Given a sample size and standard deviation, how to find the probability of the mean

Question

Say I have 24 guys sharing 123320 candy bars by a Gaussian distribution (i.e. guy number 1 and 24 get little, while the guy in the middle get the most, something like that). Given the standard deviation, how do I find (maybe an approximation) the number of candy given to the guy in the middle? Thank you :D

Answer 1

The distribution $\mathsf{Binom}(n =23, p=1/2)$ assigns probabilities to the 24 numbers $0$ through $23$ and is very well approximated by a normal distribution. Temporarily renumbering the men as 0 through 23, we find the binomial probability $p$ for each man, and multiply by cb = 123320 and round to an integer to get the number of candy bars for each man.

In R statistical software the assignment procedure is shown below. (The heavy lifting is to get the binomial probabilities. Or you could go to row number 23 of Pascal's Triangle.)

It turns out that (in spite of the rounding) exactly all of the candy bars are used. The output matrix should be self-explanatory if you ignore index numbers in brackets. Four of the men get no candy bars, but I see no way to stay nearly true to a normal distribution and give them any. (Maybe guys 12 and 13 in the middle can share.)

cb = 123320;  x = 1:24
p = dbinom(x-1, 23, 1/2);  nr = round(cb*p)
cbind(x, p, nr)  # binds three 24-vectors to make the matrix below
       x            p    nr
 [1,]  1 1.192093e-07     0
 [2,]  2 2.741814e-06     0
 [3,]  3 3.015995e-05     4
 [4,]  4 2.111197e-04    26
 [5,]  5 1.055598e-03   130
 [6,]  6 4.011273e-03   495
 [7,]  7 1.203382e-02  1484
 [8,]  8 2.922499e-02  3604
 [9,]  9 5.844998e-02  7208
[10,] 10 9.741664e-02 12013
[11,] 11 1.363833e-01 16819
[12,] 12 1.611803e-01 19877
[13,] 13 1.611803e-01 19877
[14,] 14 1.363833e-01 16819
[15,] 15 9.741664e-02 12013
[16,] 16 5.844998e-02  7208
[17,] 17 2.922499e-02  3604
[18,] 18 1.203382e-02  1484
[19,] 19 4.011273e-03   495
[20,] 20 1.055598e-03   130
[21,] 21 2.111197e-04    26
[22,] 22 3.015995e-05     4
[23,] 23 2.741814e-06     0
[24,] 24 1.192093e-07     0
sum(nr)
## 123320

Here is a figure showing that the binomial probabilities (vertical lines) are very close to a normal distribution (curve). [Its standard deviation is $\sigma = \sqrt{23/4} = 2.397916.$]