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# Three fair coins are labeled with a zero (0) on one side and

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Three fair coins are labeled with a zero (0) on one side and [#permalink]

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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)$$\sqrt{3}$$/2
(D)$$\sqrt{5}$$/2
(E) 5/4

[Reveal] Spoiler:
I don't know how to solve this question. Please provide full explanation.

Thanks
[Reveal] Spoiler: OA

Last edited by Vyshak on 30 Jul 2016, 21:28, edited 2 times in total.
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Re: Three fair coins are labeled with a zero (0) on one side and [#permalink]

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17 Jan 2008, 14:47
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variance of binomial distribution is n*p*(1-p), n = 3, p = 1/2

variance = 3*1/2*(1-1/2)= 3/4

stdev = sqrt(variance) = sqrt(3)/2

After many trials sample standard deviation is close to theoretical standard deviation = sqrt(3)/2.

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Re: Three fair coins are labeled with a zero (0) on one side and [#permalink]

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17 Jan 2008, 09:11
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Expert's post
C

$$\sigma=\sqrt{\frac{\sum(x-x_a)^2}{n}}$$

where, $$x$$ - the sum of 3 numbers
$$x_a$$ - the average sum.

probabilities for $$x$$:
$$x=0:$$ $$p_0=\frac18;$$ $$x=1:$$ $$p_1=\frac38;$$ $$x=2:$$ $$p_2=\frac38;$$ $$x=3:$$ $$p_3=\frac18$$

$$x_a=\frac32$$

$$n \to \infty$$ (n=1000 is very large number)

we can write:

$$\sigma=\sqrt{\frac18*(0-\frac32)^2+\frac38*(1-\frac32)^2+\frac38*(2-\frac32)^2+\frac18*(3-\frac32)^2}$$

$$\sigma=\sqrt{\frac18*\frac94+\frac38*\frac14+\frac38*\frac14+\frac18*\frac94}$$

$$\sigma=\sqrt{\frac{24}{32}}=\sqrt{\frac34}=\frac{\sqrt3}{2}$$

good Q +1
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Re: Three fair coins are labeled with a zero (0) on one side and [#permalink]

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04 Sep 2008, 07:16
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walker wrote:
C

$$\sigma=\sqrt{\frac{\sum(x-x_a)^2}{n}}$$

where, $$x$$ - the sum of 3 numbers
$$x_a$$ - the average sum.

probabilities for $$x$$:
$$x=0:$$ $$p_0=\frac18;$$ $$x=1:$$ $$p_1=\frac38;$$ $$x=2:$$ $$p_2=\frac38;$$ $$x=3:$$ $$p_3=\frac18$$

$$x_a=\frac32$$

$$n \to \infty$$ (n=1000 is very large number)

we can write:

$$\sigma=\sqrt{\frac18*(0-\frac32)^2+\frac38*(1-\frac32)^2+\frac38*(2-\frac32)^2+\frac18*(3-\frac32)^2}$$

$$\sigma=\sqrt{\frac18*\frac94+\frac38*\frac14+\frac38*\frac14+\frac18*\frac94}$$

$$\sigma=\sqrt{\frac{24}{32}}=\sqrt{\frac34}=\frac{\sqrt3}{2}$$

good Q +1

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]

SD = sqrt(var[x]) = 1/sqrt(x)..
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Re: Three fair coins are labeled with a zero (0) on one side and [#permalink]

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04 Sep 2008, 10:02
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Expert's post
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

SD = sqrt(var[x]) = $$\frac{sqrt{3}}{2}$$
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Re: Three fair coins are labeled with a zero (0) on one side and [#permalink]

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17 Jan 2008, 10:14
That´s a difficult solution. +1
Any other approach?

walker wrote:
C

$$\sigma=\sqrt{\frac{\sum(x-x_a)^2}{n}}$$

where, $$x$$ - the sum of 3 numbers
$$x_a$$ - the average sum.

probabilities for $$x$$:
$$x=0:$$ $$p_0=\frac18;$$ $$x=1:$$ $$p_1=\frac38;$$ $$x=2:$$ $$p_2=\frac38;$$ $$x=3:$$ $$p_3=\frac18$$

$$x_a=\frac32$$

$$n \to \infty$$ (n=1000 is very large number)

we can write:

$$\sigma=\sqrt{\frac18*(0-\frac32)^2+\frac38*(1-\frac32)^2+\frac38*(2-\frac32)^2+\frac18*(3-\frac32)^2}$$

$$\sigma=\sqrt{\frac18*\frac94+\frac38*\frac14+\frac38*\frac14+\frac18*\frac94}$$

$$\sigma=\sqrt{\frac{24}{32}}=\sqrt{\frac34}=\frac{\sqrt3}{2}$$

good Q +1
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Re: Three fair coins are labeled with a zero (0) on one side and [#permalink]

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17 Jan 2008, 14:54
Good approach. This is a tough question. How likely is to get this on the GMAT?
Q+1!!

maratikus wrote:
variance of binomial distribution is n*p*(1-p), n = 3, p = 1/2
variance = 3*1/2*(1-1/2)= 3/4

stdev = sqrt(variance) = sqrt(3)/2

After many trials sample standard deviation is close to theoretical standard deviation = sqrt(3)/2.

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Re: Three fair coins are labeled with a zero (0) on one side and [#permalink]

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20 Aug 2008, 08:31
walker wrote:
C

$$\sigma=\sqrt{\frac{\sum(x-x_a)^2}{n}}$$

where, $$x$$ - the sum of 3 numbers
$$x_a$$ - the average sum.

probabilities for $$x$$:
$$x=0:$$ $$p_0=\frac18;$$ $$x=1:$$ $$p_1=\frac38;$$ $$x=2:$$ $$p_2=\frac38;$$ $$x=3:$$ $$p_3=\frac18$$

$$x_a=\frac32$$

$$n \to \infty$$ (n=1000 is very large number)

we can write:

$$\sigma=\sqrt{\frac18*(0-\frac32)^2+\frac38*(1-\frac32)^2+\frac38*(2-\frac32)^2+\frac18*(3-\frac32)^2}$$

$$\sigma=\sqrt{\frac18*\frac94+\frac38*\frac14+\frac38*\frac14+\frac18*\frac94}$$

$$\sigma=\sqrt{\frac{24}{32}}=\sqrt{\frac34}=\frac{\sqrt3}{2}$$

good Q +1

In addition to what walker did:

no matter how many times the trail is performed, the expected value of SD doesnot remain the same.
Attachments

SD Calculation.xls [15 KiB]

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Re: Three fair coins are labeled with a zero (0) on one side and [#permalink]

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20 Aug 2008, 11:55
Is this really a question that is likely to be on the GMAT? It seems safe to say this is a 700+ question.
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Re: Three fair coins are labeled with a zero (0) on one side and [#permalink]

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03 Sep 2008, 01:58
It's quite hard. However, I got the same solution as Walker mentioned.
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Re: Three fair coins are labeled with a zero (0) on one side and [#permalink]

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03 Sep 2008, 05:22
I read somewhere that the GMAT does not expect you to know the standard deviation formula.
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Re: Three fair coins are labeled with a zero (0) on one side and [#permalink]

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03 Sep 2008, 06:06
Nerdboy wrote:
I read somewhere that the GMAT does not expect you to know the standard deviation formula.

I read the same in MGMAT math book.

I am not sure whether will get these kind of problems in real GMAT ..
If yes... then definitely it is 750+ quesiton.,

Good question!! Good discussion..
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Re: Three fair coins are labeled with a zero (0) on one side and [#permalink]

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04 Sep 2008, 10:16
Well it was a tough question..I wonder if GMAT asks such questions
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Re: Three fair coins are labeled with a zero (0) on one side and [#permalink]

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04 Sep 2008, 14:31
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

SD = sqrt(var[x]) = $$\frac{sqrt{3}}{2}$$
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Re: Three fair coins are labeled with a zero (0) on one side and [#permalink]

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28 Nov 2011, 08:16
I pray I never meet this question
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Three fair coins are labeled with a zero (0) on one side and [#permalink]

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30 Jul 2016, 20:55
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)$$\sqrt{3}$$/2
(D)$$\sqrt{5}$$/2
(E) 5/4

Last edited by Vyshak on 30 Jul 2016, 21:29, edited 1 time in total.
Topic Merged
Three fair coins are labeled with a zero (0) on one side and   [#permalink] 30 Jul 2016, 20:55
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