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 TO THE BINOMIAL A Bin(n, p) random variable X counts the number of successes in n Bernoulli trials with probability of success p on each trial. Suppose we have lots of trials. Example: Suppose the probability that a Democrat would vote for Hillary Clinton in the next presidential election is 0.7. What is the probability that among 200 Democrats, at least 150 would vote for HC? Sln: X = # Dems among 200 who would vote for HC, X ~Bin(200, 0.7). P(at least 150 among 200 would vote for HC) = P(X ≥ 150) Difficultytables do not go that far! Solution Idea: Approximate Binomial distribution by Normal distribution and use normal distribution for computation of probabilities. How does Normal distribution approximate Binomial? Look at the distributions (histograms) of Bin(n, p) for increasing n: 0.15 0.20 0.25 0.0 0.05 0.10 Bin(n=10, 0.5) 4 6 8 10 Bin(n=100, 0.5) 50 60 0.0 As n increases, Binomial distribution gets closer to the normal distribution. 0.005 0.010 40 0.015 0.0 0.020 0.02 0.025 0.04 0.030 0.06 0.08 2 700 720 740 760 780 800 Bin(n=1000, 0.75) Normal approximation for X and pˆ . Let X ~ Bin(n, p), and the observed proportion of successes p̂ = X/n. For sufficiently large n, we can approximate X by a normal distribution with mean X = np and X np(1 p). Also, for sufficiently large n, we can approximate with mean p̂ = p and pˆ p̂ by a normal distribution p(1 p) . n Note 1. Both approximations are consequences of the Central Limit Thm. Note 2. As a Rule of Thumb, for these approximations to be reasonable, we need n and p such that np≥5 and n(1-p) ≥5 ; if np≥10 and n(1-p) ≥10, the approximations are quite good. Example If a coin is tossed 100 times, what is the probability that ( A) it comes up H more than 60 times; (B) the observed proportion of H exceeds 0.6. Solution: X= # of times the coin comes up H. X ~ Bin(100, 0.5). A) Directly: P(X> 60) 100 100 100 61 39 62 38 63 37 (0.5) (0.5) (0.5) (0.5) (0.5) (0.5) = 61 62 63 We approximate X by a normal distribution with mean μX = μ=100*0.5=50 and standard deviation np(1 p) 100(0.5)(1 0.5) 5. So, P( X > 60) = P(Z > (60- 50)/5)=P(Z>2)=0.0228. B) we can approximate and p̂ by a normal distribution with mean pˆ p(1 p) 0.5(1 0.5) 0.05. n 100 P( pˆ 0.6 )=P( Z > (0.6 – 0.5)/0.05)=P( Z> 2) = 0.0228. p̂ = p =0.5 Example The admissions office at an university sends out 1000 admission letters to prospective students. The probability that an admitted student actually enrolls at that university is 0.4. Find the probability that fewer than 420 of the admitted students will enroll at this university. Solution. X = # of admitted students who enroll; X~Bin(1000, 0.4). We will approximate X with a normal distribution with mean 1000*0.4=400 and standard deviation = 1000(0.4)(0.6) 15.5. P(X ≤ 420) = P( Z ≤ (420 – 400)/15.5) = P(Z ≤ 1.29) = 0.9015.