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Normal Approximation of the Binomial Distribution.

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Presentation on theme: "Normal Approximation of the Binomial Distribution."— Presentation transcript:

1 Normal Approximation of the Binomial Distribution

2 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) = (0.5) 30 (1 – 0.5) = 0.042

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

4 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!

5

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

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

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

9 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

10 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

11 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…

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

13 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

14 = 50(0.5)(1 – 0.5) s = np(1 – p) = 3.54

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

16 z = x - x = = 1.27 P(X < 29.5) = P(z < 1.27) = s

17 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.

18 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

19 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)

20 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

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

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

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

24 Note: Discrete to Continuous pg 306


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