Mean and Standard Deviation of the Binomial Distribution

Reading

• Section 3.4.1, 3.4.2, 3.4.3

Practice Problems

3.6.4 (Page 163)

3.27, 3.28, 3.29, 3.30, 3.36

Notes

Mean and Standard Deviation for the Binomial

We have seen how to find the mean and standard deviation of combinations of variables, when those variables are independent of each other. We will now use that knowledge to find formulas for the mean and standard deviation of the binomial.

Consider a binomial setting with parameters $n$ and $p$, and denote by $X$ the number of successes.

• Denote by $S_1$ the “number of successes in the first trial”.
• Denote by $S_2$ the “number of successes in the second trial”.

and so on. Then:

• All the $S_i$ are independent of each other.

• Each $S_i$ follows the distribution given by the table:

  S	0	1
  Prob	1-p	p

• From this table we find: $$\mu_{S_i} = p$$ $$\sigma^2_{S_i} = p(1-p)$$

To see why these are true:

• The probability table for the $S_i$ is pretty easy to see, since the $i$-th trial is a trial with the two options success/failure, and we count the number of successes; so either $0$ or $1$.

• Using the table we compute:

  $$\mu_S = (1-p)\cdot 0 + p \cdot 1 = p$$ $$\sigma^2_S = (1-p)(0-p)^2 + p(1-p)^2 = p(1-p)\left[p + (1-p)\right] = p(1-p)$$

We now use the $S_i$ to say something about $X$:

• $X$ relates to the $S_i$ via: $$X = S_1 + S_2 + \cdots + S_n$$

• Using this we find formulas for the mean and standard deviation of $X$:

  $$\mu_X = np$$ $$\sigma_X = \sqrt{np(1-p)}$$

• If we define by $\hat p$ the “percent of successes”, namely

  $$\hat p = \frac{X}{n}$$

  then we have the formulas:

  $$\mu_{\hat p} = p$$ $$\sigma_{\hat p} = \frac{\sqrt{p(1-p)}}{\sqrt{n}}$$

The first part is straightforward. Because the $S_i$ are independent of each other, the formulas we learned earlier allow us to compute the mean and standard deviation of $X$ using the ones for $S_i$:

$$\mu_X = \mu_{S_1} + \mu_{S_2} + \cdots + \mu_{S_n} = p + p + \cdots + p = np$$ $$\sigma^2_X = \sigma^2_{S_1} + \sigma^2_{S_2} + \cdots + \sigma^2_{S_n} = p(1-p) + p(1-p) + \cdots + p(1-p) = np(1-p)$$

The formulas for $\hat p$ follow from the fact that it is just a linear transformation from $X$, just dividing everything by $n$.

Approximation via Normal Distribution

These formulas become useful when $n$ is large, because in that case we can approximate the binomial distribution with normal:

When $n$ “sufficiently large”, then the binomial follows an approximately normal distribution. So we have:

$$X\sim N\left(np, \sqrt{np(1-p)}\right)$$ $$\hat p\sim N\left(p, \frac{\sqrt{p(1-p)}}{\sqrt{n}}\right)$$

The rule of thumb for when $n$ is sufficiently large is that we should have:

$$np \geq 10$$ $$n(1-p) \geq 10$$

Both of these conditions should be true. We only need to check the smallest of $p$, $1-p$, since if that one results in $10$ or more then the other will do as well.

Using these we can quickly do some computations, without having to use the explicit formula for $P(X=k)$, which becomes very hard to use for large $n$.

Here’s an example:

In a multiple-choice test there are $100$ questions. We pick answers at random and there are $5$ possible answers to each question, so we have a $20\%$ chance to answer each question correctly. What are the chances, that we will get at least $25$ answers correct?

This is a binomial setting, since there is a fixed number of questions $n=5$, and for each one we either get it right with probability $p=0.2$ or we get it wrong, and since we pick answers at random the trials are independent. $X$ measures the number of correct answers.

We start by computing the mean and standard deviation of $X$. We have:

$$\mu_X = np = 100\cdot 0.2 = 20$$ $$\sigma_X = \sqrt{np(1-p)} = \sqrt{20\cdot 0.8} = 4$$

We then check the rule of thumb: We need both $np$ and $n(1-p)$ to be at least $10$. But $np$ is clearly the smallest of the two, and it is already $\geq 10$, so we are OK and can use the normal approximation.

Therefore we can approximate $X$ by $N(20, 4)$. The question therefore becomes, in $N(20, 4)$ how much data is above $x = 25$.

This is now a problem about a normal distribution, and we know well how to solve those problems. We would compute:

$$z = \frac{x-\mu}{\sigma} = \frac{25 - 20}{4} = 1.25$$

Then look that up in our table to get $p = 0.894$. Since this measures how many are below that value, we need to look at the rest, so $1-0.894=0.106$, or $10.6\%$. So there is roughly a $10\%$ chance that we would score more than $25$ points at random like that.

Continuity Correction

One important topic to discuss is that of continuity correction. This is relevant when $n$ is relatively small, like in this example.

The problem is this: We are approximating the binomial distribution with a normal distribution. But the binomial distribution corresponds to integers only, while the normal distribution allows for all numbers. So for instance according to the normal distribution there should be a number of students who scored between 24 and 25. But that is not possible. This is a discrepancy we need to somehow correct.

The fix is to divide the space between 24 and 25 in half, and count the upper half as part of 25, and the lower half as part of 24. What this means is that in these problems you want to often start halfway to the previous or next value, depending on the question.

In other words, in this instance we should be using $x=24.5$ rather than $x=25$. You can think of it as saying that we should include as part of $25$ values that would have “rounded up to 25”.

With that in mind, the computation would have been:

$$z = \frac{24.5 - 20}{4} = 1.125$$ $$p = 0.8697$$ $$1-p = 0.1303$$

So with this computation, the answer would be closer to $13\%$, rather than $10\%$. This would be a better estimate in this case.

For comparison, the perfect answer, the one that would be computed if we did the exact formula for $P(X=k)$ for all numbers from $25$ to $100$, would have given us a percent of $13.135\%$.