## What is accumulation Frequency Curve or the Ogive in Statistics

First we prepare the cumulative frequency table, then the accumulation frequencies space plotted against the upper or lower boundaries of the corresponding class intervals. By involvement the point out the curve so acquired is dubbed a accumulation frequency curve or ogive.There room two species of ogives :

less than ogive : Plot the points with the upper limits of the course as abscissae and the corresponding less 보다 cumulative frequencies as ordinates. The points room joined by free hand smooth curve to provide less 보다 cumulative frequency curve or the much less than Ogive. It is a increasing curve.

You are watching: An ogive is also called a cumulative frequency graph

better than ogive : Plot the points v the lower boundaries of the classes together abscissa and also the corresponding Greater 보다 cumulative frequencies as ordinates. Sign up with the points by a totally free hand smooth curve to get the “More 보다 Ogive”. That is a fall curve.

When the points obtained are join by straight lines, the picture obtained is dubbed cumulative frequency polygon.Less than ogive method:To build a accumulation frequency polygon and an ogive by less than method, we usage the complying with algorithm.AlgorithmStep 1 : begin with the upper boundaries of class intervals and include class frequencies to achieve the accumulation frequency distribution.Step 2 : note upper class borders along X-axis ~ above a an ideal scale.Step 3 : note cumulative frequencies follow me Y-axis ~ above a suitable scale.Step 4 : Plot the points (xi, fi) wherein xi is the upper limit that a class and also fi is corresponding cumulative frequency.Step 5 : join the points acquired in step 4 through a cost-free hand smooth curve to gain the ogive and to obtain the accumulation frequency polygon sign up with the points acquired in action 4 by line segments.

More than ogive method:To construct a accumulation frequency polygon and an ogive by an ext than method, we usage the complying with algorithm.AlgorithmStep 1 : begin with the lower borders of the course intervals and also from the complete frequencysubtract the frequency of each course to acquire the cumulative frequency distribution.Step 2 : mark the reduced class borders along X-axis on a sutiable scale.Step 3 : Mark the cumulative frequencies along Y-axis ~ above a suitable scale.Step 4 : Plot the points (xi, fi) where xi is the lower limit of a class and also fi is matching cumulative frequency.Step 5 : join the points acquired in action 4 by a cost-free hand smooth curve to obtain the ogive and to obtain the accumulation frequency polygon sign up with these clues by heat segments

## Cumulative Frequency Curve or the Ogive Example problems with Solutions

Example 1: draw a less than ogive for the adhering to frequency distribution :

 I.Q. Frequency 60 – 70 2 70 – 80 5 80 –90 12 90 – 100 31 100 – 110 39 110 – 120 10 120 – 130 4

Find the average from the curve.Solution: Let united state prepare following table reflecting the accumulation frequencies much more than the top limit.

 Class term (I. Q) Frequency (f) Cumulative frequency 60 – 70 2 2 70 – 80 5 2 + 5 = 7 80 –90 12 2 + 5 + 12 = 19 90 – 100 31 2 + 5 + 12 + 31 = 50 100 – 110 39 2 + 5 + 12 + 31 + 39 = 89 110 – 120 10 2 + 5 + 12 + 31 + 39 + 10 = 99 120 – 130 4 2 + 5 + 12 + 31 + 39 + 10 + 4 = 103

Less 보다 ogive :I.Q. Is handled the x-axis. Number of students are significant on y-axis.Points (70, 2), (80, 7), (90, 19), (100, 50), (110, 89), (120, 99), (130, 103), are plotted ~ above graph document and these points space joined by totally free hand. The curve obtained is less than ogive.

The value $$\fracN2$$ = 51.5 is significant on y-axis and also from this point a heat parallel come x-axis is drawn. This heat meets the curve at a allude P. From P attract a perpendicular PN to meet x-axis at N. N represents the median.Here mean is 100.5.Hence, the typical of offered frequency distribution is 100.5

Example 2: The adhering to table mirrors the everyday sales that 230 footpath sellers that Chandni Chowk.

 Sales in Rs. No. Of sellers 0 – 500 12 500 – 1000 18 1000 – 1500 35 1500 – 2000 42 2000 – 2500 50 2500 – 3000 45 3000 – 3500 20 3500 – 4000 8

Locate the average of the over data using only the much less than kind ogive.Solution: To draw ogive, we need to have a accumulation frequency distribution.

 Sales in Rs. No. That sellers Less than type cumulative frequency 0 – 500 12 12 500 – 1000 18 30 1000 – 1500 35 65 1500 – 2000 42 107 2000 – 2500 50 157 2500 – 3000 45 202 3000 – 3500 20 222 3500 – 4000 8 230

Less than ogive : Seles in Rs. Are taken on the y-axis and number of sellers are taken top top x-axis. For drawing less than ogive, clues (500, 12), (1000, 30), (1500, 65), (2000, 107), (2500, 157), (3000, 202), (3500, 222), (4000, 230) are plotted top top graph file and these are joined free hand to acquire the much less than ogive.

The worth $$\fracN2$$ = 115 is significant on y-axis and a heat parallel come x-axis is drawn. This heat meets the curve at a point P. Indigenous P draw a perpendicular PN to meet x-axis at median. Typical = 2000.Hence, the typical of offered frequency circulation is 2000.

Example 3: draw the 2 ogives for the adhering to frequency distribution of the weekly earnings of (less than and much more than) number of workers.

 Weekly wages Number the workers 0 – 20 41 20 – 40 51 40 – 60 64 60 – 80 38 80 – 100 7

Hence uncover the worth of median.Solution:

 Weekly wages Number the workers C.F (less than) C.F (More than) 0 – 20 41 41 201 20 – 40 51 92 160 40 – 60 64 156 109 60 – 80 38 194 45 80 – 100 7 201 7

Less than curve : Upper borders of course intervals are marked on the x-axis and less than type cumulative frequencies room taken on y-axis. For illustration less than kind curve, points (20, 41), (40, 92), (60, 156), (80, 194), (100, 201) are plotted top top the graph paper and these are joined by complimentary hand to achieve the much less than ogive.

Greater 보다 ogiveLower boundaries of course interval are marked on x-axis and greater than kind cumulative frequencies space taken on y-axis. For drawing greater than form curve, point out (0, 201), (20, 160), (40, 109), (60, 45) and also (80, 7) are plotted ~ above the graph file and these are joined by complimentary hand to achieve the higher than type ogive. From the suggest of intersection of these curves a perpendicular heat on x-axis is drawn. The allude at i beg your pardon this line meets x-axis determines the median. Here the median is 42.652.

Example 4: adhering to table offers the accumulation frequency the the age of a group of 199 teachers.Draw the much less than ogive and greater 보다 ogive and find the median.

 Age in years Cum. Frequency 20 – 25 21 25 – 30 40 30 – 35 90 35 – 40 130 40 – 45 146 45 – 50 166 50 – 55 176 55 – 60 186 60 – 65 195 65 – 70 199

Solution:

 Age in years Less than cumulative frequency Frequency Greater 보다 type 20 – 25 21 21 199 25 – 30 40 19 178 30 – 35 90 50 159 35 – 40 130 40 109 40 – 45 146 16 69 45 – 50 166 20 53 50 – 55 176 10 33 55 – 60 186 10 23 60 – 65 195 9 13 65 – 70 199 4 4

Find the end the frequencies by subtracting previous frequency native the following frequency to get an easy frequency. Currently we deserve to prepare the better than form frequency. Periods are tackled x-axis and number of teachers on y-axis.Less 보다 ogive :Plot 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 cost-free hand to obtain less than ogive.Greater than ogive : Plot the clues (20, 199), (25, 178), (30, 159), (35, 109), (40, 69), (45, 53), (50, 33), (55, 23), (60, 13), (65, 4) top top graph paper. Join these point out freehand to acquire greater 보다 ogive. Typical is the point of intersection that these two curves.

Here typical is 37.375.

Example 5: adhering to is the age circulation of a team of students. Attract the cumulative frequency polygon, accumulation frequency curve (less 보다 type) and hence obtain the typical value.

 Age Frequency 5 – 6 40 6 – 7 56 7 – 8 60 8 – 9 66 9 – 10 84 10 – 11 96 11 – 12 92 12 – 13 80 13 – 14 64 14 – 15 44 15 – 16 20 16 – 17 8

Solution: We first prepare the cumulative frequency table by less then method as given listed below :

 Age Frequency Age much less than Cumulative frequency 5 – 6 40 6 40 6 – 7 56 7 96 7 – 8 60 8 156 8 – 9 66 9 222 9 – 10 84 10 306 10 – 11 96 11 402 11 – 12 92 12 494 12 – 13 80 13 574 13 – 14 64 14 638 14 – 15 44 15 682 15 – 16 20 16 702 16 – 17 8 17 710

Other than the given course intervals, we assume a class 4-5 before the very first class interval 5-6 v zero frequency.Now, we mark the upper class boundaries (including the imagined class) follow me X-axis ~ above a an ideal scale and also the cumulative frequencies along Y-axis top top a an ideal scale.Thus, we plot the points (5, 0), (6, 40), (7, 96), (8, 156), (9, 222), (10, 306), (11, 402), (12, 494), (13, 574), (14, 638), (15, 682), (16, 702) and (17, 710).These clues are significant and join by heat segments to acquire the accumulation frequency polygon displayed in Fig.

In order to acquire the cumulative frequency curve, we draw a smooth curve passing through the points debated above. The graph (fig) shows the total variety of students together 710. The median is the age matching to $$\fracN2\,\, = \,\,\frac7102$$ = 355 students. In stimulate to find the median, we an initial located the point corresponding come 355th college student on Y-axis. Allow the point be P. From this allude draw a heat parallel come the X-axis cutting the curve in ~ Q. Indigenous this point Q attract a heat parallel to Y-axis and also meeting X-axis in ~ the suggest M. The x-coordinate of M is 10.5 (See Fig.). Hence, typical is 10.5.

Example 6: The following observations relate come the height of a group of persons. Attract the two type of accumulation frequency polygons and cumulative frequency curves and also determine the median.

 Height in cms 140–143 143–146 146–149 149–152 152–155 155–158 158–161 Frequency 3 9 26 31 45 64 78 Height in cms 161–164 164–167 167–170 170–173 173–176 176–179 179–182 Frequency 85 96 72 60 43 20 6

Solution: Less than an approach : We an initial prepare the accumulation frequency table by much less than an approach as given below :

 Height in cms Frequency Height less than Frequency 140–143 3 143 3 143–146 9 146 12 146–149 26 149 38 149–152 31 152 69 152–155 45 155 114 155–158 64 158 178 158–161 78 161 256 161–164 85 164 341 164–167 96 167 437 167–170 72 170 509 170–173 60 173 569 173–176 43 176 612 176–179 20 179 632 179–182 6 182 638

Other than the given class intervals, us assume a class interval 137-140 before the first class term 140-143 with zero frequency.Now, we note the top class borders on X-axis and also cumulative frequency follow me Y-axis top top a suitable scale.We plot the clues (140, 0), (143, 3), (146, 12), (149, 38), (152, 69), (155, 114), (158, 178), (161, 256),(164, 341), (167, 437), (170, 509), (173, 569), (176, 612),.(179, 632) and 182, 638).

These points room joined by heat segments to obtain the accumulation frequency polygon as presented in fig. And by a free hand smooth curve to attain an ogive by much less than technique as shown in fig.

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More than technique : we prepare the accumulation frequency table by an ext than method as given below :Other than the given course intervals, we assume the class interval 182-185 with zero frequency.Now, we mark the lower class limits on X-axis and the cumulative frequencies follow me Y-axis on suitable scales come plot the point out (140, 638), (143, 635), (146, 626), (149, 600), (152, 569), (155, 524), (158, 460), (161, 382), (164, 297), (167, 201), (170, 129), (173, 69), (176, 26) and also (179, 6). By joining this points by heat segments, we attain the an ext than kind frequency polygon as displayed in fig. By joining this points by a free hand curve, we obtain more than type cumulative frequency curve as points by a complimentary hand curve, we obtain more than kind cumulative frequency curve as presented in fig.We find that the two types of accumulation frequency curves crossing at point P. From suggest P perpendicular afternoon is attracted on X-axis. The value of height equivalent to M is 163.2 cm. Hence, mean is 163.2 cm.