)-P(x=2)-P(x=1)-P(x=0)
using the formula P(x=m)=(nCm)P^m(1-p)^(n-m)
=1-[(5C3)(1/4)^3(1-1/4)^2]-[(5C2)(1/4)^2(1-1/4)^3]-[(5C1)(1/4)(1-1/4)^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.
EnumaElish
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#6
Jan9-08, 01:52 PM
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Quote Quote by sara_87 View Post
for 2e:
i looked at the link but i still havent figured out what i had done wrong with my working out.
You have the exponential term -1, which shouldn't be there under either definition of the geometric distribution (referenced on the page I linked to).

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
sara_87
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#7
Jan9-08, 02:42 PM
P: 774
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(1-P)^(m-1)
and did 1-P(x=2)-P(x=1)
=1-(1/3)(2/3)-(1/3)(2/3)^0
=4/9
would that work?
EnumaElish
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#8
Jan11-08, 12:58 PM
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Shouldn't x=3 be part of your formula?
sara_87
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#9
Jan12-08, 03:09 PM
P: 774
no it should only be for x=2 and x=1 as these are the ones strictly less than three
EnumaElish
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#10
Jan14-08, 11:16 AM
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Your OP stated "P(x(>or = to)4)," doesn't that exclude x=3?


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