Here you will learn what is the formula for mean of grouped and ungrouped data and how to find mean with examples.<\/p>\n
Let’s begin –<\/p>\n
The formula for mean median and mode for grouped and ungrouped frequency distribution is given below.<\/p>\n
(i) For ungrouped distribution :\u00a0<\/strong>If \\(x_1\\), \\(x_2\\), …… \\(x_n\\) are n values of variate \\(x_i\\) then their mean \\(\\bar{x}\\) is defined as<\/p>\n \\(\\bar{x}\\) = \\(x_1 + x_2, …… + x_n \\over n\\) = \\({\\sum_{i=1}^{n}x_i}\\over n\\)<\/p>\n \\(\\implies\\) \\(\\sum x_i\\) = n\\(\\bar{x}\\)<\/p>\n<\/blockquote>\n Example<\/strong><\/span> : Neeta and her four friends secured 65, 78, 82, 94 and 71 marks in a test of mathematics. Find the average (arithmetic mean) of their marks.<\/p>\n Solution<\/strong><\/span> : Arithmetic mean or average =\u00a0 \\(65 + 78 + 82 + 94 + 71\\over 5\\) = \\(390\\over 5\\) = 78<\/p>\n Hence, arithmetic mean = 78<\/p>\n (ii) For ungrouped and grouped frequency distribution :<\/strong><\/p>\n If \\(x_1\\), \\(x_2\\), …… \\(x_n\\) are values of variate with corresponding frequencies \\(f_1\\), \\(f_2\\), …… \\(f_n\\), theb their A.M. is given by<\/p>\n \\(\\bar{x}\\) = \\(f_1x_1 + f_2x_2 + …… + f_nx_n \\over f_1 + f_2 + …… + f_n\\) = \\({\\sum_{i=1}^{n}f_ix_i}\\over N\\), where N = \\({\\sum_{i=1}^{n}f_i}\\)<\/p>\n<\/blockquote>\n Example : <\/span>Find the mean of the following freq. dist.\n\t\t\t <\/p>\n Solution : <\/span>Here N = \\(\\sum f_i\\) = 4 + 5 + 6 + 10 + 20 = 45<\/p>\n \t\t\t \\(\\sum f_ix_i\\) = 606<\/p>\n \t\t\t \\(\\therefore\\) \\(\\bar{x}\\) = \\(\\sum f_ix_i\\over N\\) = \\(606\\over 45\\) = 13.47<\/p>\n If the value of \\(x_i\\) are large, then calculation of A.M. by using mean formula is quite tedious and time consuming. In such case we take deviation of variate from an arbitrary point a.<\/p>\n Let\u00a0 \u00a0 \u00a0 \\(d_i\\) = \\(x_i\\) – a<\/p>\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.\n\t\t\t <\/p>\n Solution : <\/span>Let assumed mean a = 175<\/p>\n \t\t\tNow, a = 175 and N = \\(\\sum f_i\\) = 180<\/p>\n \t\t\t \\(\\therefore\\) \\(\\bar{x}\\) = a + (\\(\\sum f_id_i\\over N\\)) = 175 + \\((-4750)\\over 180\\) = 175 – 26.39 = 148.61<\/p>\n Sometime 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\u00a0 \u00a0\\(u_i\\) = \\(d_i\\over h\\) = \\(x_i – a\\over h\\),\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 where a is assumed mean.<\/p>\n \\(\\therefore\\)\u00a0 \\(\\bar{x}\\) = a + (\\(\\sum f_iu_i\\over N\\))h<\/p>\n<\/blockquote>\n\n
(a) By Direct Method :<\/strong><\/h4>\n
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\n \\(x_i\\)<\/td>\n 5<\/td>\n 8<\/td>\n 11<\/td>\n 14<\/td>\n 17<\/td>\n<\/tr>\n \n \\(f_i\\)<\/td>\n 4<\/td>\n 5<\/td>\n 6<\/td>\n 10<\/td>\n 20<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n (b) By short method or assumed mean method :<\/strong><\/h4>\n
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\n Class 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 \n\n
\n Class 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 \n 0-50<\/td>\n 25<\/td>\n -150<\/td>\n 17<\/td>\n -2550<\/td>\n<\/tr>\n \n 50-100<\/td>\n 75<\/td>\n -100<\/td>\n 35<\/td>\n -3500<\/td>\n<\/tr>\n \n 100-150<\/td>\n 125<\/td>\n -50<\/td>\n 43<\/td>\n -2150<\/td>\n<\/tr>\n \n 150-200<\/td>\n 175<\/td>\n 0<\/td>\n 40<\/td>\n 0<\/td>\n<\/tr>\n \n 200-250<\/td>\n 225<\/td>\n 50<\/td>\n 21<\/td>\n 1050<\/td>\n<\/tr>\n \n 250-300<\/td>\n 275<\/td>\n 100<\/td>\n 24<\/td>\n 2400<\/td>\n<\/tr>\n \n <\/td>\n <\/td>\n <\/td>\n \\(\\sum f_i\\) = 180<\/td>\n \\(\\sum f_id_i\\) = -4750<\/td>\n<\/tr>\n<\/table>\n (c) By step deviation method :<\/strong><\/h4>\n
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