Lecture 31
Qi Wang, Department of Statistics
Nov 5, 2018
If we have independent Normal random variables, then the sum(or other linear combination) of these Normal random variables is ALSO Normal
If $X_1 \sim N(\mu_1, \sigma_1), X_2 \sim N(\mu_2, \sigma_2), \cdots, X_n \sim N(\mu_n, \sigma_n)$, and $X = \sum_{i=1}^n X_i$, then
Let $X_1, X_2$ and $X_3$ be independent Normal random variables, where $$X_1 \sim N(\mu = 4, \sigma = 2), X_2 \sim N(\mu = 3.1, \sigma = 7), X_3 \sim N(\mu = 1.5, \sigma = 1.4)$$
If a Binomial distribution has a large enough combination of n and p, it behaves much like a Normal distribution, which means we can use the Normal distribution to approximate the original Binomial distribution
You may notice that Binomial is Discrete, and Normal is Continuous. This means the approximation comes at a cost of accuracy that we must try to correct. When we use the approximation, we need to perform a continuity correction:
If all conditions are satistified, find the Normal approximation to the following probability statement where $X$ follows a Binomial distribution
A class has 400 students, and each drops the course independently with probability 0.07. Let X be the number of students that finish the course