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.

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]
17 Jan 2008, 09:02

4

This post received KUDOS

1

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

5% (low)

Question Stats:

38% (05:52) correct
62% (02:18) wrong based on 27 sessions

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)sqrt3/2 D)sqrt5/2 E) 5/4

I don't know how to solve this question. Please provide full explanation.

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]

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.

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

Thanks walker... i had forgotten that x was the sum of numbers....

walker wrote:

bhushangiri wrote:

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.

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

Re: Three fair coins are labeled with a zero (0) on one side and [#permalink]
19 Sep 2013, 12:41

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________