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Chapter 2 Bernoulli Trials

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Chapter 2 Bernoulli Trials Chapter 2 Bernoulli Trials 2.1 The Binomial Distribution In Chapter 1 we learned about i.i.d. trials. In this chapter, we study a very important special case of these, namely Bernoulli trials (BT). If each trial yields has exactly two possible outcomes, then we have BT. B/c this is so important, I will be a bit redundant and explicitly present the assumptions of BT. The Assumptions of Bernoulli Trials. There are three: 1. Each trial results in one of two possible outcomes, denoted success (S) or failure (F ). 2. The probability of S remains constant from trial-to-trial and is denoted by p. Write q =1 p for the constant probability of F . − 3. The trials are independent. When we are doing arithmetic, it will be convenient to represent S by the number 1 and F by the number 0. One reason that BT are so important, is that if we have BT, we can calculate probabilities of a great many events. Our first tool for calculation, of course, is the multiplication rule that we learned in Chapter 1. For example, suppose that we have n = 5 BT with p = 0.70. The probability that the BT yield four successes followed by a failure is: P (SSSSF )= ppppq = (0.70)4(0.30) = 0.0720. Our next tool is extremely powerful and very useful in science. It is the binomial probability distribution. Suppose that we plan to perform/observe n BT. Let X denote the total number of successes in the n trials. The probability distribution of X is given by the following equation. n! − P (X = x)= pxqn x, for x =0, 1,...,n. (2.1) x!(n x)! − 17 To use this formula, recall that n! is read ‘n-factorial’ and is computed as follows. 1! = 1;2! = 2(1) = 2;3! = 3(2)(1)= 6, 4! = 4(3)(2)(1) = 24; and so on. By special definition, 0!=1. (Note to the extremely interested reader. If you want to see why Equation 2.1 is correct, go to the link to the Revised Student Study Guide on my webpage and read the More Mathematics sections of Chapters 2, 3 and 5.) I will do an extended example to illustrate the use of Equation 2.1. Suppose that n =5 and p =0.60. I will obtain the probability distribution for X. 5! P (X =0)= (0.60)0(0.40)5 =0.0102. 0!5! 5! P (X =1)= (0.60)1(0.40)4 =0.0768. 1!4! 5! P (X =2)= (0.60)2(0.40)3 =0.2304. 2!3! 5! P (X =3)= (0.60)3(0.40)2 =0.3456. 3!2! 5! P (X =4)= (0.60)4(0.40)1 =0.2592. 4!1! 5! P (X =5)= (0.60)5(0.40)0 =0.0778. 5!0! You should check the above computations to make sure you are comfortable using Equation 2.1. Here are some guidelines for this class. If n 8, you should be able to evaluate Equation 2.1 ‘by hand’ as I have done above for n =5. ≤ For n 9, I recommend using a statistical software package on a computer or the website I describe later.≥ For example, the probability distribution for X for n = 25 and p =0.50 is presented in Table 2.1. Equation 2.1 is called the binomial probability distribution with parameters n and p; it is denoted by Bin(n, p). With this notation, we see that my earlier ‘by hand’ effort was the Bin(5,0.60) and Table 2.1 is the Bin(25,0.50). Sadly, life is a bit more complicated than the above. In particular, a statistical software package should not be considered a panacea for the binomial. For example, if I direct my computer to calculate the Bin(n,0.50) for any n 1023 I get an error message; the computer program is smart enough to realize that it has messed≥ up and its answer is wrong; hence, it does not give me an answer. Why does this happen? Well, consider the computation of P (X = 1000) for the Bin(2000,0.50). This involves a really huge number (2000!) divided by the square of a really huge number (1000!), and them multiplied by a really really small positive number ((0.50)2000). Unless the computer programmer exhibits incredible care in writing the code, the result will be an overflow or an underflow or both. 18 Table 2.1: The Binomial Distribution for n = 25 and p =0.50. x P (X = x) P (X x) P (X x) ≤ ≥ 0 0.0000 0.0000 1.0000 1 0.0000 0.0000 1.0000 2 0.0000 0.0000 1.0000 3 0.0001 0.0001 1.0000 4 0.0004 0.0005 0.9999 5 0.0016 0.0020 0.9995 6 0.0053 0.0073 0.9980 7 0.0143 0.0216 0.9927 8 0.0322 0.0539 0.9784 9 0.0609 0.1148 0.9461 10 0.0974 0.2122 0.8852 11 0.1328 0.3450 0.7878 12 0.1550 0.5000 0.6550 13 0.1550 0.6550 0.5000 14 0.1328 0.7878 0.3450 15 0.0974 0.8852 0.2122 16 0.0609 0.9461 0.1148 17 0.0322 0.9784 0.0539 18 0.0143 0.9927 0.0216 19 0.0053 0.9980 0.0073 20 0.0016 0.9995 0.0020 21 0.0004 0.9999 0.0005 22 0.0001 1.0000 0.0001 23 0.0000 1.0000 0.0000 24 0.0000 1.0000 0.0000 25 0.0000 1.0000 0.0000 Total 1.0000 19 Figure 2.1: The Bin(100, 0.5) Distribution. 0.08 0.06 0.04 0.02 35 40 45 50 55 60 65 Figure 2.2: The Bin(100, 0.2) Distribution. 0.10 0.08 0.06 0.04 0.02 10 15 20 25 30 Figure 2.3: The Bin(25, 0.5) Distribution. 0.16 0.14 0.12 0.10 0.08 0.06 0.04 0.02 8 11 14 17 20 Figure 2.4: The Bin(50, 0.1) Distribution. 0.20 0.18 0.16 0.14 0.12 0.10 0.08 0.06 0.04 0.02 0 3 6 9 12 Before we condemn the programmer for carelessness or bemoan the limits of the human mind, note the following. We do not need to evaluate Equation 2.1 for large n’s b/c there is a very easy way to obtain a good approximation to the exact answer. Figures 2.1–2.4 probability histograms for several binomial probability distributions. Here is how they are drawn. The method I am going to give you works only if the possible values of the random variable are equally spaced on the number line. This definition can be modified for other situations, but we won’t need to do this. 1. On a horizontal number line, mark all possible values of X. For the binomial, these are 0, 1, 2,... n. 2. Determine the value of δ (Greek lower case delta) for the random variable of interest. The number δ is the distance between any two consecutive values of the random variable. For the binomial, δ =1, which makes life much simpler. 3. Above each x draw a rectangle, with its center at x, its base equal to δ and its height equal to P (X = x)/δ. Of course, in the current case, δ =1, so the height of each rectangle equals the probability of its center value. For a probability histogram (PH) the area of a rectangle equals the probability of its center value, b/c P (X = x) Area = Base Height = δ = P (X = x). × × δ A PH allows us to ‘see’ a probability distribution. For example, for our pictures we can see that the binomial is symmetric for p =0.50 and not symmetric for p =0.50. This is not an accident of 6 21 the four pictures I chose to present; indeed, the binomial is symmetric if, and only if, p =0.50. By the way, I am making the tacit assumption that 0 <p< 1; that is, that p is neither 1 nor 0. Trials for which successes are certain or impossible are not of interest to us. (The idea is that if p = 1, then P (X = n)=1 and its PH will consist of only one rectangle. Being just one rectangle, it is symmetric.) Here are some other facts about the binomial illustrated by our pictures. 1. The PH for a binomial always has exactly one peak. The peak can be one or two rectangles wide, but never wider. 2. If np is an integer, then there is a one-rectangle wide peak located above np. 3. If np is not an integer, then the peak will occur either at the integer immediately below or above np; or, in some cases, at both of these integers. 4. If you move away from the peak in either direction, the heights of the rectangles become shorter. If the peak occurs at either 0 or n this fact is true in the one direction away from the peak. It turns out to be very important to have ways to measure the center and spread of a PH. The center is measured by the center of gravity, defined as follows.
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