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Three fair coins

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by abhi332 » Thu Feb 25, 2010 12:27 pm
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)SQROOT(3)/2
D)SQROOT(5)/2
E)5/4

[spoiler]OA:C[/spoiler]
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by ajith » Thu Feb 25, 2010 1:41 pm
abhi332 wrote: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)SQROOT(3)/2
D)SQROOT(5)/2
E)5/4

[spoiler]OA:C[/spoiler]
1/8 times the sum is expected to be 0
1/8 time sum is expected to be 3
3/8 time sum is expected to be 1
3/8 times sum is expected to be 2

expected average = (0+3/8+3/8+6/8) = 1.5

(SD)^2 = (125(0-1.5)^2+375*(1-1.5)^2+375*(2-1.5)^2 +125(3-1.5)^2)/1000
(SD)^2 = (125(-1.5)^2+375*(0.5)^2+375*(0.5)^2 +125(1.5)^2)/1000 = 3/4
[spoiler]
SD = sqrt(3)/2
[/spoiler]
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by artistocrat » Thu Oct 20, 2011 3:10 pm
As per another thread:

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.
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