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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
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treeab291cffc65280e58ac82470ba63fbcca7805165 /149/CH23/EX23.9
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+clear
+clc
+disp('the first row of A denotes the scores of A and that of B denotes that of B')
+A(1,:)=[12 115 6 73 7 19 119 36 84 29];
+B(1,:)=[47 12 16 42 4 51 37 48 13 0];
+disp('the second row of A annd B denotes the corresponding deviation ')
+for i=1:10
+ A(2,i)= A(1,i)-51;
+ B(2,i)=B(1,i)-51;
+ end
+disp('the third row of A and B denotes the corresponding deviation square')
+for i=1:10
+ A(3,i)= A(2,i)^2;
+ B(3,i)=B(2,i)^2;
+end
+A
+B
+b=0;
+disp('sum of second row elements of A=b=')
+for i=1:10
+ b=b+A(2,i);
+ end
+ disp(b)
+ c=0;
+disp('sum of second row elements of B=c=')
+for i=1:10
+ c=c+B(2,i);
+ end
+ disp(c)
+ d=0;
+disp('sum of third row elements of A=d=')
+for i=1:10
+ d=d+A(3,i);
+ end
+ disp(d)
+ e=0;
+disp('sum of second row elements of B=e=')
+for i=1:10
+ e=e+B(3,i);
+ end
+ disp(e)
+ disp('arithmetic mean of A=')
+ f=51+b/10
+ disp('standard deviation of A=')
+ g=(d/10-(b/10)^2)^0.5
+ disp('arithmetic mean of B=')
+ h=51+c/10
+ disp('standard deviation of A=')
+ i=(e/10-(c/10)^2)^0.5
+ disp('coefficient of variation of A=')
+ (g/f)*100
+ disp('coefficient of variation of B=')
+ (i/h)*100 \ No newline at end of file