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 Binomial Distribution Click here to view the binomial tables. Conditions Fixed number of trials, n Each trial should be a success or a failure Trials should be independent of one another (i.e. one trial doesn't affect the outcome of another trial) Constant Probability of Success, pIf a distribution satisfies the above conditions then it can be modelled as a Binomial Distribution. Hence X∼B(n,p) where n is the number of trials and p is the probability of success. Modelling a Binomial Distribution We can model a situation as binomial if there is either a success or failure of the event occurring, and the other conditions (above) are met.  For example a toss of a coin can be modelled by a binomial distribution as the probability of flipping a heads is either a success or a failure (in which case a tails is tossed), and the probability is constant, 0.5 (unless the coin is biased) with each trial being independent as the last flip doesn't affect the next flip. Example 1 Suggest and specify a suitable distribution model for the following situation:  A company producing high technology hardware is under the belief that for every 10 machines they produce, 1 will be faulty.  This fits a binomial distribution (constant probability, either success or failure - faulty, independent and fixed number of trials). Let X be the number of faulty goods produced by the company:  X∼B(10, 0.1) We can find the probability that a binomial distribution will equal a certain value using a formula. The formula is nCx.px.(1-p)n-x (1-p) is sometimes labelled as q, therefore the formula is also given as nCx.px.qn-x. Where C is the choose function (this can be calculated using the nCr function of a calculator). Example 1 Therefore if we are asked to find P(X=5) where X∼B(25,0.05) we would substitute our values of x(5), p(0.05) and n(25) into our formula to calculate the answer: 25C5.0.055.0.9520 to get the answer of 0.0060 NB: Probabilities are normally given to four decimal places and always have to be less than one and greater than zero. Finding Other Probabilities If we are asked to find probabilities that are inequalities (and not equal to) then we can look up the values in our tables, but only if it is the form of P(X≤x). If it is not in that format we have to transform it using the following rules: P(Xx) = 1 - P(X≤x) P(X≥x) = 1 - P(X≤(x-1)) Once we have our probability in the correct format it is a matter of looking up the value in the tables. Example 1 X∼B(20,0.15) find P(X≤5) From the tables: 0.9327 (this means that the probability than the event X will occur less than or including 5 times is 93.27%) Example 2 X∼B(40,0.05) find P(X<7) P(X<7) = P(X≤6)  From the tables: 0.9966 Example 3 X∼B(30,0.5) find P(X>15) P(X>15) = 1 - P(X≤15) = 1 - 0.5722 (from the tables) = 0.4278 Example 4 X∼B(10, 0.4) find P(X≥8) P(X≥8) = 1 - P(X≤7) = 1 - 0.9877 (from the tables) = 0.0123 Finding the Expectation and the Variance The  expectation, E(X); average;mean, of a binomial distribution is np.  E(X) = np Example 1 Find the expectation of the distribution where X∼B(20,0.2).   E(X) = 20*0.2 E(X) = 4 (this means that for a binomial distribution of 20 items or individuals, with a 20% probability of success, we expect 4 items/individuals to be successful) Example 2 The expectation of a distribution equals 5, X∼B(n,0.5), find n. E(X) = np 5 = 0.5n n = 10 Therefore X∼B(10,0.5) to give an expectation of 5 The variance, σ2, is given as npq. Var(X) = np(1-p) Example 1 Find the variance of the distribution where X∼B(10,0.3) Var(X) = np(1-p) Var(X) = 10*0.3*0.7 Var(X) =2.1 Example 2 The variance of a distribution equals 5, X∼B(25,p), find the possible values of p. Var(X) = np(1-p) 5 = 25p(1-p) 1/5 = p - p2 p2 - p + 1/5 = 0 Solve by simultaneous equation p = 0.7236, or p = 0.2764 NB: Remember that E(aX + b) = a*E(X) + b; and that Var(aX + b) = a2Var(X). Approximating the Binomial Distribution Poisson Distribution We can approximate a binomial distribution with a Poisson distribution if the following conditions are met: n is large (greater than 50 usually) p is small (usually less than 0.3) Because Poisson looks at the mean value, we have to calculate E(X) for the binomial and this is the mean value for our Poisson distribution. If X∼B(n,p) and the suitable conditions are met then X≈Y∼Po(np) Example 1 X∼B(100, 0.1), using a suitable approximation find P(X≥6)  X≈Y∼Po(10)   P(X≥6) = 1 - P(X≤5) = 1 - 0.0671(Poisson tables) = 0.9329  Therefore the probability that X is greater than or equal to 6 is approximately 93.29% To find out more about the Poisson distribution click here.   Normal Distribution We can approximate a binomial distribution with a normal distribution if the follow conditions are met: n is large (greater than 50 usually) p is close to 0.5 We can approximate the binomial with a normal to create Y∼N(np, np(1-p)2) Hence X∼B(n,p)≈Y∼N(np, np(1-p)2), we also have to do a continuity correction, because a normal distribution is continuous whereas a binomial distribution isn't. If we have to find P(X<80) then we don't want to include 80 in our search so we find P(X<79.5), for P(X>80) we find P(X>80.5), for P(X≤80) we find P(X<80.5) and for P(X≥80) we find P(X>79.5). We would then solve normally using our normal distribution process. Example 1 X~B(100, 0.5), use a suitable approximation to find P(X≤52) X≈Y~N(50,25) and we want to find P(X≤52) Our z score is calculated using: z = (X-μ)/σ Our continuity correction means we will take x as 52.5 because we want to include 52 in our calculation. Hence z = (52.5 - 50)/5 Which gives a z value of 0.5. Looking this value up in our tables we find that P(X≤52)=0.6915     Questions If the following can be modelled as a Binomial distribution then state the distribution: a biased coin is tossed in the air, the probability of throwing a head is 0.4. If the following can be modelled as a Binomial distribution then state the distribution: a car passes a school 7 times in 1 day. X∼B(10,0.2) Find P(X=8) X∼B(15,0.1) Find P(X=20) X∼B(10,0.5) Find P(X=0) X∼B(8,0.4) Find P(X=2) X∼B(15,0.4) Find P(X=5) X∼B(20,0.15) Find P(X≤3)  X∼B(25,0.01) Find P(X≤2) X∼B(50,0.5) Find P(X≤23) X∼B(10,0.2) Find P(X≤5) X∼B(10,0.2) Find P(X≥5) X∼B(10,0.2) Find P(X>5) X∼B(10,0.2) Find P(X<5) X∼B(25, 0.08) Find P(X≥7) X∼B(30,0.1) Find P(X<9) X∼B(40,0.2) Find P(X>10)Answers X∼B(n,0.4) where X is the probability that a heads is thrown This cannot be modelled as a binomial, but can be done so as a Poisson 0.000073728 0 - 20 is outside the range 0.0010- This can be found by looking in the tables for P(X≤0) 0.2090 0.1859 0.6477 0.9980 0.3359 0.9936 0.0328 - 1-P(X≤4) 0.0064 - 1-P(X≤5) 0.9672 - P(X≤4) 0.0028 0.9980 0.1608  Page last updated on 22/09/15  