# Dice problem help

#### Cerran

##### New Member
I'm trying to get this problem set up correctly but I seem to have an issue as my calculated standard deviation seems to be off. Anyone have any suggestions? I know I have an error somewhere but I just haven't figured out where. I'm using excel to do most of the calculations.

Here is the question:

2. After one experiment where 4 dice were rolled 1,000 times, the observed distribution of averages was seen to be as follows:

Average 1 1.25 1.5 1.75 2 2.25 2.5 2.75 3 3.25 3.5
Frequency 0 5 16 16 21 35 53 95 91 111 118

Average 3.75 4 4.25 4.5 4.75 5 5.25 5.5 5.75 6
Frequency 103 92 67 71 50 29 15 8 4 0

Compute the mean and standard deviation of this distribution and compare to the estimates given by the Central Limit Theorem.

Here is where I am so far with my work. The mean I have calculated seems correct.

Mean: 3.5063

1*0/1000+1.25*5/1000+1.5*16/1000+1.75*16/1000+2*21/1000+2.25*35/1000+2.5*53/1000+2.75*95/1000+3*91/1000+3.25*111/1000+3.5*118/1000+3.75*103/1000+4*92/1000+4.25*67/1000+4.5*71/1000+4.75*50/1000+5*29/1000+5.25*15/1000+5.5*8/1000+5.75*4/1000+6*0/1000= 3.5063

Standard Deviation: 4.76519

√((1^2*0/1000+〖1.25〗^2*5/1000+〖1.5〗^2*16/1000+〖1.75〗^2*16/1000+2^2*21/1000+〖2.25〗^2*35/1000+〖2.5〗^2*53/1000+〖2.75〗^2*95/1000+3^2*91/1000+〖3.25〗^2*111/1000+〖3.5〗^2*118/1000+〖3.75〗^2*103/1000+4^2*92/100+〖4.25〗^2*67/1000+〖4.5〗^2*71/1000+〖4.75〗^2*50/1000+5^2*29/100+〖5.25〗^2*15/1000+〖5.5〗^2*8/100+〖5.75〗^2*4/1000+6^2*0/1000)-(3.5063)^2 )=4.76519

I'm stuck as to where I have gone wrong.

#### rogojel

##### TS Contributor
hi,
there is a small correction to the formula: instead of N you need to use N-1 - not the sample size but the degrees of freedom.

regards

#### Cerran

##### New Member
By degrees of freedom you mean the number of possible outcomes i.e. 6^4?

#### rogojel

##### TS Contributor
Nope, its just N-1 for the stddev calculation, where N is the sample size.

#### katxt

##### Member
The method looks fine but the answer is clearly too big. Did you remember to square root.

#### Cerran

##### New Member
I think I found the error, the formula was okay but a couple of the values entered were 100 instead of 1000 and once I fixed that the standard deviation is now down to 0.8695 which seems a bit low but certainly a lot more reasonable.

#### rogojel

##### TS Contributor
But still, you need to divide by 999 not 1000 imo. The correct value is 0.8699312.

Used this R code:

Avg=c(1 ,1.25, 1.5, 1.75, 2, 2.25, 2.5, 2.75, 3, 3.25, 3.5, 3.75, 4, 4.25, 4.5, 4.75, 5, 5.25, 5.5, 5.75, 6)
Freq=c(0, 5, 16, 16, 21, 35, 53, 95, 91, 111, 118, 103, 92, 67, 71, 50, 29, 15, 8, 4, 0)
x=rep(Avg, times=Freq)
sd(x)

#### Cerran

##### New Member
That doesn't seem to be what they want us to do when using the central limit theorem. I'm still a bit confused on the N-1. I'll put in a request to the professor and see what she says.

The problem comes from here and I've got the first question completed and just finishing up the second question.

http://media.pearsoncmg.com/aw/aw_triola_elemstats_10/cw/content/intproj/ch06/ch06.htm

For the discussion I was pointing out the fact the the results are more closely clustered around the mean for the 1000 trials just means that the distribution will be less like a standard distribution unless you did more trials in which case the standard deviation would be closer to the estimated 1.21.

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