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Three fair coins are labeled with a zero (0) on one side and [#permalink]
02 Jul 2004, 22:28

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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ash
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I'm crossing the bridge.........

Last edited by ashkg on 03 Jul 2004, 08:35, edited 1 time in total.

Re: Good SD problem [#permalink]
04 Jul 2004, 10:35

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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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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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.

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.
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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

I dont agree [or i might be missing something here] with how you mutiple sum(x-x*) 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 [N-1].