Lesson 23-STAT305-1445 (Binomial Distribution) PDF

TAIBAH UNIVERSITY ‫جامعة طيبة‬ Faculty of Science ‫كلية العلوم‬ Department of Math. ‫قسم الرياضيات‬ Probability and Statistics for Engineers STAT 301& 305 First Semester 1445 H Teacher :  Lesson 23 Bernoull...

TAIBAH UNIVERSITY ‫جامعة طيبة‬ Faculty of Science ‫كلية العلوم‬ Department of Math. ‫قسم الرياضيات‬ Probability and Statistics for Engineers STAT 301& 305 First Semester 1445 H Teacher :  Lesson 23 Bernoulli and Binomial Distribution Bernoulli Distribution: Consider a trial, or an experiment, whose outcome can be classified as either a success or a failure. If we let 𝑋 the number of the outcome is a success, then the probability mass function of 𝑋 is given by 𝑋 0 1 Total 𝑓(𝑥) = 𝑃(𝑋 = 𝑥) 𝑞 𝑃 1 Where p, 0 ≤ 𝑝 ≤ 1 is the probability that the trial is a success, and 𝑞 = 1 − 𝑝 , that means 𝑝 + 𝑞 = 1 Probability distribution function (Pdf) 𝑓(𝑥) : A random variable X is said to be Bernoulli distribution with parameter p if it’s the probability mass function is given by 𝑃 𝑖𝑓 𝑥 = 1 𝑓 𝑥 =ቊ 𝑞 𝑖𝑓 𝑥 = 0 Or 𝑓 𝑥 = 𝑝 𝑥 𝑞1−𝑥 , 𝑥 = 0,1 Where 𝑝 + 𝑞 = 1. Cumulative distribution function ( CDF). F(x) 0. 𝑥 2 ? 𝑃 𝑋 > 2 = 𝑃 𝑋 = 3 + 𝑃(𝑋 = 4) = 𝑏 3,4,0.75 + 𝑏(4,4,0.75) = 43 0.753 0.254−3 + 44 0.754 0.254−4 = (4) (0.753) (0.25) + (1) ( 0.754) (1) =0.73828 2) Find 𝑃 2 ≤ 𝑋 ≤ 4 ? 𝑃 2 ≤ 𝑋 ≤ 4 = 𝑃 𝑋 = 2 + 𝑃 𝑋 = 3 + 𝑃(𝑋 = 4) = 𝑏 2,4,0.75 + 𝑏 3,4,0.75 + 𝑏(4,4,0.75) = 42 0.752 0.254−2 + 43 0.753 0.254−3 + 44 0.754 0.254−4 = (6)(0.752) (0.252) + (4) (0.753) (0.25) + (1) ( 0.754) (1) = 0.9492 3) Find 𝑃(𝑋 ≤ 1) ? 𝑃 𝑋 ≤ 1 = 𝑃 𝑋 = 0 + 𝑃(𝑋 = 1) = 𝑏 0,4,0.75 + 𝑏(1,4,0.75) = 40 0.750 0.254−0 + 41 0.751 0.254−1 = (1) (1) (0.254) + (4) ( 0.751) (0.253) = 0.0507 2- Find the expected value (mean) and the variance of the number of components that survive? The Mean is 𝜇 = 𝐸 𝑋 = 𝑛𝑝 = 4 0.75 = 3 And the variance is 𝜎 2 = 𝑉 𝑋 = 𝑛 𝑝 𝑞 = 4 0.75 0.25 = 0.75 Also, the standard deviation is 𝜎= 𝑉(𝑋) = 0.75 = 0.866 Example 5: Let 𝑋 be a binomial random variable with probability function 𝑏 𝑥; 5,0.8 ; find (a) the values of mean and variance ? (b) 𝑃 𝑋 ≥ 2 ? Answer: a) 𝑛 = 5 ; 𝑝 = 0.8 𝑡ℎ𝑒𝑛 𝑞 = 1 − 𝑝 = 1 − 0.8 = 0.2 The mean 𝜇 = 𝐸 𝑥 = 𝑛 𝑝 = 5 0.8 = 4 The variance 𝜎 2 = 𝑉 𝑥 = 𝑛𝑝𝑞 = 5 0.8 0.2 = 0.8  𝑛 b) 𝑓 𝑥 = 𝑥 𝑝 𝑥 𝑞 𝑛−𝑥 ; 𝑥 = 0, …. , 𝑛 5 𝑓 𝑥 = (0.8)𝑥 (0.2)5−𝑥 ; 𝑥 = 0,1, …. , 5 𝑥 𝑝 𝑥 ≥ 2 = 𝑃 𝑥 = 2 + 𝑃 𝑥 = 3 + 𝑃 𝑥 = 4 + 𝑃(𝑥 = 5) 5 5 = 2 (0.8)2 (0.2)5−2 + 3 (0.8)3 (0.2)5−3 5 5 + 4 (0.8)4 (0.2)5−4 + 5 (0.8)5 (0.2)5−5 = 0.0512 + 0.2048 + 0.4096 + 0.32768 = 0.99328 Or 𝑝 𝑥 ≥ 2 = 1 − 𝑃 𝑥 < 2 = 1 − [𝑃 𝑥 = 0 + 𝑃 𝑥 = 1 ] = 1 − 50 0.8 0 0.2 5−0 + 51 0.8 1 0.2 5−1 = 1 − 0.00032 + 0.0064 = 1 − 0.00672 = 0.99328 Example 6: Let 𝑋 be a binomial random variable with probability function 𝑏 𝑥; 𝑛, 𝑝 , 𝑚𝑒𝑎𝑛 4 and variance 0.8. Find the values of 𝑛 and 𝑝. Answer: 𝜇=4 𝑎𝑛𝑑 𝑉 𝑥 = 0.8 𝜇=𝐸 𝑥 =𝑛𝑝=4 𝑉(𝑥) 0.8 0.8 𝜎2 = 𝑉 𝑥 = 𝑛𝑝𝑞 = 0.8 ֜ q = = = = 0.2 𝑛𝑝 𝑛𝑝 4 So, 𝑝 = 1 − 𝑞 = 1 − 0.2 = 0.8 𝜇 4 and 𝑛= = =5 𝑝 0.8 Example 7: If the random variable 𝑋 has a Binomial distribution , with parameters 𝑛 = 5 and 𝑝 = 0.3 Find the following: 1) 𝑃 𝑋 = 3 2) The variance 𝑉(𝑋) Answer: 5 1) 𝑓 𝑥 = 𝑥 (0.3)𝑥 (0.7)5−𝑥 𝑥 = 0,1,2,3,4,5 5 𝑃 𝑋=3 =𝑓 3 = 3 (0.3)3 (0.7)5−3 = 10 0.027 0.49 = 0.1323 2) 𝜎 2 = 𝑉 𝑋 = 𝑛𝑝𝑞 = 5 0.3 0.7 = 1.05