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The Binomial Distribution This distribution is useful for modelling situations in which the random variable (representing the outcome of an experiment) may take one of only two possible values. For example, sitting for a test, you can either have a success or a failure; if a coin is tossed, the outcome is either a Head or a Tail, etc. The following additional features characterise the Binomial: 1. The number of trials n is a known constant and it is not too large. 2. For each trial, the probability of success p is a known constant and is the same for each trial. 35 30 25 20 Frequency or e M 5. 6 4. 2 2. 8 1. 4 15 10 5 0 0 Frequency Histogram Bin Binomial (25, 0.1) Histogram 30 20 15 Frequency 10 5 Bin or e M 23 .4 21 .8 20 .2 18 .6 0 17 Frequency 25 Binomial (25, 0.9) Mean Standard Error Median Mode Standard Deviation Sample Variance Kurtosis Skewness Range Minimum Maximum Sum Count 24.19 0.281301 24 24 2.813011 7.91303 -0.45478 0.107326 13 19 32 2419 100 Binomial (40, 0.6) Mean =np = 24 S.D. = (np (1-p))1/2 = 3.09 Histogram 25 15 Frequency 10 5 Bin or e M 29 .4 26 .8 24 .2 21 .6 0 19 Frequency 20 Binomial (40, 0.6) Binomial (2500, 0.001) Histogram 30 20 15 Frequency 10 5 or e M 5. 6 4. 2 2. 8 1. 4 0 0 Frequency 25 Bin Mean Standard Error Median Mode Standard Deviation Sample Variance Kurtosis Skewness Range Minimum Maximum Sum Count 2.43 0.159706 2 3 1.597062 2.550606 -0.16406 0.494198 7 0 7 243 100 Binomial (2500, 0.001) Mean=np = 2.5 S.D. =(np (1-p))1/2 = 1.58 The Poisson Distribution: Consider the following situation. Customers come to a shop at a regular interval The average rate of arrival of customers is given (l) but the total number of customers to arrive is unknown and could be very large • The probability of getting r customers in any given interval is then given by: • P(X = r) = (e-l lr)/ r! • where e = 2.718….…. • Theory: • If x ~ Binomial (n,p), n is large, yet np is not large, then x has an approximate Poisson distribution with l = np Half a percent of 500 students in a course are likely to resort to unfair means. Find the probability that Exactly 4 students will resort to unfair means0.133836 At most 6 students 0.133602 will resort More thanmeans 4 students will to unfair Less than 4 students will resort to unfair means resort to unfair means Continuous Probability Distributions The Uniform distribution This distribution is useful for modelling situations where a random outcome may be imagined to have realizations along a straight line with equal probability Suppose that commuters waiting on a platform 50 metres long are likely to spread out evenly Suppose we call X the location chosen by a typical commuter. Then 0 X 50 and X has a uniform distribution f(X) For every value of X, f(X) = 1/50 1/50 0 25 50 X Any Expected realization of XofisXa symmetric = mode 25 TheThe distribution isValue perfectly The Median = 25 f(X) For every value of X, f(X) = 1/(b-a) 1/(b-a) 0 a (a+b)/2 b X The Expected Value of X = (a+b)/2 The distribution is perfectly 2 symmetric So the St. Dev. of X = (b-a) /12 Any realization of X is a mode The variance of X = (b-a) /12 The Median = (a+b)/2 f(X) 1/50 0 25 50 X The St. Dev.ofofXX== (50-0) (50-0)2/12 The variance /12 =14.43 =208.3 0. 24 87 10 258 .1 5 11 2 19 392 56 .9 74 29 059 .8 27 36 39 725 97 .6 99 39 26 8 M or e Frequency Histogram 14 12 10 8 6 Frequency 4 2 0 Bin Column1 Mean 24.40611 Standard Error 1.417486 Median 23.62972 Mode #N/A Standard Deviation 14.17486 Sample Variance 200.9266 Kurtosis -0.99496 Skew nessSimulation from 0.051839 Range Uniform(0, 50)49.22636 Minimum 0.334178 Maximum 49.56053 Sum 2440.611 Count 100 The Triangular distribution Can you locate the mean median and the mode? X The Triangular distribution Every normal The two shaded parts distribution is must be equal in area. symmetric about the mean. Mean-a Mean Mean +a z The two inner green areas are equal as well. Mean-a Mean Mean +a The area shaded brown is approximately 68% of the whole -1 0 1 z The area shaded orange is approximately 90% of the whole -1.645 0 +1.645 The area shaded orange is approximately 95% of the whole -2 0 +2 f(x) m Two Normal Distribution curves with same mean (m) but different standard deviation x f(z) m m+2 Two Normal Distribution curves with same standard deviation but different values of mean z Find the area shaded black Answer =0.5 +0.4452 =0.9452 The area shaded black is 0.9452 as well (by symmetry) Find the area shaded black The black shaded area below has the same area (by symmetry) Answer = 0.5+0.3944 = 0.8944 First Thisfind area Then add 0.5 this area is 0.3944 The answer is 0.675 Q3= Q1= 0.675 -0.675 Look for z so that this area is 0.25 25% 25% Q1 Q3 D9 D1= = 1.28 -1.28 Look for z so that this area is 0.4 10% 10% D1 D9 Find the first decile of the SND So Q1 for x So Q3 for x Exercise: Find the first decile of + 20 Q1 for z is –0.675 =x-0.675(5) + 20 = 0.675(5) = zs+m Q3 for2 z is 0.675 X =~ 16.625 Normal (20, 5 ) = 23.375 Find the first and the third quartile of X ~ Normal (20, 52) 35% of British men are at least 185 cm tall If I meet 200 such men on any given day, what is the probability that 100 or more of them are 185 cm or taller? Normal Approximation of the Binomial (Chapter P4 of the text) Probability(M) Prob( M = 100) = ? = P(99.5 < M < 100.5) is the normal approximation .99 100. 101. 102. 103. 104. M Prob(M 100) = Prob(M = 100) + Prob(M = This area is shaded black But it is also the area under the red polygon Prob(M 100) >=200) 99.5) 101) + Prob(M = 102) ...=+Prob(M Prob(M to the right of 99.5, or Prob(M > 99.5) Probability(M) So Prob( M = 100) = P(99.5 < M < 100.5) .99 100. 101. 102. 103. 104. M Probability(M) Find Prob( M < 103) It is also the area This area shaded under theisred black polygon or Prob(M 102.5) .99 100. 101. 102. 103. 104. M Binomial Normal Approximation Prob( M = 100) = Prob(99.5 < M <100.5) Prob(M 100) = Prob(M > 99.5) Prob( M < 103)= Prob(M 102.5) Prob( M 103) = Prob( M 103.5) We can similarly show that The approximation works if both of np and n(1-p) 5 Read page 500 (inset)