1st compliment 2nd compliment with fivee

Statistics

1. [email protected] DR. SYED IZHAR BUKHARI

2. Solution Q.No.1 Q.NO.1 Taking sample size of 2 (n=2),Find the population mean (U)of 2,4,6,8 Before solution: Population=2,4,6,8=N=4 Population size=n=2 Remember: (For probability sampling) (For Population sampling W.R.) T. Sample spaces=(N)ⁿ=(4)² =16 Required sample digit (i) = 16/4 = 4 times Required sample digit (ii) =4/4 = 1 time Sampling W.R. for Population mean(Ŭ) with Population size 2 (n=2) Sampling Lecture # 5 Samples X ( і ) Samples X ( іі ) X^=∑x/n ( і ) X^=∑x/n ( іі ) 2,2 6,2 (2+2)/2=2 (6+2)/2=4 2,4 6,4 3 5 2,6 6,6 4 6 2,8 6,8 5 7 4,2 8,2 3 5 4,4 8,4 4 6 4,6 8,6 5 7 4,8 8,8 6 8

3. Sampling W.R. for Population mean(Ŭ with Population size 2 (n=2) Sampling Lecture # 6 X^ f f(X^) = f/∑f X^ f (X^) X^² f (X^) 2 1 1/16 2/16 4/16 3 2 2/16 6/16 18/16 4 3 3/16 12/16 48/16 5 4 4/16 20/16 100/16 6 3 3/16 18/16 108/16 7 2 2/16 14/16 98/16 8 1 1/16 8/16 64/16 =∑f =16 ∑X^f(X^) = 80/16 ∑X^²f(X^) = 440/16 Population variance=µX^= ∑ X^ f(X^)=80/16= 5 Sampling variance=∂² X^= ∑ X^² f (X^)-(µX^)²=440/16-(5)² = 2.5 Samples X ( і ) Samples X ( іі ) X^=∑x/n ( і ) X^=∑x/n ( і i) 2,2 6,2 (2+2)/2=2 (6+2)/2=4 2,4 6,4 3 5 2,6 6,6 4 6 2,8 6,8 5 7 4,2 8,2 3 5 4,4 8,4 4 6 4,6 8,6 5 7 4,8 8,8 6 8

4. Sampling Lecture # 7 Samples (X) Samples (X) ² 2 4 4 16 6 36 8 64 ∑X=20 ∑X²=120 Sampling W.R. for Population mean(Ŭ with Population size 2 (n=2) Population Mean=µ= ∑ X/N = 20/4= 5 Population variance=µX^=∑X^ P(X^) =80/16= 5 Sampling variance=∂ ² X^= ∑ X^ ² P(X ^)-(µX^) ² =440/16-(5) ² =27.5-25= 2.5 * Sampling Mean= ∂²= ∑X²/N –(µ)² =120/4-(5)²=30-25= 5 (i) Population Mean= Population variance µ = µX^ 5 = 5 (ii) Sampling Mean = Sampling variance ∂² = ∂² X^ * 2.5 = 2.5 Population=2,4,6,8=N=4 Verification 5 /n=5/2=2.5 For Balancing

5. Solution(Q.No.3) Q.NO.2 4,5,9,1,5,5,3,2,4,0,0,1,4,3,1,1,6,6,6,4,8,4,3,3,1,0,8,7,3,1,3,4,5,2,4,7,3,2,6,5,1,7,5,2,5,8,2,1,6,9 0,0,0,1,1,1,1,1,1,1,1,2,2,2,2,2,3,3,3,3,3,3,3,4,4,4,4,4,4,4,5,5,5,5,5,5,6,6,6,6,6,7,7,7,8,8,8,9,9 Prepare a Frequency Distribution X Tally F III 3 1 IIII\ III 8 2 IIII\ 5 3 IIII\ II 7 4 IIII\ II 7 5 IIII\ I 6 6 IIII\ 5 7 III 3 8 III 3 9 II 2 ……. 49 Frequency Distribution Lecture # 8

6. Solution(Q.No.4) Q.NO.3 51,34,57,40,21,67,64,47,43,74 54,15,55,68,56,45,42,65,77,51 32,70,33,52,71,38,60,29,30,61 Arrangement of Data: 15,21,29,30,32,33,,34,38,40,4243,45,47,51,51,52,54,55,56,57, 60,61,64,65,67,68,70,71,74,77 Prepare a Frequency table by John Tukey’s Stem-and-Leaf display Stem (Leading digit) Leaf (Trailing digit) 1 5 2 19 3 42380 4 07352 5 1745612 6 748501 7 4701 Frequency Distribution * Stem-and-Leaf display is a way of listing the data in an array which can provide a useful description of data set and can easily be converted to a frequency table. *It is a common practice to arrange the trailing digits in each row from smallest to a highest. Lecture # 9

7. Solution Q.No.4 Q.NO.4 Practice yourself? Taking sample size of 3 (n=3),Construct the sample spaces and find Population mean of 3,6,9 Before solution: Population=3,6,9=N=0 Population size=n=0 Remember: (For probability sampling) (For Population sampling W.R.) T. Sample spaces=(N)ⁿ=( 3 )³ =27 Required sample digit (i) =00/0=9 times Required sample digit(ii)=0/0 =3 time Required sample digit (iii)=0/0 =1 time Sampling W.R. for Population mean(Ŭ) with Population size 3 (n=3) Sampling Lecture # 00 Samples X ( і ) Samples X ( іі ) Samples X (i іі ) X^ =∑x/n ( і ) X^ =∑x/n ( іі ) X^ =∑x/n ( іі i) 0,0,0 0,0,0 0,0,0

8. Sampling W.R. for Population mean(Ŭ with Population size 3 (n=3) Sampling Lecture # 00 X^ f f(X^) = f/∑f X^ f (X^) X^² f (X^) 2 1 1/16 2/16 4/16 3 2 2/16 6/16 18/16 =∑f =00 ∑X^f(X^) = 00/00 ∑X^²f(X^) = 00/00 Population variance=µX^= ∑ X^ f(X^)=00/00= Sampling variance=∂² X^= ∑ X^² f (X^)-(µX^)²=00/00-( )² = 0.0 Samples X ( і ) Samples X ( іі ) Samples X (i іі ) X^ =∑x/n ( і ) X^ =∑x/n ( іі ) X^ =∑x/n ( іі i) 0,0 0,0 0,0 (0+0)/n=0 (0+0)/n=4 (0+0)/0=4

9. Sampling Lecture # 00 Samples (X) Samples (X) ₂ ∑X=00 ∑X ₂ =000 Sampling W.R. for Population mean(Ŭ with Population size 3 (n=3) Population Mean=µ= ∑ X/N = 00/0= Population variance=µX^=∑X^ P(X^) =00/00= Sampling variance=∂ ² X^= ∑ X^ ² P(X ^)-(µX^) ² =000/00-(0) ² =00.0-00= 0.0 * Sampling Mean= ∂²= ∑X²/N –(µ)² =00/0-(0)² =00-00= (i) Population Mean= Population variance µ = µX^ = 0 (ii) Sampling Mean = Sampling variance ∂² = ∂²X^ * .0 = 0 .0 Population=N=0,0,0 Verification /n=0/0= 0.0 For Balancing

10. Solution(Q.No.5) Q.NO.5 8,9,9,6,5,4,3,2,4,0,0,0,2,16,7,6,4,6,9,6,3,8,0,1,2,5,7,8,20,3,8,1,4,6,7,3,7,3,2,6,5,1,7,5,2,0,9,8,3,2,6,8 Solution: Rearranging the data: 0,0,0, 0,0,0 , 0,0,0 , 0,0,0 , 0,0,0 , 0,0,0 , 0,0,0 Prepare a Frequency Distribution X Tally F III IIII\ III IIII\ IIII\ II IIII\ II IIII\ I IIII\ ……. Frequency Distribution Lecture # 00 Practice Yourself?

11. Solution(Q.No.6) Q.NO.6 54,37,60,43,24,70,67,50,46,77 57,18,58,71,59,48,45,68,80,54 35,73,36,55,74,41,63,32,33,64,66,85,80,50,30,70,71, Arrangement of Data: 00,00,00,00,00,00,00,00,00,00 , 00,00,00,00,00,00,00,00,00,00,00,00,00,00,00,00,00,00,00,00,00 Practice Yourself ? Prepare a Frequency table by John Tukey’s Stem-and-Leaf display Stem (Leading digit) Leaf (Trailing digit) 1 2 3 4 5 6 7 Frequency Distribution * Stem-and-Leaf display is a way of listing the data in an array which can provide a useful description of data set and can easily be converted to a frequency table. *It is a common practice to arrange the trailing digits in each row from smallest to a highest. Lecture # 00

12. Solution Q.No.7 Q.NO.7 Taking sample size of 3 (n=3),Find the population mean (U) of 2,3,4 Before solution: Population=2,3,4=N=3 Population size=n=3 Remember: (For probability sampling) (For Population sampling W.R.) T. Sample spaces=(N)ⁿ=( 3 )³ =27 Required sample digit (i)=27/3= 9times Required sample digit (ii)=9/3 = 3 time Required samples digit (iii)=3/3= 1time Sampling W.R. for Population mean(Ŭ) with Population size 3 (n=3) Sampling Lecture # 10 Samples X ( і ) Samples X ( іі ) Samples X (i іі ) X^=∑x/n ( і ) X^=∑x/n ( іі ) X^=∑x/n ( іі i) 2,2,2 3,2,2 4,2,2 2 2.3 3.3 2,2,3 3,2,3 4,2,3 2.3 2.6 3 2,2,4 3,2,4 4,2,4 2.6 3 3.3 2,3,2 3,3,2 4,3,2 2.3 2.6 3 2,3,3 3,3,3 4,3,3 2.6 3 3.3 2,3,4 3,3,4 4,3,4 3 3.3 3.6 2,4,2 3,4,2 4,4,2 2.6 3 3.3 2,4,3 3,4,3 4,4,3 3 3.3 3.6 2,4,4 3,4,4 4,4,4 3.3 3.6 4

13. Sampling W.R. for Population mean(Ŭ with Population size 3 (n=3) Sampling Lecture # 11 X^ f f(X^) = f/∑f X^ f (X^) X^² f (X^) 2 1 1/27 2/27 4/27 2.3 3 3/27 6.9/27 15.87/27 2.6 5 5/27 13/27 33.8/27 3 7 7/27 21/27 63/27 3.3 7 7/27 23.1/27 76.23/27 3.6 3 3/27 10.8/27 38.88/27 4 1 1/27 4/27 16/27 =∑f =27 ∑X^f(X^) = 80.8/27 ∑X^²f(X^) = 246.98/27 Population variance=µX^= ∑ X^ f(X^)=80.8/27= 2.99=3 Sampling variance=∂² X^= ∑ X^² f (X^)-(µX^)²=246.98/27-(2.99)² =9.15-8.94= 0.21=0.2 Samples X ( і ) Samples X ( іі ) Samples X (i іі ) X^ =∑x/n ( і ) X^ =∑x/n ( іі ) X^ =∑x/n ( іі i) 2,2,2 3,2,2 4,2,2 2 2.3 3.3 2,2,3 3,2,3 4,2,3 2.3 2.6 3 2,2,4 3,2,4 4,2,4 2.6 3 3.3 2,3,2 3,3,2 4,3,2 2.3 2.6 3 2,3,3 3,3,3 4,3,3 2.6 3 3.3 2,3,4 3,3,4 4,3,4 3 3.3 3.6 2,4,2 3,4,2 4,4,2 2.6 3 3.3 2,4,3 3,4,3 4,4,3 3 3.3 3.6 2,4,4 3,4,4 4,4,4 3.3 3.6 4

14. Sampling Lecture # 12 Samples (X) Samples (X)² 2 4 3 9 4 16 ∑X=9 ∑X²=29 Sampling W.R. for Population mean(Ŭ with Population size 3 (n=3) Population Mean=µ= ∑ X/N = 9/3= 3 Population variance=µX^=∑X^ P(X^) =80.8/27= 2.99=3 Sampling variance=∂ ² X^= ∑ X^ ² P(X ^)-(µX^) ² =246.98/27-(2.99) ² =9.15-8.94= 0.21 * Sampling Mean= ∂² = ∑X²/N –(µ)² ={29/3-( 3 )²} =9.66-9= 0.66 ( i) Population Mean= Population variance µ = µX^ 2.99 = 3 3 = 3 (ii) Sampling Mean = Sampling variance ∂² = ∂^² X^ * 0.21 = 0.22 0.2 = 0.2 Population=2,3,4=N= 3 Verification 0.66 /n=0.66/3= 0.22=0.2

15. Solution Q.No.7 Q.NO.8 Practice Yourself ? Taking sample size of 2 (n=2) find the population mean (U) of the population of 1,2,3,4 Before solution: Population=1,2,3,4=N=4 Population size=n=2 Remember: (For probability sampling) (For Population sampling W.R.) T. Sample spaces=(N)ⁿ=(4)² =16 Required samples(i) = 16/4 = 4 times Required samples(ii) =4/4 = 1 time Sampling W.R. for Population mean(Ŭ) with Population size 2 (n=2) Sampling Lecture # 13 Samples X ( і ) Samples X ( іі ) X^=∑x/n ( і ) X^=∑x/n ( іі ) 1,1 3,1 1 2 1,2 3,2 1.5 2.5 1,3 3,3 2 3 1,4 3,4 2.5 3.5 2,1 4,1 1.5 2.5 2,2 4,2 2 3 2,3 4,3 2.5 3.5 2,4 4,4 3 4

16. Sampling W.R. for Population mean(Ŭ with Population size 2 (n=2) Sampling Lecture # 14 X^ f f(X^) = f/∑f X^ f (X^) X^² f (X^) 1 1 1/16 1/16 1/16 1.5 2 2/16 3/16 4.5/16 2 3 3/16 6/16 12/16 2.5 4 4/16 10/16 25/16 3 3 3/16 9/16 27/16 3.5 2 2/16 7/16 24.5/16 4 1 1/16 4/16 16/16 =∑f =16 ∑X^f(X^) = 40/16 ∑X^²f(X^) = 110/16 Population variance=µX^= ∑ X^ f(X ^)= 40 /16= 2.5 Sampling variance=∂² X^= ∑ X^² f (X^)-(µX^) ²=110/16-( 2.5 )²= 6.875-6.25= 0.625=0.63 Samples X ( і ) Samples X ( іі ) X^=∑x/n ( і ) X^=∑x/n ( іі ) 1,1 3,1 1 2 1,2 3,2 1.5 2.5 1,3 3,3 2 3 1,4 3,4 2.5 3.5 2,1 4,1 1.5 2.5 2,2 4,2 2 3 2,3 4,3 2.5 3.5 2,4 4,4 3 4

17. Sampling Lecture # 15 Samples (X) Samples (X) ² 1 1 2 4 3 9 4 16 ∑X=10 ∑X²=30 Sampling W.R. for Population mean(Ŭ with Population size 2 (n=2) Population Mean= µ = ∑X/N = 10/4= 2.5 Population variance=µX^=∑X^ f (X^) =40/16= 2.5 Sampling variance=∂ ² X^= ∑ X^ ² f (X^)-(µX^) ² = 110 /16-( 2.5 ) ² = 0.625= * Sampling Mean= ∂² = ∑X²/N –(µ)²= 30/4-(2.5)²=7.5-6.25= 1.25 (i)Population Mean= Population variance µ = µX^ 2.5 = 2.5 (ii) Sampling Mean = Sampling variance ∂² = ∂² X^ 0.625 = 0.625 Population=1,2,3,4=N=4 Verification 1.25/n=1.25/2=0.625