Here, you are going to learn formula for mean median and mode and how to find mean, median and mode using those formula with examples.<\/p>\n
In statistics, mean<\/strong> is the average of numbers.<\/p>\n The median<\/strong> of a series is the value of middle term of the series when the values are written in ascending order. Therefore median, divided an arranged series into two equal parts.<\/p>\n In a frequency distribution the mode<\/strong> is the value of that values which have the maximum frequency.<\/p>\n The formula for mean median and mode for grouped and ungrouped frequency distribution is given below.<\/p>\n (i) For ungrouped distribution : <\/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 (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\n\n Example : <\/span>Find the mean of the following freq. dist.\n\t\t\t <\/p> <\/p>\n Solution : <\/span>Here N = \\(\\sum f_i\\) = 4 + 5 + 6 + 10 + 20 = 45 (i) For ungrouped distribution : <\/strong>Let n be the number of variate in a series then<\/p>\n Median = \\(({n + 1\\over 2})^{th}\\) term, (when n is odd)<\/p>\n Median = Mean of \\(({n\\over 2})^{th}\\) and \\(({n\\over 2} + 1)^{th}\\) terms, (where n is even)<\/p>\n<\/blockquote>\n (ii) For ungrouped frequency distribution : <\/strong>First we prepare the cumulative frequency(c.f.) column and Find value of N then<\/p>\n Median = \\(({N + 1\\over 2})^{th}\\) term, (when N is odd)<\/p>\n Median = Mean of \\(({N\\over 2})^{th}\\) and \\(({N\\over 2} + 1)^{th}\\) terms, (where n is even)<\/p>\n<\/blockquote>\n (iii) For grouped frequency distribution : <\/strong>Prepare c.f. column and find value of \\(N\\over 2\\) then find the class which contain value of c.f. is equal or just greater to N\/2, this is median class<\/p>\n Median = \\(l\\) + \\(({N\\over 2} – F)\\over f\\)\\(\\times\\)h<\/p>\n where \\(l\\) – lower limit of median class<\/p>\n f – frequency of median class<\/p>\n F – c.f. of the class preceding median class<\/p>\n h – class interval of median class<\/p>\n<\/blockquote>\n\n\n Example : <\/span>Find the median of the following frequency distribution.\n\t\t\t <\/p> <\/p>\n Solution : <\/span><\/p> <\/p>\n\t\t\t\t Here \\(N\\over 2\\) = \\(100\\over 2\\) = 50 which lies in the value of 78 of c.f. hence corresponding class of this c.f. is 20 – 30 is the median class, so (i) For ungrouped distribution : <\/strong>The value of that variate which is repeated maximum number of times.<\/p>\n (ii) For ungrouped distribution : <\/strong>The value of that variate which have maximum frequency.<\/p>\n (iii) For grouped frequency distribution : <\/strong>First we find the class which have maximum frequency, this is model class.<\/p>\n \\(\\therefore\\) Mode = (\\(l\\) + \\(f_0 – f_1\\over {2f_0 – f_1 – f_2}\\))\\(\\times\\)h<\/p>\n where \\(l\\) = lower limit of model class<\/p>\n \\(f_0\\) = freq. of model class<\/p>\n \\(f_1\\) = freq. of the class preceding model class<\/p>\n \\(f_2\\) = freq. of the class succeeding model class<\/p>\n h = class interval of model class<\/p>\n<\/blockquote>\n In a moderately asymmetric distribution the following relation between mean, median and mode of a distribution. It is known as impirical formula.<\/p>\n Mode = 3 Median – 2 Mean<\/p>\n<\/blockquote>\n Hope you learnt what is the formula for mean median and mode, learn more concepts of statistics and practice more questions to get ahead in the competition. Good luck!<\/p>\n\n\nFormula for Mean Median and Mode :<\/h2>\n
Mean : <\/h2>\n
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\n\t\t\t\t \\(x_i\\)<\/td>\n\t\t\t\t 5<\/td>\n\t\t\t\t 8<\/td>\n\t\t\t\t 11<\/td>\n\t\t\t\t 14<\/td>\n\t\t\t\t 17<\/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 4<\/td>\n\t\t\t\t 5<\/td>\n\t\t\t\t 6<\/td>\n\t\t\t\t 10<\/td>\n\t\t\t\t 20<\/td>\n\t\t\t\t <\/tr>\n\t\t\t\t <\/tbody><\/table>
\n \t\t\t \\(\\sum f_ix_i\\) = 606
\n \t\t\t \\(\\therefore\\) \\(\\bar{x}\\) = \\(\\sum f_ix_i\\over N\\) = \\(606\\over 45\\) = 13.47
<\/p>\n\n\nMedian :<\/h2>\n
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\n\t\t\t\t class<\/td>\n\t\t\t\t 0 – 10<\/td>\n\t\t\t\t 10 – 20<\/td>\n\t\t\t\t 20 – 30<\/td>\n\t\t\t\t 30 – 40<\/td>\n\t\t\t\t 40 – 50<\/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 8<\/td>\n\t\t\t\t 30<\/td>\n\t\t\t\t 40<\/td>\n\t\t\t\t 12<\/td>\n\t\t\t\t 10<\/td>\n\t\t\t\t <\/tr>\n\t\t\t\t<\/tbody><\/table> \n\t\t\t\t
\n\t\t\t\t class<\/td>\n\t\t\t\t 0 – 10<\/td>\n\t\t\t\t 10 – 20<\/td>\n\t\t\t\t 20 – 30<\/td>\n\t\t\t\t 30 – 40<\/td>\n\t\t\t\t 40 – 50<\/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 8<\/td>\n\t\t\t\t 30<\/td>\n\t\t\t\t 40<\/td>\n\t\t\t\t 12<\/td>\n\t\t\t\t 10<\/td>\n\t\t\t\t <\/tr>\n\t\t\t\t \n\t\t\t\t c.f.<\/td>\n\t\t\t\t 8<\/td>\n\t\t\t\t 38<\/td>\n\t\t\t\t 78<\/td>\n\t\t\t\t 90<\/td>\n\t\t\t\t 100<\/td>\n\t\t\t\t <\/tr>\n\t\t\t\t<\/tbody><\/table>
\n \\(l\\) = 20, f = 40, f = 38, h = 10
\n\\(\\therefore\\) Median = \\(l\\) + \\(({N\\over 2} – F)\\over f\\)\\(\\times\\)h = \n20 + \\((50 – 38)\\over 40\\)\\(\\times\\)10 = 23
<\/p>\n\n\nMode :<\/h2>\n
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Relationship Between Mean Median and Mode<\/h2>\n
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