MODULE IV BINOMIAL DISTRIBUTION A random variable X is said to follow binomial distribution with parameters n & p if P ( X ) = nc x p x q n x where x = 0, 1,2,3..n, p is the probability of success & q is the probability of failure and q = 1 p and p(x) is called the probability density function. Note n For a binomial distribution p(x) = 1 x=0 Proof n p(x) = nc 0 p 0 q n-0 + nc 1 p 1 q n- 1 + nc 2 p 2 q n-2 +..+ nc n p n q n-n x=0 = q n + nc 1 p 1 q n- 1 + nc 2 p 2 q n-2 +..+ p n = ( p + q ) n = 1 In a binomial distribution 1 ) n, the number of trials is finite. 2 ) each trial has two possible outcomes called success & failure. 3)all the trials are independent 4) p & hence q is constant for all the trials. Mean and Variance of a binomial distribution Mean = = x p(x) p(x) But for Binomial distribution p(x) = 1 = x p(x) = x nc x p x q n x = 0. nc 0 p 0 q n-0 +1. nc 1 p 1 q n- 1 +2. nc 2 p 2 q n-2 +..+ n.nc n p n q n-n = np 1 q n- 1 +n(n-1)p 2 q n-2 +..+ n p n = np(q n- 1 + (n-1)p q n-2 +..+ p n-1 ) = np( q + p ) n-1 = np Mean = = np Variance = 2 = x 2 p(x) 2 p(x) But for the binomial distribution p(x) = 1 2 = x 2 p(x) 2 = 0. nc 0 p 0 q n-0 +1. nc 1 p 1 q n- 1 +4. nc 2 p 2 q n-2 +..+ n 2.nC n p n q n-n (n 2 p 2 )
= np 1 q n- 1 +2n(n-1)p 2 q n-2 + 3 /2n(n-1)(n-2) p 3 q n-3..+ n 2 p n - n 2 p 2 = np (q n- 1 +2 (n-1)p q n-2 +3 /2 (n-1)(n-2) p 2 q n-3..+ n p n-1 )- n 2 p 2 = np[ (q n- 1 + ( n-1)p q n-2 +1 /2 (n-1)(n-2) p 2 q n-3..+ p n-1 ) +(( n-1)p q n-2 +1 (n-1)(n-2) p 2 q n-3..+ (n-1) p n-1 ) ] -n 2 p 2 = np[( q + p ) n-1 +( n-1)p (q n-2 + (n-2) pq n-3..+ p n-2 )] -n 2 p 2 = np [ ( q + p ) n-1 +( n-1)p(q + p ) n-2 ] -n 2 p 2 = np [ ( q + p ) n-1 +( n-1)p(q + p ) n-2 ] -n 2 p 2 = np ( 1+ (n-1)p) -n 2 p 2 = np + n 2 p 2 n p 2 n 2 p 2 = np(1 p) = npq Standard deviation = npq Problems 1 ) The mean and variance of a binomial variate are 16 & 8. Find i) P (X= 0) ii) P( X 2) Mean = = np = 16 Variance = 2 = npq = 8 i) P ( X = 0 ) = nc 0 p 0 q n-0 = (½) 0 (1/2) 32 = (1/2) 32 npq / np = 8/16 = 1/2 ie, q = ½ p = 1 q = ½ np = 16 ie, n = 32 ii)p( X 2) = 1 P ( X < 2) = 1 P( X = 0,1) = 1 P(X = 0) P(X = 1) = 1-33 (1/2) 32 2) Six dice are thrown 729 times.howmany times do you expect atleast 3 dice to show a 5 or 6? Here n = 6,N = 729 P(x 3) = 6Cx p x q n-x
Let p be the probability of getting 5 or 6 with 1 dice ie, p = 2/6 = 1/3 q = 1 1/3 = 2/3 P(x 3) = P ( x = 2,3,4,5,6) = p( x=3)+p(x=4)+p(x=5)+p(x=6) =0.3196 number of times = 729*0.3196 = 233 3)A basket contains 20 good oranges and 80 bad oranges. 3 oranges are drawn at random from this basket. Find the probability that out of 3 i) exactly 2 ii)atleast 2 iii)atmost 2 are good oranges. Let p be the probability of getting a good orange ie, p =80C 1 100C 1 = 0.8 q = 1-0.8 = 0.2 i ) p (x=2) = 3C2 (0.8) 2 (0.2) 1 = 0.384 ii) p(x 2) = P(2) +p(3)= 0.896 iii) p(x 2) = p(0) +p(1) + p(2) = 0.488 4) In a sampling a large number of parts manufactured by a machine, the mean number of defective in a sample of 20 is 2. Out of 1000 such samples howmany would expected to contain atleast 3 defective parts. n=20 n p =2 ie, p=1/10 q = 1-p = 9/10 p(x 3) = 1 p ( x < 3 ) = 1 p( x = 0,1,2) = 0.323 Number of samples having atleast 3 defective parts = 0.323 * 1000 = 323 Home work 1 ) Assume that on the average 1 telephone number out of 15 called between 2 pm and 3pm on a week day is busy. What is the probability that if 6 m randomly selected numbers are called i) not more than 3 will be busy ii) atleast 3 of them will be busy. 2) If 10% of bolts produced by a machine are defective. Determine the probability that out of 10 bolts, chosen at random i) one ii)none iii)atmost 2 bolts will be defective. 3) In 800 families with 5 children each, howmany families would be expected to have i) 3 boys and 2 girls ii)atmost 2 girls.iii) no girl ( Assume the probability for boys & girls tobe equal.) 4) During a war 1 ship out of 9 was sunk on an average in making a certain voyage. Find the probability that exactly 3 out of a convoy of 6 ships would arrive safely. 5) In a large consignment of electric bulbs 10 percent are defective. A random sample of 20 is taken for inspection. Find the probability that
i) all are good bulbs ii) at most there are 3 defective bulbs. iii) Exactly there are 3 defective bulbs. FITTING OF BINOMIAL DISTRIBUTION The process of determining the most appropriate values of the parameters from the given observations and writing down the probability distribution function is known as fitting of the binomial distribution. Problems 1) Fit an appropriate binomial distribution and calculate the theoretical distribution x : 0 1 2 3 4 5 f : 2 14 20 34 22 8 Here n = 5, N = 100 Mean = xi fi = 2.84 fi np = 2.84 p = 2.84/5 = 0.568 q = 0.432 p(r) = 5C r (0.568) r (0.432) 5-r, r = 0,1,2,3,4,5 Theoretical distributions are r p(r) N* p(r) 0 0.0147 1.47 = 1 1 0.097 9.7 =10 2 0.258 25.8 =26 3 0.342 34.2 =34 4 0.226 22.6 =23 5 0.060 6 =6 Total = 100 Homework 1) 5 unbiased coins are tossed and the number of heads are noted.the experiment is repeated 64 times and the following distribution is obtained No. of heads: 0 1 2 3 4 5 Frequencies: 3 6 24 26 4 1 Fit a binomial distribution and calculate theoretical frequencies 2) The following data are the no. of seeds germinating out of 10 on a damp filter paper for 80 sets of seeds. Fit a binomial distribution to the observed data.
x: 0 1 2 3 4 5 6 7 8 9 10 f: 6 20 28 12 8 6 0 0 0 0 0 3) Fit a binomial distribution to the following data x : 0 1 2 3 4 5 f : 2 14 20 34 22 8 Compare the theoretical frequency with the actual frequency. ************************************************ POISSON DISTRIBUTION If the parameters n&p of a binomial distribution are known then we can find the distribution.but when n is large and p is very small the application of binomial distribution is very difficult. Let x be any discrete random variable which can take values 0,1,2,3.. such that the probability distribution function of x, P(x)=e -λ λ x x! where λ is a positive constant, np = λ. This distribution is called the poisson distribution. Eg: 1) Number of printing mistakes on each page of a book published by a good publisher 2) Number of telephone calls arriving at a telephone switch board per minute. The poisson distribution as a limiting case of Binomial distribution RESULT Binomial distribution tends to a poisson distribution when n and p 0 such that np = λ is finite Proof Probability distribution of a binomial distribution is P ( X ) = nc x p x q n x But λ = np p = λ / n q = 1 p = 1 - λ / n p( x) = nc x (λ / n) x ( 1 - λ / n ) n-x = n(n-1)(n-2) [n-(x-1)](n-x)! λ x ( 1 - λ / n ) n x!(n-x)! n x ( 1 - λ / n ) x
= λ x n (n-1) ( n-2)(n-3) [n-(x-1)] ( 1 - λ / n ) n x! n x ( 1 - λ / n ) x = λ x n (n-1) (n-2). [n-(x-1)] ( 1 - λ / n ) n x! n n n n ( 1 - λ / n ) x = λ x ( 1-1/n) (1 2/n)..( 1 (x-1)/n) [( 1 - λ /n) n/ λ ] - λ x! ( 1 - λ / n ) x = e -λ λ x x! ( since as n each of the (x-1) factors (1-1/n), (1-2/n) 1- (x-1) /n 1 and since Lt ( 1 + 1/x) x = e) n Result For a poisson distribution p(x) = 1 x=0 Proof For a poisson distribution p(x) = e -λ λ x x=0 x! = e -λ λ 0 + e -λ λ 1 + e -λ λ 2 + e -λ λ 3 +.. 0! 1! 2! 3! = e λ [ 1 + λ/1! +λ/2!+..] = e λ e λ = 1 Mean and variance of the poisson distribution Mean = = x p(x) p(x) But for the poisson distribution p(x) = 1 Mean = = x p(x) = x e -λ λ x x! = e -λ λ x = e λ λ x (x-1)! (x-1)! = λe - λ (λ +λ 2 /2!+ λ 3 /3!+ ) = λe - λ e λ = λ
Mean = = λ Variance = 2 = x 2 p(x) 2 p(x) But for the Poisson distribution p(x) = 1 2 = x 2 p(x) 2 =( x(x-1) +x(e -λ λ x - λ 2 x! = e -λ λ x + x e -λ λ x - λ 2 (x-2 )! x! = λe - λ (λ + λ 2 / 1!+ λ 3 / 2! +..) +λ λ 2 = λ 2 +λ λ 2 = λ Variance = 2 = λ Problems 1) A poisson variate x such that p(x = 1) = 2 Px =2). Find p(x = 0). p(x = 1) = 2 Px =2). e λ λ 1 = e λ λ 2 1! 2! ie, λ = 1 p(x = 0) = e -1 1 0 = 1/e 0! 2)A Car-hire firm has two cars it hires out daily. The number of demands for a car on each day is distributed as poisson variate with mean 1.5.Obtain the proportion of days on which i) there was no demand ii) demand is refused. Here λ = 1.5 p(x = 0 ) = e 1.5 1.5 0 =0.2231 0! p(x > 2 ) = 1- p(x 2) = 1- p( x = 0,1,2) = 0.1914 3) Assuming that the probability of an individual being killed in a mine accident during a year is 1/2400. Use poisson distribution to calculate The probability that in a mine employing 200 miners there will be atleast one fatal accident in a year? Here p = 1/2400 n = 200 λ= np = 0.083 P( x 1 ) = 1 p( x < 1) = 1 p(x = 0) = 1 e 0.083 = 0.0796 Homework 1) In a certain factory that manufactures razor blades there is small chance 1/500 for any blade tobe defective. The blades are in packets of 10.Use
poisson distribution to calculate approximate no. of packets containing no defective, one defective, two defective blades respectively in a consignment of 10,000 packets. 2) If the probability of an individual suffer a bad reaction from a certain Injection is 0.001.Determine the probability that out of 200 persons i) exactly 3 ii) more than 2 iii) none suffer the bad reaction. NORMAL DISTRIBUTION This is a continuous distribution. It can be derived from the binomial distribution as a limiting case where n The no. of trials is very large.& P the probability of success is close to ½.The general equation is f(x) = 1 e -(x-μ)2 2 2п where the variable x lies between - < x<, μ & are called the parameters of the distribution.f(x) is called pdf of the normal distribution N( μ, 2 ).The graph of the normal distribution is called the normal curve.it is bell shaped and symmetric about its mean.the two tails of the curve extend to + & - The curve is unimodal.the total area under the curve is 1. Standard form of normal distribution If X is a normal random variable with mean μ & standard deviation, then the random variable Z = X-μ has the normal distribution with mean 0 and S.D 1.The pdf is f(z)= 1 / 2п e z 2 / 2, - < z< It is free from any parameters. Area Under the normal curve By taking z = x - μ the standard normal curve is formed.the total area under this curve is divided into two parts.1) z varies from - to 0 and 2) z varies from 0 to, each area is 0.5
1. A sample of 100 battery cells is tested to find the length of life, gave the following results. Mean = 12 hrs. S.D. = 3 hrs.assuming the data tobe normally distributed what % of battery cells are expected to have life i) more than 15 hrs.ii)less than 6hrs. Iii) between 10 & 14 hrs. i ) when x = 15 P( x > 15) = P( X-μ > 15 μ ) = P( z > 1) = 0.5-0.3413 = 0.1587= 16% ii) When x = 6 P( x <6) = P( X-μ < 6 μ ) = P( z < -2) = 0.5-0.4772 = 0.0228 = 2.28% iii) When P( 10< x<14) = P( 10 - μ < X-μ < 6 μ ) = P( -0.6667 < z < 0.6667) = 2*P(0< z < 0.6667) = 0.2485*2=0.497 =50% 2. In a normal distribution 31% of the items are under 45and 8% are over 64.Find the mean and S.D. Of the distribution. Area between 0 and 45-μ = 0.5-0.31= 0.19 From the table 45-μ = -0.49 Area between 0& 64-μ =0.5-0.08 = 0.42
From the table 64-μ = 1.41 45-μ = -0.49...(1) 64-μ = 1.41...(2) Solving (1) & (2) we get μ = 50 = 10 3. In a normal distribution 7% of the items are below 35 and 11% are above 63.Find the mean and S.D. Of the distribution. Area between 0 & 35-μ = 0.5-0.07= 0.43 From the table 35-μ = -1.48 Area between 0 & 63-μ = 0.5-0.11= 0.39 From the table 63-μ = 1.23 35-μ = -1.48...(1) 63-μ = 1.23...(2) solving(1) &(2) we get μ = 50 = 10. Home work 1) X is a normal variate with mean 30 and S.D. 5.Find the probability that a) 26<x<40 b) x-30 >5 2)For a normally distributed population 7%of the items have their values less than. 35 and 89% have their values less than 63.Find the mean and S.D.. ************************************************