\n\\(\\therefore\\)\u00a0 \u00a0\\(\\bar{x}\\) = a + \\(\\sum f_id_i\\over N\\), where a is assumed mean<\/p>\n<\/blockquote>\n
Example : <\/span>Find the A.M. of the following freq. dist.<\/p>\n\n\n\nClass Interval<\/td>\n | 0-50<\/td>\n | 50-100<\/td>\n | 100-150<\/td>\n | 150-200<\/td>\n | 200-250<\/td>\n | 250-300<\/td>\n<\/tr>\n |
\n\\(f_i\\)<\/td>\n | 17<\/td>\n | 35<\/td>\n | 43<\/td>\n | 40<\/td>\n | 21<\/td>\n | 24<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Solution : <\/span>Let assumed mean a = 175<\/p>\n\n\n\nClass Interval<\/td>\n | mid value \\((x_i)\\)<\/td>\n | \\(d_i\\) = \\(x_i – 175\\)<\/td>\n | frequency \\(f_i\\)<\/td>\n | \\(f_id_i\\)<\/td>\n<\/tr>\n | \n0-50<\/td>\n | 25<\/td>\n | -150<\/td>\n | 17<\/td>\n | -2550<\/td>\n<\/tr>\n | \n50-100<\/td>\n | 75<\/td>\n | -100<\/td>\n | 35<\/td>\n | -3500<\/td>\n<\/tr>\n | \n100-150<\/td>\n | 125<\/td>\n | -50<\/td>\n | 43<\/td>\n | -2150<\/td>\n<\/tr>\n | \n150-200<\/td>\n | 175<\/td>\n | 0<\/td>\n | 40<\/td>\n | 0<\/td>\n<\/tr>\n | \n200-250<\/td>\n | 225<\/td>\n | 50<\/td>\n | 21<\/td>\n | 1050<\/td>\n<\/tr>\n | \n250-300<\/td>\n | 275<\/td>\n | 100<\/td>\n | 24<\/td>\n | 2400<\/td>\n<\/tr>\n | \n\u00a0<\/td>\n | \u00a0<\/td>\n | \u00a0<\/td>\n | \\(\\sum f_i\\) = 180<\/td>\n | \\(\\sum f_id_i\\) = -4750<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Now, a = 175 and N = \\(\\sum f_i\\) = 180<\/p>\n \\(\\therefore\\) \u00a0 \\(\\bar{x}\\) = a + (\\(\\sum f_id_i\\over N\\)) = 175 + \\((-4750)\\over 180\\) = 175 – 26.39 = 148.61<\/p>\n\n Mean By Step Deviation Method<\/h2>\nSometime during the application of short method (given above) of finding the A.M. If each deviation \\(d_i\\) are divisible by a common number h(let)<\/p>\n Let \\(u_i\\) = \\(d_i\\over h\\) = \\(x_i – a\\over h\\), where a is assumed mean.<\/p>\n \n\\(\\therefore\\) \\(\\bar{x}\\) = a + (\\(\\sum f_iu_i\\over N\\))h<\/p>\n<\/blockquote>\n\n\n Example : <\/span>Find the A.M. of the following freq. dist.\n\t\t\t <\/p>\n\t\t\t\t \n\t\t\t\t \\(x_i\\)<\/td>\n\t\t\t\t | 5<\/td>\n\t\t\t\t | 15<\/td>\n\t\t\t\t | 25<\/td>\n\t\t\t\t | 35<\/td>\n\t\t\t\t | 45<\/td>\n\t\t\t\t | 55<\/td>\n\t\t\t\t <\/tr>\n\t\t\t\t | \n\t\t\t\t \\(f_i\\)<\/td>\n\t\t\t\t | 12<\/td>\n\t\t\t\t | 18<\/td>\n\t\t\t\t | 27<\/td>\n\t\t\t\t | 20<\/td>\n\t\t\t\t | 17<\/td>\n\t\t\t\t | 6<\/td>\n\t\t\t\t <\/tr>\n\t\t\t\t<\/tbody><\/table> <\/p>\n Solution : <\/span>Let assumed mean a = 35, h = 10
\n \t\t\t here N = \\(\\sum f_i\\) = 100, \\(u_i\\) = \\(x_i – 35\\over 10\\)
\n \t\t\t \\(\\sum f_iu_i\\) = (12\\(\\times\\)-3) + (18\\(\\times\\)-2) + (27\\(\\times\\)-1) + (20\\(\\times\\)0) + (17\\(\\times\\)1) + (6\\(\\times\\)2)\n \t\t\t = -70
\n \t\t\t \\(\\therefore\\) \\(\\bar{x}\\) = a + (\\(\\sum f_iu_i\\over N\\))h = 35 + \\((-70)\\over 100\\)\\(\\times\\)10 = 28
<\/p>\n\n\n | | |