\nTotal<\/td>\n | 80<\/td>\n | <\/td>\n | <\/td>\n | 430<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n From the table we may observe that<\/p>\n \nClearly, mean of this data is 35.38. It represents that on an average the age of a patient admitted to hospital was 35.38 years. \nAs we may observe that maximum class frequency is 23 belonging to class interval 35 – 45. \nSo, modal class = 35 – \u00a045 \nLower limit (l) of modal class = 35 \nFrequency (f1<\/sub>) of modal class = 23 \nClass size (h) = 10 \nFrequency (f0<\/sub>) of class preceding the modal class = 21 \nFrequency (f2<\/sub>) of class succeeding the modal class = 14 \n \nClearly mode is 36.8.It represents that maximum number of patients admitted in hospital were of 36.8 years.<\/p>\nQuestion 2. \n<\/strong>The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:<\/p>\n\n\n\nLifetimes (in hours)<\/strong><\/td>\n0 – 20<\/td>\n | 20 – 40<\/td>\n | 40 – 60<\/td>\n | 60 – 80<\/td>\n | 80 – 100<\/td>\n | 100 – 120<\/td>\n<\/tr>\n | \nFrequency<\/strong><\/td>\n10<\/td>\n | 35<\/td>\n | 52<\/td>\n | 61<\/td>\n | 38<\/td>\n | 29<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Determine the modal lifetimes of the components. \nSolution: \n<\/strong>From the data given as above we may observe that maximum class frequency is 61 belonging to class interval 60 – 80. \nSo, modal class = 60 – 80 \nLower class limit (l) of modal class = 60 \nFrequency (f1<\/sub>) of modal class = 61 \nFrequency (f0<\/sub>) of class preceding the modal class = 52 \nFrequency (f2<\/sub>) of class succeeding the modal class = 38 \nClass size (h) = 20 \n \nSo, modal lifetime of electrical components is 65.625 hours. \nQuestion 3. \n<\/strong>The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure.<\/p>\n\n\n\nExpenditure\u00a0<\/strong> \n(in Rs)<\/strong><\/td>\nNumber of families<\/strong><\/td>\n<\/tr>\n\n1000 – 1500<\/td>\n | 24<\/td>\n<\/tr>\n | \n1500 – 2000<\/td>\n | 40<\/td>\n<\/tr>\n | \n2000 – 2500<\/td>\n | 33<\/td>\n<\/tr>\n | \n2500 – 3000<\/td>\n | 28<\/td>\n<\/tr>\n | \n3000 – 3500<\/td>\n | 30<\/td>\n<\/tr>\n | \n3500 – 4000<\/td>\n | 22<\/td>\n<\/tr>\n | \n4000 – 4500<\/td>\n | 16<\/td>\n<\/tr>\n | \n4500 – 5000<\/td>\n | 7<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Solution:<\/strong> \nWe may observe from the given data that maximum class frequency is 40 belonging to 1500 – 2000 intervals. \nSo, modal class = 1500 – 2000 \nLower limit (l) of modal class = 1500 \nFrequency (f1<\/sub>) of modal class = 40 \nFrequency (f0<\/sub>) of class preceding modal class = 24 \nFrequency (f2<\/sub>) of class succeeding modal class = 33 \nClass size (h) = 500<\/p>\n\n \nNow we may find class mark as<\/p>\n
\nClass size (h) of give data = 500 \nNow taking 2750 as assumed mean (a) we may calculate di<\/sub>, ui<\/sub>\u00a0and fi<\/sub>ui<\/sub>\u00a0as following<\/p>\n\n\n\nExpenditure\u00a0<\/strong> \n(in Rs)<\/strong><\/td>\nNumber of families<\/strong>fi<\/sub><\/em><\/strong><\/td>\nxi<\/sub><\/em><\/strong><\/td>\ndi<\/sub><\/em><\/strong>\u00a0=\u00a0xi<\/sub><\/em>\u00a0– 2750<\/strong><\/td>\n<\/td>\n | fi<\/sub>ui<\/sub><\/em><\/strong><\/td>\n<\/tr>\n\n1000 – 1500<\/td>\n | 24<\/td>\n | 1250<\/td>\n | -1500<\/td>\n | -3<\/td>\n | -72<\/td>\n<\/tr>\n | \n1500 – 2000<\/td>\n | 40<\/td>\n | 1750<\/td>\n | -1000<\/td>\n | -2<\/td>\n | -80<\/td>\n<\/tr>\n | \n2000 – 2500<\/td>\n | 33<\/td>\n | 2250<\/td>\n | -500<\/td>\n | -1<\/td>\n | -33<\/td>\n<\/tr>\n | \n2500 – 3000<\/td>\n | 28<\/td>\n | 2750<\/td>\n | 0<\/td>\n | 0<\/td>\n | 0<\/td>\n<\/tr>\n | \n3000 – 3500<\/td>\n | 30<\/td>\n | 3250<\/td>\n | 500<\/td>\n | 1<\/td>\n | 30<\/td>\n<\/tr>\n | \n3500 – 4000<\/td>\n | 22<\/td>\n | 3750<\/td>\n | 1000<\/td>\n | 2<\/td>\n | 44<\/td>\n<\/tr>\n | \n4000 – 4500<\/td>\n | 16<\/td>\n | 4250<\/td>\n | 1500<\/td>\n | 3<\/td>\n | 48<\/td>\n<\/tr>\n | \n4500 – 5000<\/td>\n | 7<\/td>\n | 4750<\/td>\n | 2000<\/td>\n | 4<\/td>\n | 28<\/td>\n<\/tr>\n | \nTotal<\/td>\n | 200<\/td>\n | <\/td>\n | <\/td>\n | <\/td>\n | -35<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Now from table may observe that \n \nSo, mean monthly expenditure was Rs.2662.50.<\/p>\n Question 4. \n<\/strong>The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.<\/p>\n\n\n\nNumber of students\u00a0<\/strong> \nper teacher<\/strong><\/td>\nNumber of\u00a0<\/strong> \nstates\/U.T<\/strong><\/td>\n<\/tr>\n\n15 – 20<\/td>\n | 3<\/td>\n<\/tr>\n | \n20 – 25<\/td>\n | 8<\/td>\n<\/tr>\n | \n25 – 30<\/td>\n | 9<\/td>\n<\/tr>\n | \n30 – 35<\/td>\n | 10<\/td>\n<\/tr>\n | \n35 – 40<\/td>\n | 3<\/td>\n<\/tr>\n | \n40 – 45<\/td>\n | 0<\/td>\n<\/tr>\n | \n45 – 50<\/td>\n | 0<\/td>\n<\/tr>\n | \n50 – 55<\/td>\n | 2<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Solution:<\/strong> \nWe may observe from the given data that maximum class frequency is 10 belonging to class interval 30 – 35. \nSo, modal class = 30 – 35 \nClass size (h) = 5 \nLower limit (l) of modal class = 30 \nFrequency (f1<\/sub>) of modal class = 10 \nFrequency (f0<\/sub>) of class preceding modal class = 9 \nFrequency (f2<\/sub>) of class succeeding modal class = 3 \n \nIt represents that most of states\/U.T have a teacher \u00a0student ratio as 30.6<\/p>\n\n Now we may find class marks by using the relation<\/p>\n \nNow taking 32.5 as assumed mean (a) we may calculate di<\/sub>, ui<\/sub>\u00a0and fi<\/sub>ui<\/sub>\u00a0as following.<\/p>\n\n\n\nNumber of students<\/strong> \nper teacher<\/strong><\/td>\nNumber of states\/U.T<\/strong> \n(fi<\/sub><\/em>)<\/strong><\/td>\n <\/p>\n xi<\/sub><\/em><\/strong><\/td>\n <\/p>\n di<\/sub><\/em><\/strong>\u00a0=\u00a0xi<\/sub><\/em>\u00a0– 32.5<\/strong><\/td>\n<\/td>\n | <\/p>\n fi<\/sub>ui<\/sub><\/em><\/strong><\/td>\n<\/tr>\n\n15 – 20<\/td>\n | 3<\/td>\n | 17.5<\/td>\n | -15<\/td>\n | -3<\/td>\n | -9<\/td>\n<\/tr>\n | \n20 – 25<\/td>\n | 8<\/td>\n | 22.5<\/td>\n | -10<\/td>\n | -2<\/td>\n | -16<\/td>\n<\/tr>\n | \n25 – 30<\/td>\n | 9<\/td>\n | 27.5<\/td>\n | -5<\/td>\n | -1<\/td>\n | -9<\/td>\n<\/tr>\n | \n30 – 35<\/td>\n | 10<\/td>\n | 32.5<\/td>\n | 0<\/td>\n | 0<\/td>\n | 0<\/td>\n<\/tr>\n | \n35 – 40<\/td>\n | 3<\/td>\n | 37.5<\/td>\n | 5<\/td>\n | 1<\/td>\n | 3<\/td>\n<\/tr>\n | \n40 – 45<\/td>\n | 0<\/td>\n | 42.5<\/td>\n | 10<\/td>\n | 2<\/td>\n | 0<\/td>\n<\/tr>\n | \n45 – 50<\/td>\n | 0<\/td>\n | 47.5<\/td>\n | 15<\/td>\n | 3<\/td>\n | 0<\/td>\n<\/tr>\n | \n50 – 55<\/td>\n | 2<\/td>\n | 52.5<\/td>\n | 20<\/td>\n | 4<\/td>\n | 8<\/td>\n<\/tr>\n | \nTotal<\/td>\n | 35<\/td>\n | <\/td>\n | <\/td>\n | <\/td>\n | -23<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n \nSo mean of data is 29.2 \nIt represents that on an average teacher – student ratio was 29.2.<\/p>\n
Question 5. \n<\/strong>The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.<\/p>\n\n\n\nNumber of students<\/strong> \nper teacher<\/strong><\/td>\nNumber of\u00a0<\/strong> \nstates\/U.T<\/strong><\/td>\n<\/tr>\n\n3000 – 4000<\/td>\n | 4<\/td>\n<\/tr>\n | \n4000 – 5000<\/td>\n | 18<\/td>\n<\/tr>\n | \n5000 – 6000<\/td>\n | 9<\/td>\n<\/tr>\n | \n6000 – 7000<\/td>\n | 7<\/td>\n<\/tr>\n | \n7000 – 8000<\/td>\n | 6<\/td>\n<\/tr>\n | \n8000 – 9000<\/td>\n | 3<\/td>\n<\/tr>\n | \n9000 -10000<\/td>\n | 1<\/td>\n<\/tr>\n | \n10000 – 11000<\/td>\n | 1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Find the mode of the data.<\/p>\n Solution: \n<\/strong>From the given data we may observe that maximum class frequency is 18 belonging to class interval 4000 – 5000. \nSo, modal class = 4000 – 5000 \nLower limit (l) of modal class = 4000 \nFrequency (f1<\/sub>) of modal class = 18 \nFrequency (f0<\/sub>) of class preceding modal class = 4 \nFrequency (f2<\/sub>) of class succeeding modal class = 9 \nClass size (h) = 1000 \n \nSo mode of given data is 4608.7 runs.<\/p>\nQuestion 6. \n<\/strong>A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data:<\/p>\n\n\n\nNumber\u00a0<\/strong> \nof cars<\/strong><\/td>\n0 – 10<\/td>\n | 10 – 20<\/td>\n | 20 – 30<\/td>\n | 30 – 40<\/td>\n | 40 – 50<\/td>\n | 50 – 6<\/td>\n | 60 – 70<\/td>\n | 70 – 80<\/td>\n<\/tr>\n | \nFrequency<\/strong><\/td>\n7<\/td>\n | 14<\/td>\n | 13<\/td>\n | 12<\/td>\n | 20<\/td>\n | 11<\/td>\n | 15<\/td>\n | 8<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Solution:<\/strong> \nFrom the given data we may observe that maximum class frequency is 20 belonging to 40 – 50 class intervals. \nSo, modal class = 40 – 50 \nLower limit (l) of modal class = 40 \nFrequency (f1<\/sub>) of modal class = 20 \nFrequency (f0<\/sub>) of class preceding modal class = 12 \nFrequency (f2<\/sub>) of class succeeding modal class = 11 \nClass size = 10 \n \nSo mode of this data is 44.7 cars.<\/p>\n | | | | | | | | | | | |
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