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Central tedancy & correlation project - 2

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Central tedancy & correlation project - 2

  1. 1. 1 | P a g e Q#No.1 For the Data Give below calculates: (1) Arithmetic mean (2) Median (3) Mode (4) Harmonic mean (5) Geometric mean. Solution: (1) Arithmetic means fX X f    4265 61 X  69.918X  (2) Median 2 h n Median L c f           10 60 30.5 16 15 Median    69.67Median  (3) Mode     1 1 2 m m m f f Mode L h f f f f            15 12 60 10 15 12 15 14 Mode        67.5Mode  C-B f x fx f ( 𝟏 𝒙 ) C.f F log x 30-40 1 35 35 0.0285 1 1.5441 40-50 3 45 135 0.6666 4 4.9596 50-60 12 55 660 0.218 16 20.8844 60-70 15 65 975 0.230 31 27.1957 70-80 14 75 1050 0.1866 45 26.2508 80-90 11 85 935 0.129 56 21.2237 90-100 5 95 475 0.052 61 9.8886 61f  4265fx  1 ( ) 0.9113f x   log 111.9449f x 
  2. 2. 2 | P a g e (4)Harmonic mean: . 1 . f H M f x         61 . 0.9113 H M  . 66.93H M  (5) Geometric mean: (f logX) . logG M anti f    41.9449 . log 61 G M anti . 68.4227G M  Q#NO.2 Find the (1) Arithmetic mean (2) Median (3) Mode (4) Harmonic mean (5) Geometric mean. Solution: (1) Arithmetic mean: FX X F    18765 164 X  162.44X  C-B f x fx Flogx f ( 𝟏 𝒙 ) C.f 0-40 6 20 120 7.8062 0.3 6 40-80 15 60 900 26.6723 0.25 21 80-120 22 100 2200 44 0.22 43 120-160 30 140 4200 64.3839 0.2143 73 160-200 45 180 8100 101.4873 0.25 118 200-140 27 220 5940 63.2455 0.1228 145 240-180 13 260 3380 31.3947 0.05 15.8 280-320 6 300 1800 14.8628 0.02 164 164f  26640fx  log 353.8527f x  1 ( ) 1.427f x  
  3. 3. 3 | P a g e (2) Median 2 h n Median L c f           40 160 82 73 45 Median    168Median  (3) Mode     1 1 2 m m m f f Mode L h f f f f            45 30 160 40 45 30 45 27 Mode            45 30 160 40 45 30 45 27 Mode        15 160 40 33 Mode    178.18Mode  (4)Harmonic mean: . 1 . f H M f x         164 . 1.4271 H M  . 114.92H M  (5) Geometric mean: (f logX) . logG M anti f    353.8527 . log 164 G M anti  . log 2.1577G M anti . 143.789G M 
  4. 4. 4 | P a g e Q#No.3 Calculate the: (1) Arithmetic mean (2) Median (3) Mode (4) Harmonic mean (5) Geometric mean. Solution: (1) Arithmetic mean: fX X f    3372.5 55 X  61.318X  (2) Median 2 h n Median L c f           55 60 27.5 2 18 Median    61.18Median  (3) Mode     1 1 2 m m m f f Mode L h f f f f            18 12 60 5 18 12 18 13 Mode        30Mode  (4)Harmonic mean: . 1 . f H M f x         55 . 0.90587 H M  . 60.715H M  C-B f X fx Flogx f ( 𝟏 𝒙 ) C.f 45-50 2 47.5 94 3.3533 0.0421 2 50-55 7 52.5 367.5 12.0411 0.1333 9 55-60 12 57.5 690 21.1160 0.2086 21 60-65 18 62.5 1125 32.3258 0.288 39 65-70 13 67.5 877.5 23.7809 0.1925 52 70-75 3 72.5 217.5 5.5810 0.04137 54 55f  3372.5fx  log 98.1981f x  1 ( ) 0.90587f x  
  5. 5. 5 | P a g e (5) Geometric mean: (f logX) . logG M anti f    98.1981 . log 55 G M anti  . log 1.78542G M anti . 61.012G M  Q#No.4 For the Data Give below calculates: (1) Arithmetic mean (2) Median (3) Mode (4) Harmonic mean (5) Geometric mean. Solution: (1) Arithmetic mean: fX X f    46350 850 X  54.526X  (2) Median 2 h n Median L c f           10 60 425 420 190 Median    60.263Median  (3) Mode     1 1 2 m m m f f Mode L h f f f f            190 125 60 10 190 125 190 240 Mode          65 60 10 65 50 Mode      8.333Mode  C-B f x fx Flogx f ( 𝟏 𝒙 ) C.f 0-10 25 5 125 17.47 5 25 10-20 40 15 600 47.043 2.667 65 20-30 60 25 1500 83.876 2.4 125 30-40 75 35 2625 115.80 2.142 200 40-50 95 45 4275 157.05 2.111 295 50-60 125 55 6875 217.54 2.272 420 60-70 190 65 12350 344.45 2.923 610 70-80 240 75 18000 445.75 3.2 850 850f  46350fx  log 1428.98f x  1 ( ) 22.75f x  
  6. 6. 6 | P a g e (4) Harmonic mean: . 1 . f H M f x         850 . 22.75 H M  . 37.36H M  (5) Geometric mean: (f logX) . logG M anti f    1428.98 . log 850 G M anti . 47.98G M  Q#No.5 For the Data Give below calculates: (1) Arithmetic mean (2) Median (3) Mode (4) Harmonic mean (5) Geometric mean. Solution: (1) Arithmetic mean: fX X f    2005 72 X  27.847X  (2) Median 2 h n Median L c f           5 22.5 36 24 19 Median    25.66Median  C-B f X fx C.f F log x 1 f x       12.5-17.5 2 15 30 2 2.3522 0.4 17.5-22.5 22 20 440 24 28.6227 1.1 22.5-27.5 19 25 475 43 26.5609 0.76 27.5-32.5 14 30 420 57 20.6797 0.4667 32.5-37.5 3 35 105 60 4.6322 0.857 37.5-42.5 4 40 160 64 6.4082 0.1 42.5-47.5 6 45 270 70 9.9193 0.1333 47.5-52.5 1 50 50 71 1.6989 0.02 52.5-57.5 1 55 55 72 1.7407 0.0181 ∑f=72 ∑fx=2005 ∑flogx=102.6145 1 f x       
  7. 7. 7 | P a g e (3) Mode:     1 1 2 m m m f f Mode L h f f f f            22 2 17.5 5 22 2 22 19 Mode        20 17.5 5 20 3 Mode     21.85Mode  (4) Harmonic mean: . 1 . f H M f x         72 . 3.0837 H M  . 23.35H M  (5) Geometric mean: (f logX) . logG M anti f    102.6145 . log 72 G M anti . 26.62G M 
  8. 8. 8 | P a g e Q#No.1 Calculate the correlation co-efficient between percentage of marks scored by 12 students in statistics and economics. Solution: (1). Find the co-efficient of correlation: 2 2 2 2 ( ) ( ) n xy x y r n x x n y y                  2 2 12(23136) (750)(365) 12(47470) (750) 12(11385) (365) r              3882 7140 3395 r  710 45054900 r  3882 4923.443917 r  0.7884r  (2). Regression line: Find the regression line of following data: Y on X: Y=a+bx 2 2 ( ) yx n xy x y b n x x                2 12 23136 750 365 12 47470 (750) yxb    277632 273750 569640 562500 yxb    3882 7140 yxb  0.543yxb  X y xy 𝒙 𝟐 𝒚 𝟐 50 22 1100 2500 484 54 25 1350 2916 625 56 34 1904 3136 1156 59 25 1652 3481 784 60 26 1560 3600 678 61 30 1830 3721 900 62 32 1984 3844 1024 65 30 1950 4225 900 67 28 1876 4489 784 71 34 2414 5041 1156 71 36 2556 5041 1296 74 40 2960 5476 1600 ∑x=750 ∑y=365 ∑xy=23136 ∑𝒙 𝟐=47470 ∑𝒚 𝟐=11385
  9. 9. 9 | P a g e yx yx y b x a n         365 0.543 750 12 yxa   3.52yxa   The estimated regression line is as follows. ˆ 3.52 0.543Y X   (3). Regression line: X on Y: X=a+by 2 2 ( ) xy n xy x y b n y y                2 12 23136 750 365 12 11385 (365) xyb    277632 27375 136620 133225 xyb    3882 3395 xyb  1.143xyb  . xy x bxy y a n        750 1.143 (365) 12 xya   27.73xya  The estimated regression line is as follows. ˆ 27.73 1.143x Y 
  10. 10. 10 | P a g e Q#No.2 Calculate the Co-efficientof correlationfromthe followingdataandalsocompute RegressionlineY onX Solution: (1).Find the co-efficient of correlation: 2 2 2 2 ( ) ( ) n xy x y r n x x n y y                  2 2 (11)(1336) (110)(125) (11)(1210) (110) (11)(153) (125) r           14696 13750 (13310 12100)(16841 15625) r     946 (1210)(1216) r  946 1471360 r  946 1212.99 r  0.779r  (2). Regression line: Y on X Y a bx  2 2 ( ) yx n xy x y b n x x         2 (11)(1336) (110)(125) (11)(1210) (110) yxb    (14696) (13750) (13310) (12100) yxb    946 1210 yxb  0.781yxb  X Y XY 𝒙 𝟐 𝒚 𝟐 5 9 45 25 81 6 6 36 36 36 7 10 70 49 100 8 8 64 64 64 9 13 117 81 169 10 11 110 100 121 11 14 154 121 196 12 10 120 144 100 13 14 182 169 196 14 12 168 196 144 15 18 270 225 324 ∑X=110 ∑Y=125 ∑XY=1336 ∑𝑿 𝟐 =1240 ∑𝒀 𝟐 =1531
  11. 11. 11 | P a g e yx yx y b x a n     (125) (0.781)(110) 11 yxa   125 85.91 11 yxa   39.09 11 yxa  3.553yxa  The estimated regression line is as follows. ˆ 3.553 0.781Y x  Q#NO.3 Calculate the Co-efficient of correlation from the following data and also compute Regression line Y on X and X on Y. Solution: (1). Find the co-efficient of correlation: 2 2 2 2 ( ) ( ) n xy x y r n x x n y y                  2 2 (10)(33535) (655)(500) (10)(46059) (655) (10)(25464) (500) r           (335350) (327500) (460590 429025)(254640 250000) r     (7850) (31565)(4640) r  X Y XY 𝒙 𝟐 𝒚 𝟐 16 40 640 256 1600 72 52 3744 5184 2704 73 43 3139 5329 1849 63 49 3087 3969 2401 83 61 5063 6889 3721 80 58 4640 6400 3364 66 44 2904 4359 1936 66 58 3828 4356 3364 74 50 3700 5476 2500 62 45 2790 3844 2025 ∑X=655 ∑Y=500 ∑XY=33535 ∑𝑿 𝟐 =46059 ∑𝒀 𝟐 =25464
  12. 12. 12 | P a g e (7850) 12102.13204 r  0.6486r  (2)Regression line Y on X Y a bx  2 2 ( ) yx n xy x y b n x x         2 (10)(33535) (655)(500) (10)(46059) (655) yxb    (335350) (327500) (460590) (429025) yxb    7850 31565 yxb  0.249yxb  yx yx y b x a n     (500) (0.249)(655) 10 yxa   (500) (163.095) 10 yxa   336.905 10 yxa  33.6905yxa  The estimated regression line is as follows. ˆ 33.6905 0.249Y x  (3)Regression line X on Y: X a by  2 2 ( ) xy n xy x y b n y y         2 (10)(33535) (655)(500) (10)(25464) (500) xyb    (335350) (327500) (254640) (250000) xyb    7850 4640 xyb  1.6918103xyb  xy xy x b y a n     (655) (1.6918)(500) 10 xya   (655) (845.9) 10 xya   190.9 10 xya   19.09xya 
  13. 13. 13 | P a g e The estimated regression line is as follows. ˆ 19.09 1.6918X x   Q#NO.4 Find the co-efficient of correlation and fit the regression lines of the given data and also discuss its result. Solution: (1). Find the co-efficient of correlation: 2 2 2 2 ( ) ( ) n xy x y r n x x n y y                  2 2 8(11245) (595)(150) 8(47375) (595) 8(3038) (150) r              710 24975 1804 r  710 45054900 r  710 6712.29469 r  0.1058r  (2). Regression line Y on X Y a bx  2 2 ( ) yx n xy x y b n x x                2 8 11245 595 150 8 47375 (595) yxb    X Y XY 𝒙 𝟐 𝒚 𝟐 40 17 680 1600 289 55 19 1045 3025 361 60 23 1380 3600 529 75 15 1125 5625 225 80 18 1440 6400 324 90 17 1530 8100 289 95 11 1045 9025 121 100 30 3000 10000 900 ∑x=595 ∑y=150 ∑xy=11245 ∑𝒙 𝟐 =47375 ∑𝒚 𝟐 =3038
  14. 14. 14 | P a g e   2 89960 89250 8 47375 (595) yxb    710 37900 354025 yxb   710 24975 yxb  0.02843yxb  yx yx y b x a n         150 0.024843 595 8 yxa   16.635yxa  The estimated regression line is as follows. ˆ 16.635 0.02843Y X  (3) Regression line X on Y: X a by  2 2 ( ) xy n xy x y b n y y                2 8 11245 595 150 8 3038 (150) xyb    89960 89250 379000 22500 xyb    710 1804 xyb  0.39356xyb  . xy x bxy y a n        595 0.39356 (150) 8 xya   67.99xya  The estimated regression line is as follows. ˆ 67.99 0.39356x  
  15. 15. 15 | P a g e Q#No.5: calculate the co-efficient of correlation from given data. And also find out regression line Y on X and X on Y. Solution: x y xy 𝒙 𝟐 𝒚 𝟐 10 7 70 1000 49 15 8 120 225 64 20 3 60 400 9 25 7 175 625 49 30 18 540 900 324 35 6 210 1225 36 40 17 680 1600 289 ∑x=175 ∑y=66 ∑xy=1855 ∑𝒙 𝟐 ∑𝒚 𝟐 = 𝟖𝟐𝟎 (1). Find the co-efficient of correlation: 2 2 2 2 ( ) ( ) n xy x y r n x x n y y                             2 2 7 1855 175 66 7 5075 175 7 820 (66) r              12985 11550 4900 1384 r    1435 6781600 r  1435 2604.1505 r  0.551043r  (2) . Regression line Y on X Y a bx  2 2 ( ) yx n xy x y b n x x         1435 4900 yxb  0.2928yxb  yx y byx x a n         66 0.2928 175 7 yxa   66 51.24 7 yxa   2.1085yxa  The estimated regression line is as follows. ˆ 2.1085 0.2928x x 
  16. 16. 16 | P a g e (3)Regression line X on Y: X a by  2 2 ( ) xy n xy x y b n y y                2 7 1855 175 66 7 820 (66) xyb    1435 1384 xyb  1.0368xyb  xy x bxy y a n         175 1.036 66 7 xya   106.63 7 xya  15.2328xya  The estimated regression line is as follows. ˆ 15.2328 1.0368x y 

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