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.

In addition to what walker did:


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

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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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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First, set up each coin in a column and compute the sum of each possible trial as follows:
Coin A Coin B Coin C Sum
0 0 0 0
1 0 0 1
0 1 0 1
0 0 1 1
1 1 0 2
1 0 1 2
0 1 1 2
1 1 1 3
Now compute the average (mean) of the sums using one of the following methods:
Method 1: Use the Average Rule (Average = Sum / Number of numbers).
(0 + 1 + 1 + 1 + 2 + 2 + 2 + 3) ÷ 8 = 12 ÷ 8 = 3/2
Method 2: Multiply each possible sum by its probability and add.
(0 × 1/8) + (1 × 3/8) + (2 × 3/8) + (3 × 1/8) = 12/8 = 3/2
Method 3: Since the sums have a symmetrical form, spot immediately that the mean must be right in the middle. You have one 0, three 1’s, three 2’s and one 3 – so the mean must be exactly in the middle = 1.5 or 3/2.

Then, to get the standard deviation, do the following:
(a) Compute the difference of each trial from the average of 3/2 that was just determined. (Technically it’s “average minus trial” but the sign does not matter since the result will be squared in the next step.)
(b) Square each of those differences.
(c) Find the average (mean) of those squared differences.
(d) Take the square root of this average.
Average of Sums Sum of Each Trial Difference Squared Difference
3/2 0 3/2 9/4
3/2 1 ½ ¼
3/2 1 ½ ¼
3/2 1 ½ ¼
3/2 2 – ½ ¼
3/2 2 – ½ ¼
3/2 2 – ½ ¼
3/2 3 – 3/2 9/4
The average of the squared differences = (9/4 + ¼ + ¼ + ¼ + ¼ + ¼ + ¼ + 9/4) ÷ 8 = 6 ÷ 8 = ¾.

Finally, the square root of this average = .

The correct answer is C.
Note: When you compute averages, be careful to count all trials (or equivalently, to take probabilities into account). For instance, if you simply take each unique difference that you find (3/2, ½, –1/2 and –3/2), square those and average them, you will get 5/4, and the standard deviation as .
This is incorrect because it implies that the 3/2 and –3/2 differences are as common as the ½ and –1/2 differences. This is not true since the ½ and –1/2 differences occur three times as frequently as the 3/2 and –3/2 differences.