How Accurate is the Binomial Distribution in Predicting Coin Toss Outcomes?

[Liketothink]

1. Homework Statement
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. Homework Equations
C!/(C-xj)!(xj!)(2^C)3. The Attempt at a Solution

*
o C
o 4
o 10
o 11
3
o fj
o 0.125
o 0.3125
o 0.34375
o 0.09375
o distribution
o 0.25
o 0.043945313
o 0.080566406
o #NUM!
o Head number
o 1
o 2
o 3
o 4
Thank you for helping.

 

[Jhero]

Here is my solution:

To calculate the measured distribution function, we first count the number of occurrences for each macroscopic state. From the given outcome, we have:
- 1 heads: 4 occurrences
- 2 heads: 10 occurrences
- 3 heads: 11 occurrences
- 4 heads: 7 occurrences

Therefore, the measured distribution function is:
f1 = 4/32 = 0.125
f2 = 10/32 = 0.3125
f3 = 11/32 = 0.34375
f4 = 7/32 = 0.21875

To calculate the expected distribution function, we use the binomial distribution formula:
f(xj) = C!/(C-xj)!(xj!)(2^C)

For 1 heads: f(1) = 4!/(4-1)!(1!)(2^4) = 1/16 = 0.0625
For 2 heads: f(2) = 4!/(4-2)!(2!)(2^4) = 6/16 = 0.375
For 3 heads: f(3) = 4!/(4-3)!(3!)(2^4) = 4/16 = 0.25
For 4 heads: f(4) = 4!/(4-4)!(4!)(2^4) = 1/16 = 0.0625

Therefore, the expected distribution function is:
f1 = 0.0625
f2 = 0.375
f3 = 0.25
f4 = 0.0625

Comparing the measured and expected distribution functions, we can see that they are similar but not exactly the same. This is expected as there will always be some degree of variation between the measured and expected results in any experiment. However, the overall trend and shape of the two distributions are similar, indicating that the results are consistent with the expected outcome based on the binomial distribution.