{"id":12361,"date":"2022-10-22T23:10:08","date_gmt":"2022-10-22T17:40:08","guid":{"rendered":"https:\/\/mathemerize.com\/?p=12361"},"modified":"2022-10-22T23:10:13","modified_gmt":"2022-10-22T17:40:13","slug":"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","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/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\/","title":{"rendered":"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 :"},"content":{"rendered":"\n
Number of cars<\/td> | 0 – 10<\/td> | 10 – 20<\/td> | 20 – 30<\/td> | 30 – 40<\/td> | 40 – 50<\/td> | 50 – 60<\/td> | 60 – 70<\/td> | 70 – 80<\/td><\/tr> |
Frequency<\/td> | 7<\/td> | 14<\/td> | 13<\/td> | 12<\/td> | 20<\/td> | 11<\/td> | 15<\/td> | 8<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\nSolution :<\/h2>\n\n\n\nThe class (40 – 50) has the maximum frequency frequency. Therefore, this is the modal class.<\/p>\n\n\n\n Here, l = 40, h = 10, \\(f_1\\) = 20, \\(f_0\\) = 12 and \\(f_2\\) = 11<\/p>\n\n\n\n Now, let us substitute these values in the formula <\/p>\n\n\n\n Mode = l + (\\(f_1 – f_0\\over 2f_1 – f_0 – f_2\\)) \\(\\times\\) h<\/p>\n\n\n\n Mode = 40 + (\\(20 – 12\\over 40 – 12 – 11\\)) \\(\\times\\) 10<\/p>\n\n\n\n Mode = 40 + 4.7 = 44.7<\/p>\n\n\n\n Hence, mode = 44.7 cars<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":" Question : Number of cars 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 Frequency 7 14 13 12 20 11 15 8 Solution : The class (40 – 50) has the maximum frequency frequency. Therefore, this is the modal …<\/p>\n |