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1 Normal Distribution 1 Normal Distribution. 1.1 Introduction A Bernoulli trial is simple random experiment that ends in success or failure. A Bernoulli trial can be used to make a new random experiment by repeating the Bernoulli trial and recording the number of successes. Now repeating a Bernoulli trial a large number of times has an irritating side e¤ect. Suppose we take a tossed die and look for a 3 to come up, but we do this 6000 times. 1 This is a Bernoulli trial with a probability of success of 6 repeated 6000 times. What is the probability that we will see exactly 1000 success? This is de…nitely the most likely possible outcome, 1000 successes out of 6000 tries. But it is still very unlikely that an particular experiment like this will turn out so exactly. In fact, if 6000 tosses did produce exactly 1000 successes, that would be rather suspicious. The probability of exactly 1000 successes in 6000 tries almost does not need to be calculated. whatever the probability of this, it will be very close to zero. It is probably too small a probability to be of any practical use. It turns out that it is not all that bad, 0:014. Still this is small enough that it means that have even a chance of seeing it actually happen, we would need to repeat the full experiment as many as 100 times. All told, 600,000 tosses of a die. When we repeat a Bernoulli trial a large number of times, it is unlikely that we will be interested in a speci…c number of successes, and much more likely that we will be interested in the event that the number of successes lies within a range of possibilities. Instead of looking for the probability that exactly 1000 successes occur, we are more likely interested in the number of success being around 1000, say o¤ by no more than 15 in either direction. What is the probability that the number of successes is between 9985 and 1015? It turns out that this is a more reasonable 0:41: Unfortunately to do this calculation precisely, you need a pretty powerful computer or calculator. It would seem that the calculations needed to deal with a Bernoulli trial repeated a large number of times would make such probability models too dif- …cult to deal with. However, we noticed something about the graphs of these probability models. When we take any probability p and n large enough, all the graphs of the probability models look very similar. As it turns out, they are more than similar, they are all basically the same. There is one universal proba- bility distribution that will approximate all repeated Bernoulli experiments with enough repetitions. What this means is: Remark 1 In a repeated Bernoulli experiment with enough repetitions, there is an easy method of calculating the probabilities of events of the form "The number of successes is less than or equal to k0;" "The number of successes is at most k1 and at least k0, and "The number of successes is greater than or equal 1 to k0: That is to say, we can quickly compute p(k k0) p(k0 k k1) p(k k1) for any success-probability p when n is any number large enough. The trick to this shortcut is to use one common scale on all our probability models. To …nd this common scale we look back at our measures of the center of a data set and the variation in that data set. We compute a mean value for our probability model, an variance for the model, and a standard deviation for the model. It turns out that there is one table of values that will approximate all sorts of tables obtained from Bernoulli trials with success probability p and large enough n: That table, however, gives probabilities in terms of standard deviations from the mean. To use it in any particular example, we must convert the example probability model to a scale of standard deviations from the mean. 1.2 The Mean, Variance and Standard Deviation of a Re- peated Bernoulli Trial Our …rst step is to consider the mean, variance and standard deviation of a repeated Bernoulli trial. Suppose we have a Bernoulli trial with probability of success p0 and repeat it n times. We …rst commute the mean, variance and standard deviation of its probability model as a set of weighted data. We will then convert the probability model tabulated in terms of the number of successes to a model tabulated by standard deviations from the mean. The result will look weird at …rst, but it is the key to making later calculations much easier. As usual we start with a small n and use a very nice probability of success 1.2.1 Example For example, start with the simplest example: a Bernoulli trial with probability 1 of success 2 and n = 2: Thus the probability model we obtain is # Success 0 1 2 1 1 1 p(#) 4 2 4 The mean of this is computed by multiplying each number of success by its probability and taking the sum. We keep with tradition and use the Greek letter mu, , to stand for this mean. 1 1 1 = 0 + 1 + 2 = 1. 2 4 2 4 We compute the variance by multiplying the square of the distance from the mean by the probability of each number of success. Again tradition dictates 2 we denote this by 2. Then 1 1 1 1 2 = (0 1)2 + (1 1)2 + (2 1)2 = . 2 4 2 4 2 Once we have a variance, we bring it back to normal units by taking the square root. The result is the standard deviation, and we denote it with the Greek letter sigma, . 1 2 = 0:70711: p2 ' We convert each number of success to a scale based on standard deviations from the mean. Thus 0 successes is 1 success from the mean of 1. Each 1 standard deviation measures as 0:70711 successes. So 1 success is 0:70711 = 1:4142 standard deviations. We start our chart as # Success 0 1 2 Standard Deviations 1:4142 1 1 1 p(#) 4 2 4 Next 1 successes is 0 successes from the mean of 1. Each standard deviation measures as 0:70711 successes. So 0 success is 0 standard deviations. We continue our chart as # Success 0 1 2 Standard Deviations 1:4142 0 1 1 1 p(#) 4 2 4 Then 2 successes is +1 successes from the mean of 1. Each standard deviation 1 measures as 0:70711 successes. So +1 success is 0:70711 = 1:4142 standard deviations. We …nish our chart with # Success 0 1 2 Standard Deviations 1:4142 0 1:4142 1 1 1 p2(#) 4 2 4 Notice that the formula that converts the number of success k0 into x0 the number of standard deviations from the mean is k0 x = : 0 All we do next, is drop the old scale and keep the new scale x0 1:4142 0 1:4142 1 1 1 p2(x = x0) 4 2 4 3 1.2.2 Example Just to be thorough, we will do one more small example explicitly; n = 3: Thus the probability model we obtain is # Success 0 1 2 3 1 3 3 1 p(#) 8 8 8 8 We compute the mean as before: 1 3 3 1 12 3 = 0 + 1 + 2 + 3 = = . 3 8 8 8 8 8 2 We compute the variance by multiplying the square of the distance from the mean by the probability of each number of success: 3 1 3 3 3 3 3 1 2 = (0 )2 + (1 )2 + (2 )2 + (0 )2 3 2 8 2 8 2 8 2 8 9 3 3 9 = + + + 32 32 32 32 24 3 = = . 32 4 Once we have a variance, we also have the standard deviation: p3 3 = 0:86603: 2 ' We convert each number of success to a scale based on standard deviations 3 3 from the mean. Thus 0 successes is 2 success from the mean of 2 . Each 3 standard deviation measures as 0:86603 successes. So 2 = 1:5 success is 1:5 1 0:86603 = 1: 732 standard deviations. Next 1 success is 2 success from the 3 1 0:5 mean of 2 . So 2 = 0:5 success is 0:86603 = 0:57735 standard deviations. 1 3 1 Then 2 successes is 2 success from the mean of 2 . So 2 = 0:5 success is 0:5 3 0:86603 = 0:57735 standard deviations. And …nally 3 successes is 2 success from 3 3 1:5 the mean of . So = 1:5 success is = 1: 732 standard deviations. 2 2 0:86603 # Success 0 1 2 3 Standard Deviations 1: 732 0:57735 0:57735 1: 732 1 3 3 1 p3(#) 8 8 8 8 Cleaning up, we get x0 1: 732 0:57735 0:57735 1: 732 1 3 3 1 p3(x = x0) 8 8 8 8 where x is the number of standard deviations from the mean. These are just examples of a small number of repetitions while we are inter- ested cases where there are a large number of repetitions. It would seem that calculating the mean, variance and standard deviations for a probability with a large number of values would make the steps above too lengthy to carry out. Fortunately, we have a shortcut. 4 Theorem 2 Suppose we have a Bernoulli trial with a probability of success given by p.
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