{"id":4033,"date":"2021-08-14T20:32:21","date_gmt":"2021-08-14T20:32:21","guid":{"rendered":"https:\/\/mathemerize.com\/?p=4033"},"modified":"2021-11-29T18:38:43","modified_gmt":"2021-11-29T13:08:43","slug":"mean-square-deviation-formula","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/mean-square-deviation-formula\/","title":{"rendered":"Mean Square Deviation Formula and Example"},"content":{"rendered":"
Here you will learn mean square deviation formula and relation between mean square deviation and variance with example.<\/p>\n
Let’s begin –<\/p>\n
The mean square deviation of a distribution is the mean of the square of deviations of variate from assumed mean. It is denoted by \\(S^2\\).<\/p>\n
\nHence \\(S^2\\) = \\(\\sum{x_i – a}^2\\over n\\) = \\(\\sum{d_i}^2\\over n\\)\u00a0 \u00a0 \u00a0 (for ungrouped dist.)<\/p>\n
\\(S^2\\) = \\(\\sum{x_i – a}^2\\over N\\) = \\(\\sum{f_id_i}^2\\over N\\)\u00a0 \u00a0 (for frequency dist.),\u00a0 \u00a0 where \\(d_i\\) = \\(x_i – a\\)<\/p>\n<\/blockquote>\n
Relation between variance and mean square deviation<\/h2>\n
\\(\\because\\)\u00a0 \u00a0 \\({\\sigma}^2\\) = \\(\\sum{f_id_i}^2\\over N\\) – \\(({\\sum f_i{d_i}\\over N})^2\\)<\/p>\n
\n\\(\\implies\\)\u00a0 \u00a0 \\({\\sigma}^2\\) = \\(s^2\\) – \\(d^2\\),\u00a0 \u00a0 where d = \\(\\bar{x} – a\\) = \\({\\sum f_i{d_i}\\over N}\\)<\/p>\n
\\(\\implies\\)\u00a0 \u00a0 \\(s^2\\) = \\({\\sigma}^2\\) + \\(d^2\\),\u00a0 \u00a0 \\(\\implies\\) \\(s^2\\) \\(\\geq\\) \\({\\sigma}^2\\)<\/p>\n<\/blockquote>\n
Hence the variance is the minimum value of mean square deviation of a distribution.<\/p>\n\n\n
Example : <\/span>Find the variance of the following freq. dist.\n <\/p>
\n
\n class<\/td>\n 0 – 2<\/td>\n 2 – 4<\/td>\n 4 – 6<\/td>\n 6 – 8<\/td>\n 8 – 10<\/td>\n 10 – 12<\/td>\n <\/tr>\n \n \\(f_i\\)<\/td>\n 2<\/td>\n 7<\/td>\n 12<\/td>\n 19<\/td>\n 9<\/td>\n 1<\/td>\n <\/tr>\n <\/tbody><\/table> <\/p>\n
Solution : <\/span>Let a = 7 and h = 2\n <\/p>
\n
\n class<\/td>\n \\(x_i\\)<\/td>\n \\(f_i\\)<\/td>\n \\(u_i\\) = \\(x_i – a\\over h\\)<\/td>\n \\(f_iu_i\\)<\/td>\n \\(f_iu_i^2\\)<\/td>\n <\/tr>\n \n 0 – 2<\/td>\n 1<\/td>\n 2<\/td>\n -3<\/td>\n -6<\/td>\n 18<\/td>\n <\/tr>\n \n 2 – 4<\/td>\n 3<\/td>\n 7<\/td>\n -2<\/td>\n -14<\/td>\n 28<\/td>\n <\/tr>\n \n 4 – 6<\/td>\n 5<\/td>\n 12<\/td>\n -1<\/td>\n -12<\/td>\n 12<\/td>\n <\/tr>\n \n 6 – 8<\/td>\n 7<\/td>\n 19<\/td>\n 0<\/td>\n 0<\/td>\n 0<\/td>\n <\/tr>\n \n 8 – 10<\/td>\n 9<\/td>\n 9<\/td>\n 1<\/td>\n 9<\/td>\n 9<\/td>\n <\/tr>\n \n 10 – 12<\/td>\n 11<\/td>\n 1<\/td>\n 2<\/td>\n 2<\/td>\n 4<\/td>\n <\/tr>\n \n <\/td>\n <\/td>\n N = 50<\/td>\n <\/td>\n \\(\\sum{f_iu_i}\\) = -21<\/td>\n \\(\\sum{f_iu_i^2}\\) = 71<\/td>\n <\/tr>\n <\/tbody><\/table> <\/p>\n
\\(\\because\\) \\({\\sigma^2}\\) = \\(h^2\\)[\\(\\sum \n f_i{u_i}^2\\over n\\) – \\(({\\sum f_i{u_i}\\over n})^2\\)]
\n = 4[\\(71\\over 50\\) – (\\({-21\\over 50}^2\\))]
\n = 4[1.42 – 0.1764] = 4.97
\n <\/p>\n\n\nMathematical Properties of Variance<\/h2>\n
(i) Var.\\((x_i + p\\)) = Var.(\\(x_i\\))<\/p>\n
(ii) Var.\\((px_i\\)) = \\(p^2\\)Var.(\\(x_i\\))<\/p>\n
(iii) Var\\((ax_i + b\\)) = \\(a^2\\).Var(\\(x_i\\))<\/p>\n
where p, a, b are constants.<\/p>\n\n\n