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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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.

C is the answer.

That´s a difficult solution. +1
Any other approach?

Good approach. This is a tough question. How likely is to get this on the GMAT?
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.

Is this really a question that is likely to be on the GMAT? It seems safe to say this is a 700+ question.

It's quite hard. However, I got the same solution as Walker mentioned.

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..

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)..

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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Well it was a tough question..I wonder if GMAT asks such questions

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

I pray I never meet this question

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

Bunuel: Isn't this out of scope for GMAT?
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Yes, it's out of scope. So, you can ignore it and move on.
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How did these values come about, mentioned below? The rest is just formula * probability, so I am clear about it.
probabilities for xx:
x=0:x=0: p0=1/8;p0=1/8; x=1:x=1: p1=3/8;p1=3/8; x=2:x=2: p2=3/ 8;p2=3/8; x=3:x=3: p3=1/8 p3=1/8

xa=3/2

Hi Guys can you provide some theory for the mean , variance and standard deviation for probability distribution