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Three fair coins are labeled with a zero (0) on one side and [#permalink]
17 Jan 2008, 09:02

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Three fair coins are labeled with a zero (0) on one side and a one (1) on the other side. Jimmy flips all three coins at once and computes the sum of the numbers displayed. He does this over 1000 times, writing down the sums in a long list. What is the expected standard deviation of the sums on this list? A) 1/2 B)3/4 C)sqrt3/2 D)sqrt5/2 E) 5/4

I don't know how to solve this question. Please provide full explanation.

Walker, can you point out the flaw in my approach. I know from binomial dist parameters that my approach is wrong... but i have forgotten why this approach is worng.

For the trial, prob mass funtion for all outcomes = 1/2 so E[x] = 0*1/2 + 1*1/2 ... first moment E[x^2] = 0^2 * 1/2 + 1^2 *1/2 ... second moment

E[x^2] - (E[x])^2 = 1/4 ... second central moment = var[x]

Walker, can you point out the flaw in my approach. I know from binomial dist parameters that my approach is wrong... but i have forgotten why this approach is worng.

I did not dig deep, but maybe you forgot that x is "the sum of the numbers" and E(x)=3C0*0*1/2^3+3C1*1*1/2^3+3C2*2*1/2^3+3C3*3*1/2^3=0+3/8+6/8+3/8=12/8=3/2 E(x^2)=3C0*0*1/2^3+3C1*1*1/2^3+3C2*4*1/2^3+3C3*9*1/2^3=0+3/8+12/8+9/8=24/8=3 var[x]=E[x^2] - (E[x])^2=3-(3/2)^2=3/4

Thanks walker... i had forgotten that x was the sum of numbers....

walker wrote:

bhushangiri wrote:

Walker, can you point out the flaw in my approach. I know from binomial dist parameters that my approach is wrong... but i have forgotten why this approach is worng.

I did not dig deep, but maybe you forgot that x is "the sum of the numbers" and E(x)=3C0*0*1/2^3+3C1*1*1/2^3+3C2*2*1/2^3+3C3*3*1/2^3=0+3/8+6/8+3/8=12/8=3/2 E(x^2)=3C0*0*1/2^3+3C1*1*1/2^3+3C2*4*1/2^3+3C3*9*1/2^3=0+3/8+12/8+9/8=24/8=3 var[x]=E[x^2] - (E[x])^2=3-(3/2)^2=3/4

Re: Three fair coins are labeled with a zero (0) on one side and [#permalink]
19 Sep 2013, 12:41

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