# Confidence intervals confusion

I'm looking at a problem where the number of samples, $$N$$, is 2 million. Each sample represents a profit or loss. Let $$X_n$$ be the dollar payoff to the n-th game. $$E[X_n] = 0$$. Let the total payoff of $$D(N)$$ be $$D(N)=\sum_{n=1}^N X_n$$, then $$E[D(n)] = 0$$. The standard deviation of the D(N) is 7141. Then it states that, there is roughly a 5% chance that the gain/loss exceeds 1.96 standard deviations (i.e., a 5% chance of gains or losses beyond +/- 1.96*7141=13,997).

I plugged in these numbers on https://www.socscistatistics.com/confidenceinterval/default3.aspx, and got the numbers in the below image.

Can someone explain to me why the confidence interval here was computed using the $$S_M$$ (standard error) instead of the standard deviation?

• Right I think I understand this. In this particular example that I saw, the standard deviation of $D(N)$ is proportional to $\sqrt{N}$ so it actually grows as the sample size gets larger. It's not intuitive to me why the example used the standard deviation to compute the confidence interval here Apr 5, 2020 at 14:56
• No, I'm talking about standard deviation in this comment. The standard deviation that the example derived is $\sqrt{N\frac{51}{2}}$ Apr 5, 2020 at 15:04