pbinom review

Suppose we take a poll of a random sample of 1000 people, and ask them who they will vote for in November. Suppose that in the population P(DEM) = 0.51, i.e., 51% of people in the population plan to vote DEM. What is the probability that in our sample, less than 50% of people will say they plan to vote DEM?

The normal approximation to pbinom

Recall that we can approximate a \(Bin(\theta, n)\) distribution using \(\mathcal{N}(\theta \times n, n\times \theta \times (1- \theta))\). Compute the probability that fewer than 50% of people will say DEM

Intervals

What is the probability that between 48% and 51% will say DEM? Use pbinom

Transforming random normal variables

Suppose \(X\sim\mathcal{N}(\mu, \sigma^2)\). We can “shift” \(X\) by subtracting \(\mu\).

\((X-\mu) \sim \mathcal{N}(0, \sigma^2)\)

Using that, how can we compute pnorm(q = 1, mean = 0.5, sigma = 1.5) another way?

We can also divide a variable by a constant, “shrinking” it, assuming its mean is 0:

If \(Y\sim \mathcal{N}(0, \sigma^2)\), then \(Y/a = \sim \mathcal{N}(0, \sigma^2/a^2)\)

That means that \((X-\mu)/\sigma \sim \mathcal{N}(0, 1)\). (N.B., (0, 1^2) is the “standard normal distribution”).

Again, let’s recompute pnorm(q = 1, mean = 0.5, sigma = 1.5).

Average of i.i.d. (independently identically distributed) X’s

If \(X_i\sim \mathcal{N}(\mu, \sigma^2)\), then

\(\bar{X} = \frac{X_1 + X_2 + ... + X_n}{n} \sim \mathcal{N}(\mu, \sigma^2/n)\)