\n120 \u2013 130<\/td>\n 4<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nFind the median from the curve. \nSolution: \u00a0 \u00a0\u00a0<\/strong>Let us prepare following table showing the cumulative frequencies more than the upper limit.<\/p>\n\n\n\nClass interval (I. Q)<\/strong><\/td>\nFrequency (f)<\/strong><\/td>\nCumulative frequency<\/strong><\/td>\n<\/tr>\n\n60 \u2013 70<\/td>\n 2<\/td>\n 2<\/td>\n<\/tr>\n \n70 \u2013 80<\/td>\n 5<\/td>\n 2 + 5 = 7<\/td>\n<\/tr>\n \n80 \u201390<\/td>\n 12<\/td>\n 2 + 5 + 12 = 19<\/td>\n<\/tr>\n \n90 \u2013 100<\/td>\n 31<\/td>\n 2 + 5 + 12 + 31 = 50<\/td>\n<\/tr>\n \n100 \u2013 110<\/td>\n 39<\/td>\n 2 + 5 + 12 + 31 + 39 = 89<\/td>\n<\/tr>\n \n110 \u2013 120<\/td>\n 10<\/td>\n 2 + 5 + 12 + 31 + 39 + 10 = 99<\/td>\n<\/tr>\n \n120 \u2013 130<\/td>\n 4<\/td>\n 2 + 5 + 12 + 31 + 39 + 10 + 4 = 103<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nLess than ogive :<\/strong> \nI.Q. is taken on the x-axis. Number of students are marked on y-axis. \nPoints (70, 2), (80, 7), (90, 19), (100, 50), (110, 89), (120, 99), (130, 103), are plotted on graph paper and these points are joined by free hand. The curve obtained is less than ogive. \n \nThe value\u00a0\\(\\frac{N}{2}\\) = 51.5 is marked on y-axis and from\u00a0 this point a line parallel to x-axis is drawn. This line meets the curve at a point P. From P draw a perpendicular PN to meet x-axis at N. N represents the median. \nHere median is 100.5. \nHence, the median of given frequency distribution is 100.5<\/p>\nExample 2:<\/strong>\u00a0 \u00a0 The following table shows the daily sales of 230 footpath sellers of Chandni Chowk.<\/p>\n\n\n\nSales in Rs.<\/strong><\/td>\nNo. of sellers<\/strong><\/td>\n<\/tr>\n\n0 \u2013 500<\/td>\n 12<\/td>\n<\/tr>\n \n500 \u2013 1000<\/td>\n 18<\/td>\n<\/tr>\n \n1000 \u2013 1500<\/td>\n 35<\/td>\n<\/tr>\n \n1500 \u2013 2000<\/td>\n 42<\/td>\n<\/tr>\n \n2000 \u2013 2500<\/td>\n 50<\/td>\n<\/tr>\n \n2500 \u2013 3000<\/td>\n 45<\/td>\n<\/tr>\n \n3000 \u2013 3500<\/td>\n 20<\/td>\n<\/tr>\n \n3500 \u2013 4000<\/td>\n 8<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nLocate the median of the above data using only the less than type ogive. \nSolution: \u00a0 \u00a0\u00a0<\/strong>To draw ogive, we need to have a cumulative frequency distribution.<\/p>\n\n\n\nSales in Rs.<\/strong><\/td>\nNo. of sellers<\/strong><\/td>\nLess than type cumulative frequency<\/strong><\/td>\n<\/tr>\n\n0 \u2013 500<\/td>\n 12<\/td>\n 12<\/td>\n<\/tr>\n \n500 \u2013 1000<\/td>\n 18<\/td>\n 30<\/td>\n<\/tr>\n \n1000 \u2013 1500<\/td>\n 35<\/td>\n 65<\/td>\n<\/tr>\n \n1500 \u2013 2000<\/td>\n 42<\/td>\n 107<\/td>\n<\/tr>\n \n2000 \u2013 2500<\/td>\n 50<\/td>\n 157<\/td>\n<\/tr>\n \n2500 \u2013 3000<\/td>\n 45<\/td>\n 202<\/td>\n<\/tr>\n \n3000 \u2013 3500<\/td>\n 20<\/td>\n 222<\/td>\n<\/tr>\n \n3500 \u2013 4000<\/td>\n 8<\/td>\n 230<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nLess than ogive : <\/strong> \nSeles in Rs. are taken on the y-axis and number of sellers are taken on x-axis. For drawing less than ogive, points (500, 12), (1000, 30), (1500, 65), (2000, 107), (2500, 157), (3000, 202), (3500, 222), (4000, 230) are plotted on graph paper and these are joined free hand to obtain the less than ogive. \n \nThe value \\(\\frac{N}{2}\\) = 115 is marked on y-axis and a line parallel to x-axis is drawn. This line meets the curve at a point P. From P draw a perpendicular PN to meet x-axis at median. Median = 2000. \nHence, the median of given frequency distribution is 2000.<\/p>\nExample 3:<\/strong>\u00a0 \u00a0 Draw the two ogives for the following frequency distribution of the weekly wages of (less than and more than) number of workers.<\/p>\n\n\n\nWeekly wages<\/strong><\/td>\nNumber of workers<\/strong><\/td>\n<\/tr>\n\n0 \u2013 20<\/td>\n 41<\/td>\n<\/tr>\n \n20 \u2013 40<\/td>\n 51<\/td>\n<\/tr>\n \n40 \u2013 60<\/td>\n 64<\/td>\n<\/tr>\n \n60 \u2013 80<\/td>\n 38<\/td>\n<\/tr>\n \n80 \u2013 100<\/td>\n 7<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nHence find the value of median. \nSolution: \u00a0 \u00a0\u00a0<\/strong><\/p>\n\n\n\nWeekly wages<\/strong><\/td>\nNumber of workers<\/strong><\/td>\nC.F (less than)<\/strong><\/td>\nC.F (More than)<\/strong><\/td>\n<\/tr>\n\n0 \u2013 20<\/td>\n 41<\/td>\n 41<\/td>\n 201<\/td>\n<\/tr>\n \n20 \u2013 40<\/td>\n 51<\/td>\n 92<\/td>\n 160<\/td>\n<\/tr>\n \n40 \u2013 60<\/td>\n 64<\/td>\n 156<\/td>\n 109<\/td>\n<\/tr>\n \n60 \u2013 80<\/td>\n 38<\/td>\n 194<\/td>\n 45<\/td>\n<\/tr>\n \n80 \u2013 100<\/td>\n 7<\/td>\n 201<\/td>\n 7<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nLess than curve :\u00a0\u00a0 <\/strong> \nUpper limits of class intervals are marked on the x-axis and less than type cumulative frequencies are taken on y-axis. For drawing less than type curve, points (20, 41), (40, 92), (60, 156), (80, 194), (100, 201) are plotted on the graph paper and these are joined by free hand to obtain the less than ogive. \n \nGreater than ogive<\/strong> \nLower limits of class interval are marked on x-axis and greater than type cumulative frequencies are taken on y-axis. For drawing greater than type curve, points (0, 201), (20, 160), (40, 109), (60, 45) and (80, 7) are plotted on the graph paper and these are joined by free hand to obtain the greater than type ogive. From the point of intersection of these curves a perpendicular line on x-axis is drawn. The point at which this line meets x-axis determines the median. Here the median is 42.652.<\/p>\nExample 4:<\/strong>\u00a0 \u00a0 Following table gives the cumulative frequency of the age of a group of 199 teachers. \nDraw the less than ogive and greater than ogive and find the median.<\/p>\n\n\n\nAge in years<\/strong><\/td>\nCum. Frequency<\/strong><\/td>\n<\/tr>\n\n20 \u2013 25<\/td>\n 21<\/td>\n<\/tr>\n \n25 \u2013 30<\/td>\n 40<\/td>\n<\/tr>\n \n30 \u2013 35<\/td>\n 90<\/td>\n<\/tr>\n \n35 \u2013 40<\/td>\n 130<\/td>\n<\/tr>\n \n40 \u2013 45<\/td>\n 146<\/td>\n<\/tr>\n \n45 \u2013 50<\/td>\n 166<\/td>\n<\/tr>\n \n50 \u2013 55<\/td>\n 176<\/td>\n<\/tr>\n \n55 \u2013 60<\/td>\n 186<\/td>\n<\/tr>\n \n60 \u2013 65<\/td>\n 195<\/td>\n<\/tr>\n \n65 \u2013 70<\/td>\n 199<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nSolution: \u00a0 \u00a0\u00a0<\/strong><\/p>\n\n\n\nAge in years<\/strong><\/td>\nLess than cumulative frequency<\/strong><\/td>\nFrequency<\/strong><\/td>\nGreater than type<\/strong><\/td>\n<\/tr>\n\n20 \u2013 25<\/td>\n 21<\/td>\n 21<\/td>\n 199<\/td>\n<\/tr>\n \n25 \u2013 30<\/td>\n 40<\/td>\n 19<\/td>\n 178<\/td>\n<\/tr>\n \n30 \u2013 35<\/td>\n 90<\/td>\n 50<\/td>\n 159<\/td>\n<\/tr>\n \n35 \u2013 40<\/td>\n 130<\/td>\n 40<\/td>\n 109<\/td>\n<\/tr>\n \n40 \u2013 45<\/td>\n 146<\/td>\n 16<\/td>\n 69<\/td>\n<\/tr>\n \n45 \u2013 50<\/td>\n 166<\/td>\n 20<\/td>\n 53<\/td>\n<\/tr>\n \n50 \u2013 55<\/td>\n 176<\/td>\n 10<\/td>\n 33<\/td>\n<\/tr>\n \n55 \u2013 60<\/td>\n 186<\/td>\n 10<\/td>\n 23<\/td>\n<\/tr>\n \n60 \u2013 65<\/td>\n 195<\/td>\n 9<\/td>\n 13<\/td>\n<\/tr>\n \n65 \u2013 70<\/td>\n 199<\/td>\n 4<\/td>\n 4<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nFind out the frequencies by subtracting previous\u00a0 frequency from the next frequency to get simple frequency. Now we can prepare the greater than type frequency. Ages are taken on x-axis and number of teachers on y-axis. \nLess than ogive :<\/strong> \nPlot the points (25, 21), (30, 40), (35, 90), (40, 130), (45, 146), (50, 166), (55, 176), (60, 186), (65, 195), (70, 199) on graph paper. Join these points free hand to get less than ogive. \nGreater than ogive : <\/strong> \nPlot the points (20, 199), (25, 178), (30, 159), (35, 109), (40, 69), (45, 53), (50, 33), (55, 23), (60, 13), (65, 4) on graph paper. Join these points freehand to get greater than ogive. Median is the point of intersection of these two curves. \n \nHere median is 37.375.<\/p>\nExample 5:<\/strong>\u00a0 \u00a0 Following is the age distribution of a group of students. Draw the cumulative frequency polygon, cumulative frequency curve (less than type) and hence obtain the median value.<\/p>\n\n\n\nAge<\/strong><\/td>\nFrequency<\/strong><\/td>\n<\/tr>\n\n5 \u2013 6<\/td>\n 40<\/td>\n<\/tr>\n \n6 \u2013 7<\/td>\n 56<\/td>\n<\/tr>\n \n7 \u2013 8<\/td>\n 60<\/td>\n<\/tr>\n \n8 \u2013 9<\/td>\n 66<\/td>\n<\/tr>\n \n9 \u2013 10<\/td>\n 84<\/td>\n<\/tr>\n \n10 \u2013 11<\/td>\n 96<\/td>\n<\/tr>\n \n11 \u2013 12<\/td>\n 92<\/td>\n<\/tr>\n \n12 \u2013 13<\/td>\n 80<\/td>\n<\/tr>\n \n13 \u2013 14<\/td>\n 64<\/td>\n<\/tr>\n \n14 \u2013 15<\/td>\n 44<\/td>\n<\/tr>\n \n15 \u2013 16<\/td>\n 20<\/td>\n<\/tr>\n \n16 \u2013 17<\/td>\n 8<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nSolution: \u00a0 \u00a0\u00a0<\/strong>We first prepare the cumulative frequency table by less then method as given below :<\/p>\n\n\n\nAge<\/strong><\/td>\nFrequency<\/strong><\/td>\nAge less than<\/strong><\/td>\nCumulative frequency<\/strong><\/td>\n<\/tr>\n\n5 \u2013 6<\/td>\n 40<\/td>\n 6<\/td>\n 40<\/td>\n<\/tr>\n \n6 \u2013 7<\/td>\n 56<\/td>\n 7<\/td>\n 96<\/td>\n<\/tr>\n \n7 \u2013 8<\/td>\n 60<\/td>\n 8<\/td>\n 156<\/td>\n<\/tr>\n \n8 \u2013 9<\/td>\n 66<\/td>\n 9<\/td>\n 222<\/td>\n<\/tr>\n \n9 \u2013 10<\/td>\n 84<\/td>\n 10<\/td>\n 306<\/td>\n<\/tr>\n \n10 \u2013 11<\/td>\n 96<\/td>\n 11<\/td>\n 402<\/td>\n<\/tr>\n \n11 \u2013 12<\/td>\n 92<\/td>\n 12<\/td>\n 494<\/td>\n<\/tr>\n \n12 \u2013 13<\/td>\n 80<\/td>\n 13<\/td>\n 574<\/td>\n<\/tr>\n \n13 \u2013 14<\/td>\n 64<\/td>\n 14<\/td>\n 638<\/td>\n<\/tr>\n \n14 \u2013 15<\/td>\n 44<\/td>\n 15<\/td>\n 682<\/td>\n<\/tr>\n \n15 \u2013 16<\/td>\n 20<\/td>\n 16<\/td>\n 702<\/td>\n<\/tr>\n \n16 \u2013 17<\/td>\n 8<\/td>\n 17<\/td>\n 710<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nOther than the given class intervals, we assume a class 4-5 before the first class interval 5-6 with zero frequency. \nNow, we mark the upper class limits (including the imagined class) along X-axis on a suitable scale and the cumulative frequencies along Y-axis on a suitable scale. \nThus, we plot the points (5, 0), (6, 40), (7, 96),\u00a0 (8, 156), (9, 222), (10, 306), (11, 402), (12, 494), (13, 574), (14, 638), (15, 682), (16, 702) and (17, 710). \nThese points are marked and joined by line segments to obtain the cumulative frequency polygon shown in Fig. \n \nIn order to obtain the cumulative frequency curve, we draw a smooth curve passing through the points discussed above. The graph (fig) shows the total number of students as 710. The median is the age corresponding to \\(\\frac{N}{2}\\,\\, = \\,\\,\\frac{{710}}{2}\\)\u00a0= 355 students. In order to find the median, we first located the point corresponding to 355th student on Y-axis. Let the point be P. From this point draw a line parallel to the X-axis cutting the curve at Q. From this point Q draw a line parallel to Y-axis and meeting X-axis at the point M. The x-coordinate of M is 10.5 (See Fig.). Hence, median is 10.5. \n <\/p>\n
Example 6:<\/strong>\u00a0 \u00a0 The following observations relate to the height of a group of persons. Draw the two type of cumulative frequency polygons and cumulative frequency curves and determine the median.<\/p>\n\n\n\nHeight in cms<\/td>\n 140\u2013143<\/td>\n 143\u2013146<\/td>\n 146\u2013149<\/td>\n 149\u2013152<\/td>\n 152\u2013155<\/td>\n 155\u2013158<\/td>\n 158\u2013161<\/td>\n<\/tr>\n \nFrequency<\/td>\n 3<\/td>\n 9<\/td>\n 26<\/td>\n 31<\/td>\n 45<\/td>\n 64<\/td>\n 78<\/td>\n<\/tr>\n \nHeight in cms<\/td>\n 161\u2013164<\/td>\n 164\u2013167<\/td>\n 167\u2013170<\/td>\n 170\u2013173<\/td>\n 173\u2013176<\/td>\n 176\u2013179<\/td>\n 179\u2013182<\/td>\n<\/tr>\n \nFrequency<\/td>\n 85<\/td>\n 96<\/td>\n 72<\/td>\n 60<\/td>\n 43<\/td>\n 20<\/td>\n 6<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nSolution: \u00a0 \u00a0\u00a0<\/strong>Less than method :\u00a0 We first prepare the cumulative frequency table by less than method as given below :<\/p>\n\n\n\nHeight in cms<\/strong><\/td>\nFrequency<\/strong><\/td>\nHeight less than<\/strong><\/td>\nFrequency<\/strong><\/td>\n<\/tr>\n\n140\u2013143<\/td>\n 3<\/td>\n 143<\/td>\n 3<\/td>\n<\/tr>\n \n143\u2013146<\/td>\n 9<\/td>\n 146<\/td>\n 12<\/td>\n<\/tr>\n \n146\u2013149<\/td>\n 26<\/td>\n 149<\/td>\n 38<\/td>\n<\/tr>\n \n149\u2013152<\/td>\n 31<\/td>\n 152<\/td>\n 69<\/td>\n<\/tr>\n \n152\u2013155<\/td>\n 45<\/td>\n 155<\/td>\n 114<\/td>\n<\/tr>\n \n155\u2013158<\/td>\n 64<\/td>\n 158<\/td>\n 178<\/td>\n<\/tr>\n \n158\u2013161<\/td>\n 78<\/td>\n 161<\/td>\n 256<\/td>\n<\/tr>\n \n161\u2013164<\/td>\n 85<\/td>\n 164<\/td>\n 341<\/td>\n<\/tr>\n \n164\u2013167<\/td>\n 96<\/td>\n 167<\/td>\n 437<\/td>\n<\/tr>\n \n167\u2013170<\/td>\n 72<\/td>\n 170<\/td>\n 509<\/td>\n<\/tr>\n \n170\u2013173<\/td>\n 60<\/td>\n 173<\/td>\n 569<\/td>\n<\/tr>\n \n173\u2013176<\/td>\n 43<\/td>\n 176<\/td>\n 612<\/td>\n<\/tr>\n \n176\u2013179<\/td>\n 20<\/td>\n 179<\/td>\n 632<\/td>\n<\/tr>\n \n179\u2013182<\/td>\n 6<\/td>\n 182<\/td>\n 638<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nOther than the given class intervals, we assume a class interval 137-140 prior to the first class interval 140-143 with zero frequency. \nNow, we mark the upper class limits on X-axis and cumulative frequency along Y-axis on a suitable scale. \nWe plot the points (140, 0), (143, 3), (146, 12),\u00a0 (149, 38), (152, 69), (155, 114), (158, 178), (161, 256), \n(164, 341), (167, 437), (170, 509), (173, 569), (176, 612),.(179, 632) and 182, 638). \n \nThese points are joined by line segments to obtain the cumulative frequency polygon as shown in fig. and by a free hand smooth curve to obtain an ogive by less than method as shown in fig. \n \nMore than method : We prepare the cumulative frequency table by more than method as given below : \nOther than the given class intervals, we assume the class interval 182-185 with zero frequency. \nNow, we mark the lower class limits on X-axis and the cumulative frequencies along Y-axis on suitable scales to plot the points (140, 638), (143, 635), (146, 626),\u00a0 (149, 600), (152, 569), (155, 524), (158, 460), (161, 382),\u00a0 (164, 297), (167, 201), (170, 129), (173, 69), (176, 26) and (179, 6). By joining these points by line segments, we obtain the more than type frequency polygon as shown in fig. By joining these points by a free hand curve, we obtain more than type cumulative frequency curve as points by a free hand curve, we obtain more than type cumulative frequency curves as shown in fig. \nWe find that the two types of cumulative frequency curves intersect at point P. From point P perpendicular PM is drawn on X-axis. The value of height corresponding to M is 163.2 cm. Hence, median is 163.2 cm.<\/p>\n","protected":false},"excerpt":{"rendered":"
What is Cumulative Frequency Curve or the Ogive\u00a0in Statistics First we prepare the cumulative frequency table, then the cumulative frequencies are plotted against the upper or lower limits of the corresponding class intervals. By joining the points the curve so obtained is called a cumulative frequency curve or ogive. There are two types of ogives […]<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_genesis_hide_title":false,"_genesis_hide_breadcrumbs":false,"_genesis_hide_singular_image":false,"_genesis_hide_footer_widgets":false,"_genesis_custom_body_class":"","_genesis_custom_post_class":"","_genesis_layout":"","footnotes":""},"categories":[5],"tags":[1183,1186,1185,1108,1184],"yoast_head":"\n
What is Cumulative Frequency Curve or the Ogive in Statistics - A Plus Topper<\/title>\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n