Find the mean deviation from the mean for the following data :
(i) Classes0−10010−200200−300300−400400−500500−600600−700700−800Frequencies489107543
(ii) Classes95−105105−115115−125125−135135−145145−155Frequencies91326263012
(iii) Classes0−1010−2020−3030−4040−5050−60Frequencies68141642
Find the mean deviation from the mean for the following data :
(i) Classes0−10010−200200−300300−400400−500500−600600−700700−800Frequencies489107543
(ii) Classes95−105105−115115−125125−135135−145145−155Frequencies91326263012
(iii) Classes0−1010−2020−3030−4040−5050−60Frequencies68141642
ClassfiMid value (xi)di=xi−AhA=350,h=100fidi|xi−¯x|fi|xi−¯x|0−100450−3−123081232100−2008150−2−162081664200−3009250−1−9108972300−4001035000880400−50074501792644500−6005550210192960600−70046502122921168700−80037504123921176Total∑fi=5047896
Mean ¯x=A+∑fidi∑fi×h=350+450×100
=350+8=¯x=358
Mean deviation about the mean =∑fi|xi−¯x|∑fi=789650=157.92
ClassfiMid value (xi)di=xi−AhA=130,h=10fidi|xi−¯x|fi|xi−¯x|95−1059100−3−2725.3227.7105−11513110−2−2615.3198.9115−12526120−1−265.3137.8125−13530130004.7141.0135−1451214011214.7176.4145−1551015022024.7247.0Total100−471128.8
Mean ¯x=A+∑fidi∑fi×h=130+(−47)100×10
=130−4.7=125.3
Mean deviation about the mean =∑fi|xi−¯x|∑fi=1128.8100=11.288
ClassfiCFMid value (xi)|xi−M|fi|xi−M|0−10665|5−27.86|=22.86137.1610−208(14)C15|15−27.86|=12.86102.8820−30(14)2825|25−27.86|=2.8640.0430−40164435|35−27.86|=7.14114.2440−5044845|45−27.86|=17.1468.5650−6025055|55−27.86|=27.1454.28∑fi=50
N2=502=25
⇒C=14,f=14,l=20,h=10
Median M=l+N2−Cf×h=20+25−1414×10
=20+11×1014
=20+7.86−27.86
∴ Mean deviation about the mean =∑fi|xi−M|∑fi=517.1650=10.34
ClassfiMid value (xi)di=xi−AhA=350,h=100fidi|xi−¯x|fi|xi−¯x|0−100450−3−123081232100−2008150−2−162081664200−3009250−1−9108972300−4001035000880400−50074501792644500−6005550210192960600−70046502122921168700−80037504123921176Total∑fi=5047896
Mean ¯x=A+∑fidi∑fi×h=350+450×100
=350+8=¯x=358
Mean deviation about the mean =∑fi|xi−¯x|∑fi=789650=157.92
ClassfiMid value (xi)di=xi−AhA=130,h=10fidi|xi−¯x|fi|xi−¯x|95−1059100−3−2725.3227.7105−11513110−2−2615.3198.9115−12526120−1−265.3137.8125−13530130004.7141.0135−1451214011214.7176.4145−1551015022024.7247.0Total100−471128.8
Mean ¯x=A+∑fidi∑fi×h=130+(−47)100×10
=130−4.7=125.3
Mean deviation about the mean =∑fi|xi−¯x|∑fi=1128.8100=11.288
ClassfiCFMid value (xi)|xi−M|fi|xi−M|0−10665|5−27.86|=22.86137.1610−208(14)C15|15−27.86|=12.86102.8820−30(14)2825|25−27.86|=2.8640.0430−40164435|35−27.86|=7.14114.2440−5044845|45−27.86|=17.1468.5650−6025055|55−27.86|=27.1454.28∑fi=50
N2=502=25
⇒C=14,f=14,l=20,h=10
Median M=l+N2−Cf×h=20+25−1414×10
=20+11×1014
=20+7.86−27.86
∴ Mean deviation about the mean =∑fi|xi−M|∑fi=517.1650=10.34
