i<\/sub>=150<\/td>\n| <\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Now,\u00a0\\(N=150\\\\ \\Longrightarrow \\frac { N }{ 2 } =75 \u00a0\\)
\nThe cumulative frequency just greater than 75 is 120 and corresponding class is 200 – 300.
\nThus, the median class is 200 – 300
\nl = 200, h = 100, f = 48
\nc = C.F. preceding median class = 72 and\u00a0\\(\\frac { N }{ 2 } \u00a0 \\)= 75
\n![\"RS](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/RS-Aggarwal-Solutions-Class-10-Chapter-9-Mean-Median-Mode-of-Grouped-Data-Ex-9B-3.1.gif\")
\nHence the median of daily wages is Rs. 206.25.
\nQuestion 4:<\/strong><\/span>
\nWe prepare the frequency table, given below:<\/p>\n\n\n\nClass<\/strong><\/td>\nFrequency\u00a0fi<\/sub><\/strong><\/td>\nC.F.<\/strong><\/td>\n<\/tr>\n\n| 5\u00a0– 10<\/td>\n | 5<\/td>\n | 5<\/td>\n<\/tr>\n | \n| 10 –\u00a015<\/td>\n | 6<\/td>\n | 11<\/td>\n<\/tr>\n | \n| 15\u00a0–\u00a020<\/td>\n | 15<\/td>\n | 26<\/td>\n<\/tr>\n | \n| 20 –\u00a025<\/td>\n | 10<\/td>\n | 36<\/td>\n<\/tr>\n | \n| 25\u00a0–\u00a030<\/td>\n | 5<\/td>\n | 41<\/td>\n<\/tr>\n | \n| 30\u00a0–\u00a035<\/td>\n | 4<\/td>\n | 45<\/td>\n<\/tr>\n | \n| 35\u00a0–\u00a040<\/td>\n | 2<\/td>\n | 47<\/td>\n<\/tr>\n | \n| 40\u00a0–\u00a045<\/td>\n | 2<\/td>\n | 49<\/td>\n<\/tr>\n | \n| <\/td>\n | N=\u2211 fi<\/sub>=49<\/td>\n| <\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Now, \\(N=49\\\\ \\Longrightarrow \\frac { N }{ 2 } =24.5 \u00a0\\)
\nThe cumulative frequency just greater than 24.5 is 26 and corresponding class is 15 – 20.
\nThus, the median class is 15 – 20
\nl = 15, h = 5, f = 15
\nc = CF preceding median class = 11 and\u00a0\\(\\frac { N }{ 2 } \u00a0 \\)=24.5
\n![\"RS](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/RS-Aggarwal-Solutions-Class-10-Chapter-9-Mean-Median-Mode-of-Grouped-Data-Ex-9B-4.1.gif\")
\nMedian of frequency distribution is 19.5
\nQuestion 5:<\/strong><\/span>
\nWe prepare the cumulative frequency table as given below:<\/p>\n\n\n\nConsumption<\/strong><\/td>\nFrequency\u00a0fi<\/sub><\/strong><\/td>\nC.F.<\/strong><\/td>\n<\/tr>\n\n| 65\u00a0– 85<\/td>\n | 4<\/td>\n | 4<\/td>\n<\/tr>\n | \n| 85\u00a0–\u00a0105<\/td>\n | 5<\/td>\n | 9<\/td>\n<\/tr>\n | \n| 105\u00a0–\u00a0125<\/td>\n | 13<\/td>\n | 22<\/td>\n<\/tr>\n | \n| 125\u00a0–\u00a0145<\/td>\n | 20<\/td>\n | 42<\/td>\n<\/tr>\n | \n| 145\u00a0–\u00a0165<\/td>\n | 14<\/td>\n | 56<\/td>\n<\/tr>\n | \n| 165\u00a0–\u00a0185<\/td>\n | 7<\/td>\n | 63<\/td>\n<\/tr>\n | \n| 185\u00a0–\u00a0205<\/td>\n | 4<\/td>\n | 67<\/td>\n<\/tr>\n | \n| <\/td>\n | N=\u2211 fi<\/sub>=67<\/td>\n| <\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Now, \\(N=67\\\\ \\Longrightarrow \\frac { N }{ 2 } =33.5 \u00a0\\)
\nThe cumulative frequency just greater than 33.5 is 42 and the corresponding class 125 – 145.
\nThus, the median class is 125 – 145
\nl = 125, h = 20, and c = CF preceding the median class = 22, \\(\\frac { N }{ 2 } \u00a0 \\)= 33.5
\n![\"RS](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/RS-Aggarwal-Solutions-Class-10-Chapter-9-Mean-Median-Mode-of-Grouped-Data-Ex-9B-5.1.gif\")
\nHence median of electricity consumed is 136.5
\nQuestion 6:<\/strong><\/span>
\nFrequency table is given below:<\/p>\n\n\n\nHieght<\/strong><\/td>\nFrequency\u00a0fi<\/sub><\/strong><\/td>\nC.F.<\/strong><\/td>\n<\/tr>\n\n| 135 – 140<\/td>\n | 6<\/td>\n | 6<\/td>\n<\/tr>\n | \n| 140 – 145<\/td>\n | 10<\/td>\n | 16<\/td>\n<\/tr>\n | \n| 145 – 150<\/td>\n | 18<\/td>\n | 34<\/td>\n<\/tr>\n | \n| 150 – 155<\/td>\n | 22<\/td>\n | 56<\/td>\n<\/tr>\n | \n| 155 – 160<\/td>\n | 20<\/td>\n | 76<\/td>\n<\/tr>\n | \n| 160 – 165<\/td>\n | 15<\/td>\n | 91<\/td>\n<\/tr>\n | \n| 165 – 170<\/td>\n | 6<\/td>\n | 97<\/td>\n<\/tr>\n | \n| 170 – 175<\/td>\n | 3<\/td>\n | 100<\/td>\n<\/tr>\n | \n| <\/td>\n | N=\u2211 fi<\/sub>=100<\/td>\n| <\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n \\(N=100\\\\ \\Longrightarrow \\frac { N }{ 2 } =50 \u00a0\\)
\nThe cumulative frequency just greater than 50 is 56 and the corresponding class is 150 – 155
\nThus, the median class is 150 – 155
\nl = 150, h = 5, f = 22, c = C.F.preceding median class = 34
\n![\"RS](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/RS-Aggarwal-Solutions-Class-10-Chapter-9-Mean-Median-Mode-of-Grouped-Data-Ex-9B-6.1.gif\")
\nHence, Median = 153.64
\nQuestion 7:<\/strong><\/span>
\nThe frequency table is given below. Let the missing frequency be x.<\/p>\n\n\n\nClass<\/strong><\/td>\nFrequency\u00a0fi<\/sub><\/strong><\/td>\nC.F.<\/strong><\/td>\n<\/tr>\n\n| 0 – 10<\/td>\n | 5<\/td>\n | 5<\/td>\n<\/tr>\n | \n| 10 – 20<\/td>\n | 25<\/td>\n | 30<\/td>\n<\/tr>\n | \n| 20 – 30<\/td>\n | x<\/td>\n | 30+x<\/td>\n<\/tr>\n | \n| 30 – 40<\/td>\n | 18<\/td>\n | 48+x<\/td>\n<\/tr>\n | \n| 40 – 50<\/td>\n | 7<\/td>\n | 55+x<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Median = 24
\nMedian class is 20 – 30
\n![\"RS](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/RS-Aggarwal-Solutions-Class-10-Chapter-9-Mean-Median-Mode-of-Grouped-Data-Ex-9B-7.1.gif\")
\nl = 20, h = 10, f = x, c = C.F. preceding median class = 30
\n![\"RS](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/RS-Aggarwal-Solutions-Class-10-Chapter-9-Mean-Median-Mode-of-Grouped-Data-Ex-9B-7.2.jpg\")
\nHence, the missing frequency is 25.
\nQuestion 8:<\/strong><\/span>
\nLet f1<\/sub>\u00a0and f2<\/sub>\u00a0be the frequencies of classes 20 – 30 and 40 – 50 respectively, then
\n![\"RS](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-10-Chapter-9-Mean-Median-Mode-of-Grouped-Data-Ex-9B-100.jpg\")
\nMedian is 35, which lies in 30 – 40, so the median class is 30 – 40.
\nl = 30, h = 10, f = 40, N = 170 and c = 10 + 20 +f1<\/sub>\u00a0= (30 + f1<\/sub>)
\n![\"RS](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/RS-Aggarwal-Solutions-Class-10-Chapter-9-Mean-Median-Mode-of-Grouped-Data-Ex-9B-8.1.jpg\")
\n![\"RS](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/RS-Aggarwal-Solutions-Class-10-Chapter-9-Mean-Median-Mode-of-Grouped-Data-Ex-9B-8.2.gif\")
\nQuestion 9:<\/strong><\/span>
\nLet f1<\/sub>\u00a0and f2\u00a0<\/sub>be the frequencies of class intervals 0 – 10 and 40 – 50
\n![\"RS](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/RS-Aggarwal-Solutions-Class-10-Chapter-9-Mean-Median-Mode-of-Grouped-Data-Ex-9B-9.1.gif\")
\nMedian is 32.5 which lies in 30 – 40, so the median class is 30 – 40
\nl = 30, h = 10, f = 12, N = 40 and\u00a0c = f1<\/sub>\u00a0+5+9= (14 + f1<\/sub>)
\n![\"RS](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/RS-Aggarwal-Solutions-Class-10-Chapter-9-Mean-Median-Mode-of-Grouped-Data-Ex-9B-9.2.gif\")
\nQuestion 10:<\/strong><\/span>
\nThe given series is of inclusive form. Converting it into exclusive form and preparing the cumulative frequency table, we get<\/p>\n\n\n\nClass<\/strong><\/td>\nFrequency\u00a0fi<\/sub><\/strong><\/td>\nC.F.<\/strong><\/td>\n<\/tr>\n\n| 18.5\u00a0–\u00a025.5<\/td>\n | 35<\/td>\n | 35<\/td>\n<\/tr>\n | \n| 25.5\u00a0–\u00a032.5<\/td>\n | 96<\/td>\n | 131<\/td>\n<\/tr>\n | \n| 32.5 – 39.5<\/td>\n | 68<\/td>\n | 199<\/td>\n<\/tr>\n | \n| 39.5 –\u00a046.5<\/td>\n | 102<\/td>\n | 301<\/td>\n<\/tr>\n | \n| 46.5 –\u00a053.5<\/td>\n | 35<\/td>\n | 336<\/td>\n<\/tr>\n | \n| 53.5 – 60.5<\/td>\n | 4<\/td>\n | 340<\/td>\n<\/tr>\n | \n| <\/td>\n | N=\u2211 fi<\/sub>=340<\/td>\n| <\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n \\(N=340\\\\ \\Longrightarrow \\frac { N }{ 2 } =170 \u00a0\\)
\nThe cumulative frequency just greater than 170 is 199 and the corresponding class is 32.5 – 39.5.
\nMedian class is 32.5 – 39.5
\nl = 32.5, h = 7, f = 68, c = C.F. of preceding median class = 131
\n![\"RS](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/RS-Aggarwal-Solutions-Class-10-Chapter-9-Mean-Median-Mode-of-Grouped-Data-Ex-9B-10.1.gif\")
\nHence median is 36.5 years
\nQuestion 11:<\/strong><\/span>
\nGiven series is in inclusive form converting it into exclusive form and preparing the cumulative frequency table, we get<\/p>\n\n\n\n| Wages per day <\/strong><\/p>\n (in Rs.)<\/strong><\/td>\nFrequency\u00a0fi<\/sub><\/strong><\/td>\nC.F.<\/strong><\/td>\n<\/tr>\n\n| 60.5 –\u00a070.5<\/td>\n | 5<\/td>\n | 5<\/td>\n<\/tr>\n | \n| 70.5\u00a0–\u00a080.5<\/td>\n | 15<\/td>\n | 20<\/td>\n<\/tr>\n | \n| 80.5 –\u00a090.5<\/td>\n | 20<\/td>\n | 40<\/td>\n<\/tr>\n | \n| 90.5 –\u00a0100.5<\/td>\n | 30<\/td>\n | 70<\/td>\n<\/tr>\n | \n| 100.5 –\u00a0110.5<\/td>\n | 20<\/td>\n | 90<\/td>\n<\/tr>\n | \n| 110.5 –\u00a0120.5<\/td>\n | 8<\/td>\n | 98<\/td>\n<\/tr>\n | \n| <\/td>\n | N=\u2211 fi<\/sub>=98<\/td>\n| <\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n \\(N=98\\\\ \\Longrightarrow \\frac { N }{ 2 } =49 \u00a0\\)
\nThe cumulative frequency just greater than 49 is 70 and corresponding class is 90.5 – 100.5.
\nmedian class is 90.5 – 100.5
\nl = 90.5, h = 10, f = 30, c = CF preceding median class = 40
\n![\"RS](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/RS-Aggarwal-Solutions-Class-10-Chapter-9-Mean-Median-Mode-of-Grouped-Data-Ex-9B-11.1.gif\")
\nHence, Median = Rs 93.50
\nQuestion 12:<\/strong><\/span>
\nThe given series is converted from inclusive to exclusive form and preparing the cumulative frequency table, we get<\/p>\n\n\n\nMarks<\/strong><\/td>\nFrequency\u00a0fi<\/sub><\/strong><\/td>\nC.F.<\/strong><\/td>\n<\/tr>\n\n| 10.5\u00a0–\u00a015.5<\/td>\n | 2<\/td>\n | 2<\/td>\n<\/tr>\n | \n| 15.5 – 20.5<\/td>\n | 3<\/td>\n | 5<\/td>\n<\/tr>\n | \n| 20.5 –\u00a025.5<\/td>\n | 6<\/td>\n | 11<\/td>\n<\/tr>\n | \n| 25.5 –\u00a030.5<\/td>\n | 7<\/td>\n | 18<\/td>\n<\/tr>\n | \n| 30.5 –\u00a035.5<\/td>\n | 14<\/td>\n | 32<\/td>\n<\/tr>\n | \n| 35.5 -40.5<\/td>\n | 12<\/td>\n | 44<\/td>\n<\/tr>\n | \n| 40.5 -45.5<\/td>\n | 4<\/td>\n | 48<\/td>\n<\/tr>\n | \n| 45.5 -50.5<\/td>\n | 2<\/td>\n | 50<\/td>\n<\/tr>\n | \n| <\/td>\n | N=\u2211 fi<\/sub>=50<\/td>\n| <\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n \\(N=50\\\\ \\Longrightarrow \\frac { N }{ 2 } =25 \u00a0\\)
\nThe cumulative frequency just greater than 25 is 32.
\nThe corresponding class is 30.5 – 35.5.
\nThus, the median class is 30.5 – 35.5
\nl = 30.5, h = 5, f = 14, c = C.F preceding median class = 18
\n![\"RS](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/RS-Aggarwal-Solutions-Class-10-Chapter-9-Mean-Median-Mode-of-Grouped-Data-Ex-9B-12.1.gif\")
\nHence, Median = 33
\nQuestion 13:<\/strong><\/span>
\nThe given series is converted from inclusive to exclusive form and preparing the cumulative frequency table, we get<\/p>\n\n\n\nMarks<\/strong><\/td>\nFrequency\u00a0fi<\/sub><\/strong><\/td>\nC.F.<\/strong><\/td>\n<\/tr>\n\n| 0.5 –\u00a05.5<\/td>\n | 7<\/td>\n | 7<\/td>\n<\/tr>\n | \n| 5.5\u00a0–\u00a010.5<\/td>\n | 10<\/td>\n | 17<\/td>\n<\/tr>\n | \n| 10.5\u00a0–\u00a015.5<\/td>\n | 16<\/td>\n | 33<\/td>\n<\/tr>\n | \n| 15.5\u00a0–\u00a020.5<\/td>\n | 32<\/td>\n | 65<\/td>\n<\/tr>\n | \n| 20.5\u00a0–\u00a025.5<\/td>\n | 24<\/td>\n | 89<\/td>\n<\/tr>\n | \n| 25.5 –\u00a030.5<\/td>\n | 16<\/td>\n | 105<\/td>\n<\/tr>\n | \n| 30.5 –\u00a035.5<\/td>\n | 11<\/td>\n | 116<\/td>\n<\/tr>\n | \n| 35.5 –\u00a040.5<\/td>\n | 5<\/td>\n | 121<\/td>\n<\/tr>\n | \n| 40.5 –\u00a045.5<\/td>\n | 2<\/td>\n | 123<\/td>\n<\/tr>\n | \n| <\/td>\n | N=\u2211 fi<\/sub>=123<\/td>\n| <\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n \\(N=123\\\\ \\Longrightarrow \\frac { N }{ 2 } =61.5\u00a0 \\)
\nThe cumulative frequency just greater than 61.5 is 65.
\nThe corresponding median class is 15.5 – 20.5.
\nThen the median class is 15.5 – 20.5
\nl = 15.5, h = 5, f = 32, c = C.F. preceding median class = 33
\n![\"RS](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/RS-Aggarwal-Solutions-Class-10-Chapter-9-Mean-Median-Mode-of-Grouped-Data-Ex-9B-13.1.gif\")
\nHence, Median = 19.95
\nQuestion 14:<\/strong><\/span><\/p>\n\n\n\nMarks<\/strong><\/td>\nFrequency\u00a0fi<\/sub><\/strong><\/td>\nC.F.<\/strong><\/td>\n<\/tr>\n\n| 0 – 10<\/td>\n | 12<\/td>\n | 12<\/td>\n<\/tr>\n | \n| 10 – 20<\/td>\n | 20<\/td>\n | 32<\/td>\n<\/tr>\n | \n| 20 – 30<\/td>\n | 25<\/td>\n | 57<\/td>\n<\/tr>\n | \n| 30 – 40<\/td>\n | 23<\/td>\n | 80<\/td>\n<\/tr>\n | \n| 40 – 50<\/td>\n | 12<\/td>\n | 92<\/td>\n<\/tr>\n | \n| 50 – 60<\/td>\n | 24<\/td>\n | 116<\/td>\n<\/tr>\n | \n| 60 – 70<\/td>\n | 24<\/td>\n | 116<\/td>\n<\/tr>\n | \n| 70 – 80<\/td>\n | 36<\/td>\n | 200<\/td>\n<\/tr>\n | \n| <\/td>\n | N=\u2211 fi<\/sub>=200<\/td>\n| <\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n \\(N=200\\\\ \\Longrightarrow \\frac { N }{ 2 } =100 \u00a0\\)
\nThe cumulative frequency just greater than 100 is 116 and the corresponding class is 50 – 60.
\nThus the median class is 50 – 60
\nl = 50, h = 10, f = 24, c = C.F. preceding median class = 92, \\(\\frac { N }{ 2 } \u00a0 \\)= 100
\n![\"RS](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/RS-Aggarwal-Solutions-Class-10-Chapter-9-Mean-Median-Mode-of-Grouped-Data-Ex-9B-14.1.gif\")
\nHence, Median = 53.33<\/p>\n Hope given RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9B<\/a> are helpful to complete your math homework.<\/p>\nIf you have any doubts, please comment below. A Plus Topper<\/a> try to provide online math tutoring for you.<\/p>\n","protected":false},"excerpt":{"rendered":" | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | ![\"RS](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/RS-Aggarwal-Solutions-Class-10-Chapter-9-Mean-Median-Mode-of-Grouped-Data-Ex-9B-1.1.jpg\")
![\"RS](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2020\/12\/RS-Aggarwal-Solutions-Class-10-Chapter-9-Mean-Median-Mode-of-Grouped-Data-Ex-9B-2.1.gif\")