{"id":12324,"date":"2022-10-17T21:29:43","date_gmt":"2022-10-17T15:59:43","guid":{"rendered":"https:\/\/mathemerize.com\/?p=12324"},"modified":"2022-10-17T21:29:48","modified_gmt":"2022-10-17T15:59:48","slug":"the-following-data-gives-the-information-on-the-observed-lifetimes-in-hours-of-225-electrical-components-determine-the-modal-lifetime-of-the-components","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/the-following-data-gives-the-information-on-the-observed-lifetimes-in-hours-of-225-electrical-components-determine-the-modal-lifetime-of-the-components\/","title":{"rendered":"The following data gives the information on the observed lifetimes (in hours) of 225 electrical components. Determine the modal lifetime of the components"},"content":{"rendered":"\n
Lifetimes (in hours)<\/td> | 0 – 20<\/td> | 20 – 40<\/td> | 40 – 60<\/td> | 60 – 80<\/td> | 80 – 100<\/td> | 100 – 120<\/td><\/tr> |
Frequency<\/td> | 10<\/td> | 35<\/td> | 52<\/td> | 61<\/td> | 38<\/td> | 29<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\nSolution :<\/h2>\n\n\n\nThe following (60 – 80) has the maximum frequency. Therefore, this is the modal class.<\/p>\n\n\n\n Here, l = 60, h = 20, \\(f_1\\) = 61, \\(f_0\\) = 52 and \\(f_2\\) = 38<\/p>\n\n\n\n Now, let us substitute these value 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 = 60 + (\\(61 – 52\\over 122 – 52 – 38\\))\\(\\times\\) 20<\/p>\n\n\n\n = 60 + \\(9\\over 32\\) \\(\\times\\) 20<\/p>\n\n\n\n = 60 + 5.625 = 65.625<\/p>\n\n\n\n Thus, the modal lifetimes of the components is 65.625<\/strong> hours.<\/p>\n","protected":false},"excerpt":{"rendered":" Question : Lifetimes (in hours) 0 – 20 20 – 40 40 – 60 60 – 80 80 – 100 100 – 120 Frequency 10 35 52 61 38 29 Solution : The following (60 – 80) has the maximum frequency. Therefore, this is the modal class. Here, l = 60, h = 20, \\(f_1\\) …<\/p>\n |