# PHysics lab question

1. Sep 1, 2009

### GreyGus

1. The problem statement, all variables and given/known data
For this exercise, four coins are tossed 32 times and the number of heads are recorded for each toss. Each toss falls into one of the following macroscopic states; 0 heads, 1 heads, 2 heads, 3 heads and 4 heads. Suppose the 32 tosses result in the following outcome: 3,2,3,2,2,4,0,3,0,2,0,4,4,2,3,1,2,0,1,3,2,3,1,3,3,2,3,2,3,3,2 and 1. Your task is to count the number of times when 1 heads, 2 heads, ... appears, and to calculate the measured and expected distribution functions.
To calculate the measured distribution function, if nj is the number of counts for jth heads for N trials, then the experimental distribution function is fj=nj/N. For example, the number of counts with zero heads is 4 giving f0=4/32=0.125.
The expected distribution for such an experiment follows a binomial distribution function and is given by
C!/(C-xj)!(xj!)(2^C)
where C is the total number of coins, xj is the number of heads. Thus for the case of 0 heads, f0=4!(4−0)!0!2^4=1/16=0.0625.

2. Relevant equations
C!/(C-xj)!(xj!)(2^C)

3. The attempt at a solution
• C
• 4
• 10
• 11
3
• fj
• 0.125
• 0.3125
• 0.34375
• 0.09375
• distribution
• 0.25
• 0.043945313
• 0.080566406
• #NUM!
• 1
• 2
• 3
• 4

For the last one the factorial is for a negative number for it doesn't work.

Last edited: Sep 1, 2009
2. Sep 1, 2009

### Staff: Mentor

How are you getting the factorial of a negative number. C is 4 and xj runs from 0 to 4 so C-xj runs from 4 to 0. There are no negative numbers involved.

3. Sep 1, 2009

### Staff: Mentor

That gives the probability of a toss having xj heads (or tails). Thus 0 heads has a probability of 1/16. If you toss the coins 32 times, the expected number of 0 head tosses is (1/16)(32) = 2. Compare that expected value with the actual number seen in a particular series of tosses.

(I'm not sure how you got a negative number anywhere.)

4. Sep 1, 2009

### GreyGus

I had a negative number because for the last one I applied the formula and got
(3!)/(3-4)!(4)!(2^3).

I might have posted the problem wrong, for that I apologize. But it's suppose to be four columns, one for the different C's I counted, fj's, expected distribution and the head number I'm suppose to count. Like if it's 2 ok, so how many 2's are there. So that's how I did it and got a negative number.

Last edited: Sep 1, 2009
5. Sep 2, 2009

### Staff: Mentor

This makes no sense. Here you have C = 3, meaning you are flipping 3 coins. And you have xj = 4, meaning you are looking for 4 heads!

As far as I can see, you are always flipping four coins at a time, so C is always 4. xj varies from 0 to 4. No negatives.