Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

This topic is locked. If you want to discuss this question please re-post 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? _________________

ash
________________________
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. _________________

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 _________________

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.
__ __ __

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].