\n90-100<\/td>\n \n9<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
Solution: \u00a0 \u00a0<\/strong>In Fig. a histogram and a frequency polygon (in dotted lines) are drawn on the same scale. \n <\/p>\nExample 2: \u00a0 \u00a0<\/strong>Construct a frequency polygon for the following data : \n \nSolution: \u00a0 \u00a0<\/strong>First we obtain the class marks as given in the following table. \n \n <\/p>\nExample 3: \u00a0 \u00a0<\/strong>Prepare frequency polygon with the help of histogram from the following data :<\/p>\n\n\n\n\nClasses<\/strong><\/p>\n<\/td>\n0-6<\/td>\n 6-12<\/td>\n 12-18<\/td>\n 18-24<\/td>\n 24-30<\/td>\n 30-36<\/td>\n<\/tr>\n \nFrequencies <\/strong><\/td>\n4<\/td>\n 8<\/td>\n 15<\/td>\n 20<\/td>\n 12<\/td>\n \n6<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
Solution: \u00a0 \u00a0<\/strong>We plot the classes (-6-0), (0-6), (6-12), (12-18), (18-24), (24-30), (30-36) and (36-42) along the x-axis. The frequencies of these classes are respectively, 0, 4, 8, 15, 20, 12, 6 and 0. These frequencies are plotted along the y-axis. \n \nAfter drawing the histogram for the data, we join the mid-points of the top sides of the rectangles of the histogram. The frequency polygon is made with dotted line segments.<\/p>\nExample 4: \u00a0 \u00a0<\/strong>Draw frequency polygon for the data given below, without drawing histogram :<\/p>\n\n\n\n\nClasses<\/strong><\/p>\n<\/td>\n140-150<\/td>\n 150-160<\/td>\n 160-170<\/td>\n 170-180<\/td>\n 180-190<\/td>\n 190-200<\/td>\n<\/tr>\n \nFrequencies <\/strong><\/td>\n5<\/td>\n 10<\/td>\n 20<\/td>\n 9<\/td>\n 6<\/td>\n \n2<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
Solution: \u00a0 \u00a0<\/strong>We have the classes\u00a0130-140, 140-150, 150-160, 160-170, 170-180, 180-190, 190-200, 200-210 having frequencies 0, 5, 10, 20, 9, 6, 2, 0 respectively. \nThe class marks of the classes are \n\\(\\frac { 130+140 }{ 2 } =135,\\frac { 140+150 }{ 2 } =145,….,\\frac { 200+210 }{ 2 } =205\\) \nrespectively. \nWe take class marks along x-axis and the frequencies along y-axis. \nWe plot the points (135, 0), (145, 5), (155, 10), (165, 20), (175, 9), (185, 6), (195, 2) and (205, 0). Now, we joint these points and get the required frequency polygon ABCDEFGH as in figure. \n <\/p>\nExample 5: \u00a0 \u00a0<\/strong>Ages (in years) of the members of two sports clubs were recorded and the data collected is as under.\u00a0 \u00a0\u00a0\u00a0 <\/strong><\/p>\n\n\n\n\nAge (in years)<\/strong><\/p>\n<\/td>\nNumber of members of Club A<\/strong><\/td>\nNumber of members of Club B<\/strong><\/td>\n<\/tr>\n\n\n20-29<\/p>\n
30-39<\/p>\n
40-49<\/p>\n
50-59<\/p>\n
60-69<\/p>\n
70-79<\/p>\n<\/td>\n
5<\/p>\n10<\/p>\n
15<\/p>\n
10<\/p>\n
20<\/p>\n
5<\/td>\n
\n10<\/p>\n
15<\/p>\n
10<\/p>\n
5<\/p>\n
10<\/p>\n
15<\/p>\n<\/td>\n<\/tr>\n
\nTotal<\/td>\n 65<\/td>\n \n65<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
Represent the ages of the members of the sports clubs A and B on the same graph by two frequency polygons. Compare the pattern of the two clubs according to ages of the members. \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/strong><\/p>\nSolution: \u00a0 \u00a0<\/strong>The class marks of the classes\u00a020-29, 30-39, 40-49, 50-59, 60-69, 70-79 are 24.5, 34.5, 44.5, 54.5, 64.5, 74.5 respectively. The corresponding frequencies for clubs A and B are 5, 10, 15, 10, 20, 5 ; 10, 15, 10, 5, 10, 15 respectively. \nFor the club A, the frequency polygon is drawn by joining the points (14.5, 0), (24.5, 5), (34.5, 10), (44.5, 15), (54.5, 10), (64.5, 20), (74.5, 5), (84.5, 0). \nFor the club B, the frequency polygon is drawn by joining the points (14.5, 0), (24.5, 10), (34.5, 15), (44.5, 10), (54.5, 5), (64.5, 10), (74.5, 15,), (84.5, 0). \n \nWe observe that the club B is more youthful in comparison to club A.<\/p>\nExample 6: \u00a0 \u00a0<\/strong>The following table gives the distribution of students of two sections according to the marks obtained by them. \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0 <\/strong><\/p>\n\n\n\n\nSection A<\/strong><\/p>\n<\/td>\nSection B<\/strong><\/td>\n<\/tr>\n\n\nMarks<\/strong><\/p>\n<\/td>\nFrequency<\/strong><\/td>\nMarks<\/strong><\/td>\n\nFrequency<\/strong><\/p>\n<\/td>\n<\/tr>\n\n0-10<\/p>\n10-20<\/p>\n
20-30<\/p>\n
30-40<\/p>\n
40-50<\/td>\n
3<\/p>\n9<\/p>\n
17<\/p>\n
12<\/p>\n
9<\/td>\n
0-10<\/p>\n10-20<\/p>\n
20-30<\/p>\n
30-40<\/p>\n
40-50<\/td>\n
\n5<\/p>\n
19<\/p>\n
15<\/p>\n
10<\/p>\n
1<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
Represent the marks of the students of both the sections on the same graph by two frequency polygons. From the two polygons compare the performance of the two section. \nSolution: \u00a0 \u00a0<\/strong>Class marks for section A are : 5, 15, 25, 35, 45 and corresponding frequencies as 3, 9, 17, 12, 9 respectively. \nIts frequency polygon is the join of the points (by line segments) (-5, 0), (5, 3), (15, 9), (25, 17), (35, 12), (45, 9) and (60, 0). \nSimilarly for the section B, the frequency polygon is the join of the points (-5, 0), \n(5, 5), (15, 19), (25, 15), (35, 10), (45, 1) and (60, 0). \nWe draw the two frequency polygon : \n <\/p>\nExample 7: \u00a0 \u00a0<\/strong>Make a histogram and a frequency polygon from the given data :<\/p>\n\n\n\nMarks<\/strong><\/td>\nNo. of students<\/strong><\/td>\n<\/tr>\n\n30-40<\/td>\n 12<\/td>\n<\/tr>\n \n40-50<\/td>\n 18<\/td>\n<\/tr>\n \n50-60<\/td>\n 26<\/td>\n<\/tr>\n \n60-70<\/td>\n 16<\/td>\n<\/tr>\n \n70-80<\/td>\n 10<\/td>\n<\/tr>\n \n80-90<\/td>\n 6<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nSolution: \u00a0 \u00a0<\/strong>The frequency polygon has been shown by dotted lines \n <\/p>\n","protected":false},"excerpt":{"rendered":"What is the Frequency Polygon A frequency polygon is the polygon obtained by joining the mid-points of upper horizontal sides of all the rectangles in a histogram. Construction of a frequency polygon with Histogram. Obtain the frequency distribution and draw a histogram representing it. Obtain the mid-points of the upper horizontal side of each rectangle. […]<\/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":[1119,1118],"yoast_head":"\n
What is the Frequency Polygon - 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