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Normal Approximation of the Binomial Distribution

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What is the Probability of getting exactly 30 tails if a coin is tossed 50 times? X = tails, x = 30 n = 50, p = 0.5 P(X = x) = 50 30 (0.5) 30 (1 – 0.5) 50 - 30 = 0.042

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Find the probability that tails will occur less than 30 times…. The magnitude of the calculation motivates us to determine an easier way…

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Notice: The most frequent outcome of flipping a coin should be 25 tails, then 24/26, then 23/27 and so on…. This structure has already been modeled and studied as a normal distribution!

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Therefore Under certain conditions, we may use the Normal Model to approximate probabilities from a Binomial Distribution!!

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Conditions that must be met 1. np > 5 2. nq > 5

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When we will use the Normal distribution to approximate the Binomial distribution, we have to create a z score

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z = x - x = np(1 – p) x = E(x) = np x = given value, but it must be corrected by 0.5 either way (if it is discrete…) s s

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Ex: If you toss a coin 50 times, estimate the probability that you will get tails less than 30 times. To approximate, first we will check to see of the conditions are met

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1. Is np > 5 ? 50(0.5) > 5 ? 25 > 5 YES! 2. Is nq > 5? 50(0.5) > 5 ? 25 > 5 YES! If these conditions are met, that means there were enough trials so a comparable distribution was created…

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Now we need to determine the z score z = x - x s

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Ex: If you toss a coin 50 times, estimate the probability that you will get tails less than 30 times. Success: tails n = 50 p = 0.5 E = x = 50(0.5) = 25

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= 50(0.5)(1 – 0.5) s = np(1 – p) = 3.54

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Less than 30 tails.. x < 30 Imagine the 30 bin 30 29.530.5

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z = x - x = 29.5 - 25 3.54 = 1.27 P(X < 29.5) = P(z < 1.27) = 0.8980 s

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A bank found that 24% of it’s loans become delinquent. If 200 loans are made, find the probability that at least 60 are delinquent.

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We may analyze this situation as a binomial model because: 1. A loan is either paid back or not 2. The loans are all independent

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Check to see if we can approximate… 1.np = (200)(0.24) = 48 (which is greater than 5, so yes) 2. nq = (200)(0.76) = 152 (which is greater than 5, so yes)

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E(X) = x = np = (200)(0.24) = 48 s = (npq) 1/2 =[(200)(0.24)(0.76)] 1/2 = 6.04 X = 60, then adjust to 59.5

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z = 59.5 – 48 6.04 z = 1.90 (97.13% from the chart) Therefore, the number of delinquent loans is 100% - 97.13% = 2.87% Sketch this to confirm!

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If we were to actually calculate the probability directly, it would come out to 3.07%

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Page 449 1,2[odd] 3,5,6, 8,10

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Note: Discrete to Continuous pg 306

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