{"id":12359,"date":"2022-10-22T22:59:54","date_gmt":"2022-10-22T17:29:54","guid":{"rendered":"https:\/\/mathemerize.com\/?p=12359"},"modified":"2022-10-22T22:59:59","modified_gmt":"2022-10-22T17:29:59","slug":"the-given-distribution-shows-the-number-of-runs-scored-by-some-batsmen-of-the-world-in-one-day-international-cricket-matches-find-the-mode-of-the-data","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/the-given-distribution-shows-the-number-of-runs-scored-by-some-batsmen-of-the-world-in-one-day-international-cricket-matches-find-the-mode-of-the-data\/","title":{"rendered":"The given distribution shows the number of runs scored by some batsmen of the world in one-day international cricket matches. Find the mode of the data."},"content":{"rendered":"\n
Runs scored<\/td> | Number of batsmen<\/td><\/tr> |
3000 – 4000<\/td> | 4<\/td><\/tr> |
4000 – 5000<\/td> | 18<\/td><\/tr> |
5000 – 6000<\/td> | 9<\/td><\/tr> |
6000 – 7000<\/td> | 7<\/td><\/tr> |
7000 – 8000<\/td> | 6<\/td><\/tr> |
8000 – 9000<\/td> | 3<\/td><\/tr> |
9000 – 10000<\/td> | 1<\/td><\/tr> |
10000 – 11000<\/td> | 1<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\nSolution :<\/h2>\n\n\n\nThe class (4000 – 5000) has the maximum frequency. Therefore, this is the modal class.<\/p>\n\n\n\n Here, l = 4000, h = 1000, \\(f_1\\) = 18, \\(f_0\\) = 4 and \\(f_2\\) = 9<\/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 = 4000 + \\(18 – 4\\over 36 – 4 – 9\\)\\(\\times\\) 1000 <\/p>\n\n\n\n Mode = 4000 + 608.695 = 4608.695<\/p>\n\n\n\n Thus, the mode of the given data is 4608.7<\/strong> runs.<\/p>\n","protected":false},"excerpt":{"rendered":" Question : Runs scored Number of batsmen 3000 – 4000 4 4000 – 5000 18 5000 – 6000 9 6000 – 7000 7 7000 – 8000 6 8000 – 9000 3 9000 – 10000 1 10000 – 11000 1 Solution : The class (4000 – 5000) has the maximum frequency. Therefore, this is the modal …<\/p>\n |