# Why am I able to divide the sum of observation values by the number of observations for the mean and the probability of heads? [closed]

Day to day, we use the arithmetic mean to calculate the sample mean to estimate the mean of a population, for example, a normal distribution. And also, we compute the mean to find out the probability of heads landing for coin tosses. Here, I am wondering how we know that we can divide the sum of heads(HEADS = 1, TAILS = 0) by the number of coin tosses to get the probability of heads after a trial or a sequence of trials. I have been taught the way we compute the mean but I am wondering how we found the way to compute the mean and the probability of heads after a trial or a sequence of trials. I just want to know what mathematical backgrounds are there for this.

• The (unaccepted, but valuable) answer to stats.stackexchange.com/questions/441911/… may help. Commented Mar 10 at 3:14
• @jbwman Thank you for the reference and I will go through it. Commented Mar 10 at 3:25
• Only the original user who posted the question can accept an answer (not the community at large). Thus, an outstanding answer can go unaccepted if: $1)$ that user forgets about Stack Exchange and never logs back in $2)$ that user just doesn’t feel like it $3)$ that user wants to keep from “closing out” the question in case someone has an even better answer they would not be inclined to post when the original question appears to have been resolved by an “accepted” answer $4)$ etc. A lack of acceptance does not imply a lack of quality.
– Dave
Commented Mar 10 at 13:12
• @Dave Thank you for the kind explanation. Commented Mar 10 at 15:20
• @xabzakabecd The method of moments should give you the same estimator Commented Mar 11 at 10:14