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NORMAL APPROXIMATION TO THE BINOMIAL

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.

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NORMAL APPROXIMATION TO THE BINOMIAL

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  1. 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) Difficulty- tables do not go that far! Solution Idea: Approximate Binomial distribution by Normal distribution and use normal distribution for computation of probabilities.

  2. How does Normal distribution approximate Binomial? • Look at the distributions (histograms) of Bin(n, p) for increasing n: Bin(n=10, 0.5) Bin(n=100, 0.5) As n increases, Binomial distribution gets closer to the normal distribution. Bin(n=1000, 0.75)

  3. Normal approximation for X and Let X ~ Bin(n, p), and the observed proportion of successes = X/n. For sufficiently large n, we can approximate X by a normal distribution with mean = np and . Also, for sufficiently large n, we can approximate by a normal distribution with mean = p and . 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.

  4. 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) = We approximate X by a normal distribution with mean μX = μ=100*0.5=50 and standard deviation So, P( X > 60) = P(Z > (60- 50)/5)=P(Z>2)=0.0228. B) we can approximate by a normal distribution with mean = p =0.5 and P( )=P( Z > (0.6 – 0.5)/0.05)=P( Z> 2) = 0.0228.

  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 = P(X ≤ 420) = P( Z ≤ (420 – 400)/15.5) = P(Z ≤ 1.29) = 0.9015.

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