Problem Solving

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

sanju09 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)sqrt3/2
(D)sqrt5/2
(E) 5/4

Answer is (D).

There will be four cases for flipping all three coins:
(1) All "Zero" => sum=0
(2) One "One" and two "Zero" => sum=1
(3) Two "One" and one "Zero" => sum=2
(4) All "One" => sum=3

The expected result is:

250 times for each situation for 1000 times.

Thus the question asks us to find the standard derivation for a series where there are 1000 integers, including 250 "0", 250 "1", 250 "2" and 250 "3)

From these information, we can find the average = 1.5 and standard derivation = sqrt(5)/2

billzhao wrote:

sanju09 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)sqrt3/2
(D)sqrt5/2
(E) 5/4

Answer is (D).

There will be four cases for flipping all three coins:
(1) All "Zero" => sum=0
(2) One "One" and two "Zero" => sum=1
(3) Two "One" and one "Zero" => sum=2
(4) All "One" => sum=3

The expected result is:

250 times for each situation for 1000 times.

Thus the question asks us to find the standard derivation for a series where there are 1000 integers, including 250 "0", 250 "1", 250 "2" and 250 "3)

From these information, we can find the average = 1.5 and standard derivation = sqrt(5)/2

so sorry, D is not the OA!

How can you anticipate each happens exactly 250 times, it is not the possible answer.

naaga wrote:How can you anticipate each happens exactly 250 times, it is not the possible answer.

True

Sum will be 0-3. Mean is 1.5

Standard deviation will be ((0-1.5)^2 + (1-1.5)^2 + (2-1.5)^2 + (3-1.5)^2)/4

Answer is 5/4 E

Please post the answer in the OG.


Thanks.

Please post the answer in the OG.


Thanks.

Please post the answer in the OG.


Thanks.

i think its A; sd= 1/2....(most probable).....

OA?

It is C = root3/2 ..

The 4 possible sums are 0, 1, 2, 3

The probability of it being 0 is 1/8
The probability of it being 1 is 3/8
The probability of it being 2 is 3/8
The probability of it being 3 is 1/8

So the mean is (0*1+1*3+2*3+3*1)/8 = 1.5

The variance = (0-1.5)^*1/8 + (1-1.5)^2*3/8+(2-1.5)^2*3/8+(3-1.5)^2*1/8 = 3/4

Standard deviation = root (variance) = (root 3)/2

Gabriel,

I did not understand the part where you computed the probab. of the {0,1,2,3}. Could you please kindly explain?

Best regards

gabriel wrote:It is C = root3/2 ..

The 4 possible sums are 0, 1, 2, 3

The probability of it being 0 is 1/8
The probability of it being 1 is 3/8
The probability of it being 2 is 3/8
The probability of it being 3 is 1/8

So the mean is (0*1+1*3+2*3+3*1)/8 = 1.5

The variance = (0-1.5)^*1/8 + (1-1.5)^2*3/8+(2-1.5)^2*3/8+(3-1.5)^2*1/8 = 3/4

Standard deviation = root (variance) = (root 3)/2

Agree with you.

sanju09 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)sqrt3/2
(D)sqrt5/2
(E) 5/4

The answer will be (C)

The possible values for the sum will be 0,1,2,3

The probability of getting sum 0 is 1/8, sum 1 as well as sum 2 each will have a probability of 3/8 and the probability to get a sum 3 will be 1/8.

Folks, since the expected SD remains same whether it is for 8 observations or 1000 observations, let us consider 8 observations only.

So out of these 8 observations we will have one 0, three 1's , three 2's and one 3.

Hence the mean of these numbers is 3/2.

From this data we can calculate the sum of the squares of the deviations of each observation from the mean which will turn out to be 6.

So SD = sqrt(6/8) = sqrt(3)/2

Big applauds to both gabriel and sureshbala!

see more of it


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.

So C is the answer.