# Intuition on the Expected Value and Standard Deviation of a binomial

Intro:

I was originally going to ask two separate questions, one on the interpretation of the expected value, and another on the interpretation of the standard deviation, but I feel like they're connected enough to ask them as one.

I'm using the grey lines to separate the first part from the second. Let's begin:

Expected Value:

Say we have a binomial random variable with $$n$$ trials and a probability of success of $$p$$ on each trial.

Its expected value is $$np$$ - the number of trials multiplied by the probability of success on each trial.

I interpret the expected value in the following way:

If we were to carry out the experiment which this random variable corresponds to $$n$$ times, we should expect the number of successes to be "around" $$np$$.

That's because by the law of large numbers, the ratio of the number of successes to the number of trials should approach the probability of success on any single trial as the number of trials, $$n$$, tends towards infinity.

(I can't really formalize why the law of large numbers works, although it makes intuitive sense...)

If the total number of successes is around $$np$$, then...

$$\frac{np}{n} = p$$

...the law of large numbers is satisfied.

Is this interpretation correct?

Standard Deviation:

The standard deviation of the random variable is $$\sqrt{n(p)(1-p)}$$.

But...I'm lost on how I should interpret this.

Does it mean that I if I run the experiment $$n$$ times, I should expect the actual number of successes to be...

1. Less than $$\sqrt{n(p)(1-p)}$$ away from the expected value?

2. Around $$\sqrt{n(p)(1-p)}$$ away from the expected value?

3. In between $$\sqrt{n(p)(1-p)}$$ and $$2\sqrt{n(p)(1-p)}$$ away from the expected value?

4. Does it depend on the number of trials?

The first three interpretations all seem to agree with the law of large numbers, as...

$$\lim_{n\rightarrow\infty}\frac{np \pm k\sqrt{n(p)(1-p)}}{n} = p$$

...regardless of the value of some integer $$k$$, the ratio of the number of successes to the total number of trials goes to what we defined to be the "probability" of a success on each trial.

I also feel like the number of trials and the probability should matter...but I'm not sure how....

Thanks!

• Consider the experiment of tossing a biased coin $n$ times. The number of successes is $X$. If you repeat this experiment a large number of times, you would predict that the average value of $X$ (over all your repeated experiments) is approximately $np$. You would predict that the average value of $(X - np)^2$ (over all your repeated experiments) is approximately $np(1 - p)$. – littleO Aug 22 at 19:07
• If n is large, the distribution tends to be normal, so 1 happens about 67% of the time. – herb steinberg Aug 22 at 21:32

For one trial where $$x=0$$ or $$x=1$$, we get $$x^2=x$$ so the $$E(x^2)=E(x)=p$$
The variance for one trial is $$V=E(x^2)-(E(x))^2 = p-p^2= p(1-p)$$
The variance for $$n$$ trials will be $$V=np(1-p)$$ therefore, the standard deviation is $$\delta =\sqrt {np(1-p)}$$