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Three fair coins are labeled with a zero (0) on one side and [#permalink]
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02 Jul 2004, 23:28
This topic is locked. If you want to discuss this question please repost it in the respective forum. 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?
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Last edited by ashkg on 03 Jul 2004, 09:35, edited 1 time in total.



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Was anyone able to solve this ? I would like to see some other solution than mine, which was a bit lengthy.
I got an answer of sqrt(3/4).
 ash
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Guys give this a shot !!!!!!!!!!!
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Intern
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I think it should be the standard deviation of
0, 1, 1, 1, 2, 2, 2, 3
Ans : is sqrt(3)/2



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Joined: 20 May 2004
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Re: Good SD problem [#permalink]
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04 Jul 2004, 11:35
ashkg 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?
The simple answer to this question is, it is impossible to predict the S.D from the data given.The possible sum that a person will get after flipping the coins are, 3, 2, 1, 0. The person flips for 1000 times. 1000 times is relatively large number. If you plot the result, you should get almost a bell shaped curve. The S.D. has to be almost zero, however, we cannot say for sure without real data.
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gmat2004



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This has to do with knowing about probability distribution
The expected average is .5
If we plot this on a bell curve, where we can assume a normal distribution, we have .5, the mean, in the center and 0 and 1 at the extremities as it is very unlikely that you will get 100% either 0 or 1. For a normal distribution, there are about 68% of data which fall within 1 standard deviation of the mean. This means that the 1 st. dev. value is .68.5 = .18
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Joined: 31 Dec 1969
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WE: Supply Chain Management (Energy and Utilities)

Paul wrote: This has to do with knowing about probability distribution The expected average is .5 If we plot this on a bell curve, where we can assume a normal distribution, we have .5, the mean, in the center and 0 and 1 at the extremities as it is very unlikely that you will get 100% either 0 or 1. For a normal distribution, there are about 68% of data which fall within 1 standard deviation of the mean. This means that the 1 st. dev. value is .68.5 = .18
Paul,
I will agree with you, if only one coin was used in the experiment. Here, three coins with a value of 1 for one side and another side 0 is used. Therefore, I feel, the average expected is 1.5 (3 X 0.5). Another point is the question asks for a value for standard deviation. We cannot presume for first and second standard deviation. Therefore, we cannot determine the standard deviation for the experiment. If we presume that 1000 calculations is large then we have to get an S.D. zero.



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heres my solution.
BTW the ans choices are 1/2, 3/4, 5/4 , (sqrt3)/2, (sqrt5)/2
Sorry that I didnt give it earlier.
There coins are thrown.
__ __ __
In a throw, total no of outcomes = 2^3 = 8
Probability of getting sum as 0 = 1/8
Probability of getting sum as 1 = 3/8
Probability of getting sum as 2 = 3/8
Probability of getting sum as 3 = 1/8
In 1000 throws,
Probability of getting sum as 0 = 1/8 * 1000 = 125
Probability of getting sum as 1 = 3/8 * 1000 = 375
Probability of getting sum as 2 = 3/8 * 1000 = 375
Probability of getting sum as 3 = 1/8 * 1000 = 125
total sum of 1000 numbers = 125*0 +375*1 +375*2 + 125*3 = 1500
mean =1500/1000 = 1.5
variance = (
[ (1.50)^2 ] * 125 +
[ (1.51)^2 ] * 375 +
[ (1.52)^2 ] * 375 +
[ (1.53)^2 ] * 125 ) / 1000
Dont do all intermediate calculations........lots of numbers(eg. 125) will cancel out.
SD = sqrt(variance) = sqrt(3/4) which is one of the answer options.
 ash.
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sathya76 wrote: I think it should be the standard deviation of
0, 1, 1, 1, 2, 2, 2, 3
Ans : is sqrt(3)/2
this seems to be the trick......the SD of the 1000 throws will be the same as the SD of the set of numbers shown by sathya above.
I knew I was doing it the lengthy way( my soln is in prev post), but didnt know any shorter way
thanks sathya !
 ash
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nice solution ashk. I did make some unwarranted assumption with the 68% of value under 1 st. dev., value of which we don't know.
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Joined: 14 Jul 2004
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Satya76 & Ashkg:
Questions for you guys.
1) In your variance calcuation arent you guys forgetting to divide by the (N1), the denominator of the variance formulae?
2) Also i calculated the SD for the # 0,1,1,1,2,2,2,3 and this doesnt work out to sqrt(3)/2.
3) Also i dont understand why you are saying the variance of the 1000 tries is
variance = (
[ (1.50)^2 ] * 125 +
[ (1.51)^2 ] * 375 +
[ (1.52)^2 ] * 375 +
[ (1.53)^2 ] * 125 ) / 1000
variance = sqr of SD.
SD = sqrt([sum(xx*)2]/[N1])
I dont agree with how you mutiple sum(xx*) with the # of times this event can occur in 1000 tries.



Director
Joined: 14 Jul 2004
Posts: 698

Satya76 & Ashkg:
Questions for you guys.
1) In your variance calcuation arent you guys forgetting to divide by the (N1), the denominator of the variance formulae?
2) Also i calculated the SD for the # 0,1,1,1,2,2,2,3 and this doesnt work out to sqrt(3)/2.
3) Also i dont understand why you are saying the variance of the 1000 tries is
variance = (
[ (1.50)^2 ] * 125 +
[ (1.51)^2 ] * 375 +
[ (1.52)^2 ] * 375 +
[ (1.53)^2 ] * 125 ) / 1000
variance = sqr of SD.
SD = sqrt([sum(xx*)2]/[N1])
I dont agree [or i might be missing something here] with how you mutiple sum(xx*) with the # of times this event can occur in 1000 tries.



Director
Joined: 20 Jul 2004
Posts: 592

gmataquaguy wrote: 2) Also i calculated the SD for the # 0,1,1,1,2,2,2,3 and this doesnt work out to sqrt(3)/2.
Take a look at my workaround...
Attachments
SD.JPG [ 28.27 KiB  Viewed 720 times ]



Director
Joined: 14 Jul 2004
Posts: 698

gmataquaguy wrote: Satya76 & Ashkg:
Questions for you guys.
1) In your variance calcuation arent you guys forgetting to divide by the (N1), the denominator of the variance formulae?
2) Also i calculated the SD for the # 0,1,1,1,2,2,2,3 and this doesnt work out to sqrt(3)/2.
3) Also i dont understand why you are saying the variance of the 1000 tries is
variance = ( [ (1.50)^2 ] * 125 + [ (1.51)^2 ] * 375 + [ (1.52)^2 ] * 375 + [ (1.53)^2 ] * 125 ) / 1000
variance = sqr of SD.
SD = sqrt([sum(xx*)2]/[N1])
I dont agree [or i might be missing something here] with how you mutiple sum(xx*) with the # of times this event can occur in 1000 tries.
Did some more research......The standard deviation has 2 different formulas based on whether elements are a "sample size" of a population or you have the entire population.
If you have the entire population as we do in this case we divide by [N], the # of elements in the sample (which in this case is 8 or 1000 depending on Ash's method or satya's method). If this was a "sample size" then you would divide by [N1].



Joined: 31 Dec 1969
Location: Russian Federation
Concentration: Entrepreneurship, International Business
GMAT 3: 740 Q40 V50 GMAT 4: 700 Q48 V38 GMAT 5: 710 Q45 V41 GMAT 6: 680 Q47 V36 GMAT 9: 740 Q49 V42 GMAT 11: 500 Q47 V33 GMAT 14: 760 Q49 V44
WE: Supply Chain Management (Energy and Utilities)

hardworker_indian wrote: gmataquaguy wrote: 2) Also i calculated the SD for the # 0,1,1,1,2,2,2,3 and this doesnt work out to sqrt(3)/2. Take a look at my workaround...
hardworker_indian,
where is your workaround? I didn't see it in this thread..Please let me know



Director
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Anonymous wrote: hardworker_indian wrote: gmataquaguy wrote: 2) Also i calculated the SD for the # 0,1,1,1,2,2,2,3 and this doesnt work out to sqrt(3)/2. Take a look at my workaround... hardworker_indian, where is your workaround? I didn't see it in this thread..Please let me know
I attached it as a .jpg and it should be simply visible on the thread.
Not sure, but you may have to be a registered user to view files.










