initial commit / add all books

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diff --git a/331/CH6/EX6.1/Example_6_1.sce b/331/CH6/EX6.1/Example_6_1.sce
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+//Caption:Probability of occurrence of 'A' given that 'B' has occurred

+//P(A/B)

+//Example6.1

+//Page 171

+clear;

+clc;

+//A be the event that an employee's monthly salary is more than Rs.15,000

+//B be the event that the employee is a regular taker of Alpha Brand Tea

+n = 200; //total number of employee's

+A_intersec_B = 20; //Alpha Brand tesa takers both in A and B

+P_AinterB = A_intersec_B/n;//Probability that Alpha brand tea takers both in A&B

+B = 120; //Number of Alpha Brand tea takers

+P_B = B/n; // Probility of number of Alpha brand tea takers

+P_AB = P_AinterB/P_B;

+disp(P_AinterB,'Probability that Alpha brand tea takers both in A&B P(A intersection B)=')

+disp(P_B,'Probility of number of Alpha brand tea takers P(B)=')

+disp('The probability that employee is having monthly salary more than')

+disp(P_AB,'Rs.15,000, if the employee is a regular taker of Alpha brand Tea P(A/B)=')

+//Result

+//Probability that Alpha brand tea takers both in A&B P(A intersection B)=   

+//    0.1  

+//Probility of number of Alpha brand tea takers P(B)=   

+//    0.6  

+//The probability that employee is having monthly salary more than   

+//Rs.15,000, if the employee is a regular taker of Alpha brand Tea P(A/B)=   

+//   0.1666667  
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diff --git a/331/CH6/EX6.2/Example_6_2.sce b/331/CH6/EX6.2/Example_6_2.sce
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+//Caption: Cumulative distribution of the binomial distribution

+//F(X,n,p)

+//Example6.2

+//Page 176

+clear;

+clc;

+//(a). Probability of 0 head

+X1 = 0;// 0 head

+n = 10; //number of trials

+Pr = 0.5; //Probability of success in each binomial trial

+Ompr = 1-Pr; //Probability of failure in each binomial trial

+[P1,Q1]=cdfbin("PQ",X1,n,Pr,Ompr);

+disp(P1,'Probability of zero head P(X=0,10,0.5)=')

+//(b). Probability of 3 head

+X2 =3;// 3 head

+P2 = (factorial(n)/(factorial(X2)*factorial(n-X2)))*(Pr^X2)*(Ompr^(n-X2));

+disp(P2,'Probability of three heads P(X=3,10,0.5)=')

+//(c). at most 2 heads

+X3 = 2;

+[P3,Q3]=cdfbin("PQ",X3,n,Pr,Ompr);

+disp(P3,'Probability of atmost 2 heads P(X<=2,10,0.5)=')

+//(d). at least 3 heads

+disp(1-P3,'Probability of atleat 3 heads P(X>=3,10,0.5)=')

+//Result

+//Probability of zero head P(X=0,10,0.5)=   

+//    0.0009766  

+//Probability of three heads P(X=3,10,0.5)=   

+//    0.1171875  

+//Probability of atmost 2 heads P(X<=2,10,0.5)=   

+//    0.0546875  

+//Probability of atleat 3 heads P(X>=3,10,0.5)=   

+//    0.9453125 
\ No newline at end of file
diff --git a/331/CH6/EX6.3/Example_6_3.sce b/331/CH6/EX6.3/Example_6_3.sce
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+//Caption: Cumulative distribution of the binomial distribution

+//F(X,n,p)

+//Example6.3

+//Page176

+clear;

+clc;

+//(a). Probability of nil project in time

+X1 = 0;// nil project

+n = 5; //number of trials

+Pr = 0.9; //Probability of success in each binomial trial

+Ompr = 1-Pr; //Probability of failure in each binomial trial

+[P1,Q1]=cdfbin("PQ",X1,n,Pr,Ompr);

+disp(P1,'Probability of zero head P(X=0,10,0.5)=')

+//(b). two projects in time

+X2 = 2;// two projects

+P2 = (factorial(n)/(factorial(X2)*factorial(n-X2)))*(Pr^X2)*(Ompr^(n-X2));

+disp(P2,'Probability of three heads P(X=2,10,0.5)=')

+//(c). at most one project in time

+X3 = 1;

+[P3,Q3]=cdfbin("PQ",X3,n,Pr,Ompr);

+disp(P3,'Probability of atmost 2 heads P(X<=1,10,0.5)=')

+//(d). at least two projects in time

+disp(1-P3,'Probability of atleat 3 heads P(X>=2,10,0.5)=')

+//Result

+//Probability of zero head P(X=0,10,0.5)=   

+//     0.00001  

+//Probability of three heads P(X=2,10,0.5)=   

+//     0.0081  

+//Probability of atmost 2 heads P(X<=1,10,0.5)=   

+//     0.00046  

+//Probability of atleat 3 heads P(X>=2,10,0.5)=   

+//     0.99954  
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diff --git a/331/CH6/EX6.4/Example_6_4.sce b/331/CH6/EX6.4/Example_6_4.sce
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+//Caption: Poisson distribution

+//Example6.4

+//Page179

+clear;

+clc;

+//(a): Exactly 0 customer will arrive in 10 minutes interval

+X1= 0; //no. of customer

+Mean = 4;//mean arrival rate of 4 per 10 minutes

+[P1,Q1]=cdfpoi("PQ",X1,Mean)

+disp(P1,'Exactly 0 customer will arrive P(X=0,4) is =')

+//(b): Exactly 2 customers will arrive in 10 minutes interval

+X2 =2; //no. of customer

+P2 = exp(-Mean)*(Mean^X2)/(factorial(2))

+disp(P2,'Exactly 2 customer will arrive P(X=2,4) is =')

+//(c): at most 2 customers will arrive in 10 minutes interval

+[P3,Q3]=cdfpoi("PQ",X2,Mean)

+disp(P3,'Atmost 2 customer will arrive P(X<=2,4) is =')

+//(d): at least 3 customers will arrive in 10 minutes interval

+X3 =3;

+P4 = 1-P3

+disp(P4,'At least 3 customer will arrive P(X>=3,4) is=')

+//Result

+//Exactly 0 customer will arrive P(X=0,4) is =   

+// 

+//    0.0183156  

+// 

+// Exactly 2 customer will arrive P(X=2,4) is =   

+// 

+//    0.1465251  

+// 

+// Atmost 2 customer will arrive P(X<=2,4) is =   

+// 

+//    0.2381033  

+// 

+// At least 3 customer will arrive P(X>=3,4) is=   

+// 

+//    0.7618967  
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diff --git a/331/CH6/EX6.5/Example_6_5.sce b/331/CH6/EX6.5/Example_6_5.sce
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+//Caption: Poisson Distribution

+//Example6.5

+//Page179

+clear;

+clc;

+//(a): Probability no piece in the sample is defective

+X1= 0; //nil  defective

+p= 0.04;//probability that an inspected piece will be defective

+n = 25; //number of sample units

+Mean = n*p;//mean of the poisson distribution

+[P1,Q1]=cdfpoi("PQ",X1,Mean)

+disp(P1,'No piece will be defective P(X=0,1) is =')

+

+//(b): Probability 3 pieces in the sample will be defective

+X2 = 3; //3 pieces in the sample will be defective

+P2 = exp(-Mean)*(Mean^X2)/(factorial(X2))

+disp(P2,'Probability 3 pieces will be defective P(X=3,1) is =')

+

+//(c): at most 2 pieces will be defective

+X3 = 2;

+[P3,Q3]=cdfpoi("PQ",X3,Mean)

+disp(P3,'Atmost 2 pieces will be defective P(X<=2,1) is =')

+

+//(d): at least 3 pieces will be defective

+P4 = 1-P3

+disp(P4,'At least 3 pieces will be defective P(X>=3,1) is=')

+//Result

+//  

+// No piece will be defective P(X=0,1) is =   

+// 

+//    0.3678794  

+// 

+// Probability 3 pieces will be defective P(X=3,1) is =   

+// 

+//    0.0613132  

+// 

+// Atmost 2 pieces will be defective P(X<=2,1) is =   

+// 

+//    0.9196986  

+// 

+// At least 3 pieces will be defective P(X>=3,1) is=   

+// 

+//    0.0803014 

diff --git a/331/CH6/EX6.6/Example_6_6.sce b/331/CH6/EX6.6/Example_6_6.sce
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+//Caption: Uniform Distribution

+//Example6.6

+//Page181

+clear;

+clc;

+a = 230; // lower limit

+b = 450; //upper limit

+P = (1/(b-a)); //uniform distribution of daily demand for packed meals

+Demand = 0.8; //service level of satisfying the demand of the canteen

+Q = (1/P)*Demand+a;

+disp(Q,'The demand of the product which can be satisfied w.r.to the given service level is=')

+//Result

+//The demand of the product which can be satisfied w.r.to the given service level is=   

+// 

+//    406. 
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diff --git a/331/CH6/EX6.7/Example_6_7.sce b/331/CH6/EX6.7/Example_6_7.sce
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+//Caption: Exponential Distribution

+//Example6.7

+//Page183

+clear;

+clc;

+//Example6.7(a): The probability that the service time of the terminal Less than 0.5 hr

+Mean = 30; //Service rate of the terminal per day

+Mean = Mean/24; //Service rate of the terminal per hour

+X1 = 0.5;

+F = 1-exp(-Mean*X1); 

+disp(F,'The cumulative probability distribution P(X<=0.5) is =');

+//Example6.7(b): The probability that the service time of the terminal greate than 0.75 hr

+X2 = 0.75;

+F = 1-exp(-Mean*X2);

+disp(1-F,'The cumulative probility distribution P(X>=0.75) is =' );

+//Result

+//The cumulative probability distribution P(X<=0.5) is = 0.4647386  

+// 

+//The cumulative probility distribution P(X>=0.75) is =  0.3916056  
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diff --git a/331/CH6/EX6.8/Example_6_8.sce b/331/CH6/EX6.8/Example_6_8.sce
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+//Caption: Exponential Distribution

+//Example6.8

+//Page183

+clear;

+clc;

+//Example6.8(a): The probability that the execution time of programs < 4 minutes

+T = 5; //Average execution time of the programs 

+X1 = 4;

+F = 1-exp(-X1/T); 

+disp(F,'The probability that the execution time of programs < 4 minutes P(X<=4) is =');

+//Example6.8(b): The probability that the execution time of programs > 6 minutes

+X2 =6;

+F = 1-exp(-X2/T);

+disp(1-F,'The probability that the execution time of programs > 6 minutes P(X>=6) is =');

+//Result

+//The probability that the execution time of programs < 4 minutes P(X<=4) is =   

+//     0.5506710  

+//The probability that the execution time of programs > 6 minutes P(X>=6) is =   

+//    0.3011942 
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diff --git a/331/CH6/EX6.9/Example_6_9.sce b/331/CH6/EX6.9/Example_6_9.sce
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+//Caption: Normal Distribution

+//Example6.9

+//Page185

+clear;

+clc;

+//Example6.9(a):Probability that the monthly income < Rs.11,000

+X = 11000;

+Mean = 10000;

+Std = 2000;

+n = 200; //sample number of respondents

+[P,Q]=cdfnor("PQ",X,Mean,Std)//Cumulative normal distribution

+disp(P,'The probability that the monthly income is < Rs.11,000 is =')

+disp(n*P,'The number of respondents having income is less than Rs.11,000 =')

+

+//Example6.9(b): Probability that the monthly income > Rs.12,000

+X = 12000;

+[P,Q]=cdfnor("PQ",X,Mean,Std)//Cumulative normal distribution

+disp(Q,'The probability that the monthly income is > Rs12,000 is =')

+disp(n*Q,'The number of respondents having income is greater than Rs.12,000 =')

+

+//Example6.9(c): Probability that the monthly income is in between Rs.7,000 and

+//Rs.11,200

+X1 = 11200;

+X2 = 7000;

+[P1,Q1]=cdfnor("PQ",X1,Mean,Std);

+[P2,Q2]=cdfnor("PQ",X2,Mean,Std);

+disp(P1-P2,'The probability that the monthly income in between Rs.7,000 & Rs.11,200 is =');

+disp(n*(P1-P2),'The number of respondents having income in between Rs.7,000 & Rs.11,200 is =')

+//Result

+//The probability that the monthly income is < Rs.11,000 is =   

+// 

+//    0.6914625  

+// 

+// The number of respondents having income is less than Rs.11,000 =   

+// 

+//    138.29249  

+// 

+// The probability that the monthly income is > Rs12,000 is =   

+// 

+//    0.1586553  

+// 

+// The number of respondents having income is greater than Rs.12,000 =   

+// 

+//    31.731051  

+// 

+// The probability that the monthly income in between Rs.7,000 & Rs.11,200 is =   

+// 

+//    0.6589397  

+// 

+// The number of respondents having income in between Rs.7,000 & Rs.11,200 is =   

+// 

+//    131.78794
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