Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Normal approximation of Binomial probabilities Normal approximation of Binomial probabilities Recall binomial experiment: Identical trials Two outcomes: success and failure Probability for success and failure consistent Independent trials Normal approximation of Binomial probabilities Binomial probability: Given n and p, n k nk P( X k ) p (1 p) k Normal approximation of Binomial probabilities When n is large, finding the above probability becomes increasingly burdensome if k is in the middle of n, since n k P( X k ) P ( X k ) and P( X k ) P ( X k ), i 0 Or i i i ik i i i P( X k ) 1 P( X k ) In this case, using complement may not help either. Normal approximation of Binomial probabilities We have talked about using Poisson to approximate binomial for n*p is less than 5. or n*p<5. Then how about n*p > 5? Here we can actually use normal distribution to approximate binomial. Normal approximation of Binomial probabilities How to approximate: Given any random variable X~BIN(n, p) 1. Find its mean (n*p) and variance (n*p*(1-p)) 2. Set =n*p, and 2 = n*p*(1-p). 3. Then we can consider X as X~N(n*p, n*p*(1-p) ) 4. To find probabilities, we can standardize X into Z and look them up in the Z/Normal probability table. Normal approximation of Binomial probabilities However, there is something we have missed. Previously, when we use Poisson to approximate Binomial, they are both DISCRETE. But now, Binomial distribution is for DISCRETE random variables and Normal distribution is for CONTINUOUS random variables. Normal approximation of Binomial probabilities What is the potential problem? Simply, how do we calculate P(X=k)? Recall that: For a discrete random variable, we just calculate it directly, there is no problem. For a continuous random variable, we know that for ANY INDIVIDUAL VALUE OF X, P(X=k)=0!!! Normal approximation of Binomial probabilities We have to make some corrections. There is a technique called “continuity correction”. All we need to do is to add a “continuity correction factor”. Under normal approximation, P(X=k)=P(k-0.5 < X < k+0.5) By this correction, we are computing the probability over an interval instead at a single point. Example A hotel has 100 rooms and the probability a room is occupied on any given night is 0.6. Assume the conditions of the binomial are met for the number of occupied rooms on any given night. Example 1. Find the probability that there are 50 rooms occupied at a given night using the exact distribution. 2. Find the probability that there are 50 rooms occupied at a given night using normal approximation. Example 3. Find the probability that there are at least 50 rooms are occupied at a given night using the exact distribution. 4. Find the probability that there are at least 50 rooms occupied at a given night using normal approximation.