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Published byLogan Caldwell Modified about 1 year ago

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The Binomial Distribution ► Arrangements ► Remember the binomial theorem?

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Expanding using arrangements (a+b) 4 =aaaa + aaab + aaba + abaa + baaa + aabb + abab + abba + baab + baba + bbaa + abbb + babb + bbab + bbba + bbbb Arrangements of 4 As Arrangements of 3 As and 1 B Arrangements of 2 As and 2 Bs Arrangements of 1 A and 3 Bs Arrangements of 4 Bs The 2 nd line contains terms corresponding to a 3 b so coefficient is The 3 rd line contains terms corresponding to a 2 b 2 so coefficient is The 4 th line contains terms corresponding to ab 3 so coefficient is The 1 st line contains terms corresponding to a 4 so coefficient is The 5 th line contains terms corresponding to b 4 so coefficient is

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Arrangements How many ways are there of arranging these? A A A B B B B B B n = 9r = 3

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Example – using a calculator How many ways are there of arranging these? A A A B B B B B B n = 9r = 3 To calculate this, type “9” followed by “ n C r ” followed by “3” and press equals? Use your calculator to work out Explain your answer.

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Binomial Theorem A general rule for any expansion is… A special case occurs when a=1 and b=x …

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The Binomial Distribution The binomial distribution is a discrete distribution defined by 2 parameters the number of trials - n the probability of a success - p WRITTEN : … which means the discrete random variable X is binomially distributed

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Binomial distribution - example ► Testing for defects “with replacement” Have many light bulbs Pick one at random, test for defect, put it back If there are many light bulbs, do not have to replace

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Binomial distribution ► Consider 3 trials n=3 ► p is the probability of picking a good bulb so (1-p) is the probability of picking a defect bulb ► the random variable X is the measure of the number of good bulbs ► If we want P(X=0): Can happen one way: defect-defect-defect (1-p)(1-p)(1-p) (1-p) 3

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Binomial distribution ► If we want P(X=1): Can happen three ways: 100, 010, 001 p(1-p)(1-p)+(1-p)p(1-p)+(1-p)(1-p)p 3p(1-p) 2 For simplicity: 1 - good bulb 0 - defect bulb

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Binomial distribution ► If we want P(X=2): Can happen three ways: 110, 011, 101 pp(1-p)+(1-p)pp+p(1-p)p 3p 2 (1-p) 1 - good bulb 0 - defect bulb

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Binomial distribution ► We want P(X=3): Can happen one way: 111 ppp p3p3 1 - good bulb 0 - defect bulb

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Binomial distribution P(X=0): (1-p) 3 P(X=3): p 3 P(X=1):3p(1-p) 2 P(X=2): 3p 2 (1-p) r is the number of good bulbs

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Binomial distribution function In general, the binomial distribution function is given by: or q = 1 -p

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Binomial distribution - example ► Testing for defects “with replacement” Suppose 10 bulbs were tested The probability of a good bulb is 0.7 What is the probability of there being 8 good bulbs in the test? n = 10 p = 0.7 r = 8

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Binomial distribution - example n = 10p = 0.9r = 8 = (3 d.p.)

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Binomial distribution ► Typical shape of binomial: Symmetric Mean and Mode approx = p*n r P

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Binomial distribution - expected value ► For the binomial distribution The mean value of X is given by np … this is also the expected value of X - E(X)

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Binomial distribution - example ► Testing for defects “with replacement” Suppose 10 bulbs were tested The probability of a good bulb is 0.7 What is the probability of there being 8 good bulbs in the test? The mean (expected value)? = (3 d.p.) The mean value of X is given by np = 10 x 0.7 = 7

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