Variance for the Binomial distribution is similarly calculated using the *n* and *p* parameters. Let’s use the 10 fair coin flips example to try to understand how variance is calculated. Each coin flip has a certain probability of landing as heads or tails: 0.5 and 0.5, respectively.

The variance of a single coin flip will be the probability that the success happens times the probability that it does not happen: *p·(1-p)*, or 0.5 x 0.5. Because we have *n = 10* number of coin flips, the variance of a single fair coin flip is multiplied by the number of flips. Thus we get the equation:

```
$\begin{aligned}
Variance(\#\;of\;Heads) = Var(X) = n \times p \times (1-p) \\
Variance(\#\;of\;Heads) = 10 \times 0.5 \times (1 - 0.5) = 2.5
\end{aligned}$
```

Let’s consider our 20 multiple choice quiz again. The variance around getting an individual question correct would be *p·(1-p)*, or 0.25 x 0.75. We then multiply this variance for all 20 questions in the quiz and get:

`$Variance(\#\;of\;Correct\;Answers) \&= 20 \times 0.25 \times (1 - 0.25) = 3.75$`

We would expect to get 5 correct answers, but the overall variance of the probability distribution is 3.75.

### Instructions

**1.**

A certain basketball player has an 85% chance of making a given free throw. Uncomment `variance_baskets`

and set it equal to the variance of free throws made from 20 shots.

Then print `variance_baskets`

.

**2.**

Let’s say a student has a 98% chance of arriving on time to class. Uncomment `variance_late`

and set it equal to the variance of days the student will arrive late to class throughout the 180 school days in a school year.

Then print `variance_late`

.