Register to reply

Binomial and geometric distributions

by sara_87
Tags: binomial, distributions, geometric
Share this thread:
sara_87
#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
Phys.Org News Partner Science news on Phys.org
Security CTO to detail Android Fake ID flaw at Black Hat
Huge waves measured for first time in Arctic Ocean
Mysterious molecules in space
HallsofIvy
#2
Dec29-07, 04:16 PM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,345
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
#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
#4
Jan2-08, 04:03 PM
Sci Advisor
HW Helper
EnumaElish's Avatar
P: 2,482
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
#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
#6
Jan9-08, 01:52 PM
Sci Advisor
HW Helper
EnumaElish's Avatar
P: 2,482
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
#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
#8
Jan11-08, 12:58 PM
Sci Advisor
HW Helper
EnumaElish's Avatar
P: 2,482
Shouldn't x=3 be part of your formula?
sara_87
#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
#10
Jan14-08, 11:16 AM
Sci Advisor
HW Helper
EnumaElish's Avatar
P: 2,482
Your OP stated "P(x(>or = to)4)," doesn't that exclude x=3?


Register to reply

Related Discussions
How is the negative binomial the inverse of the binomial distribution? Set Theory, Logic, Probability, Statistics 1
Advice on Exponential, Binomial, & Normal Distributions Set Theory, Logic, Probability, Statistics 3
Geometric series/geometric progression Precalculus Mathematics Homework 2
Is this a geometric progression? or binomial expansion i'm not sure how they got 10^9 Calculus & Beyond Homework 2
Pdf of the sum of two distributions Calculus & Beyond Homework 0