{"id":4057,"date":"2021-08-15T16:41:48","date_gmt":"2021-08-15T16:41:48","guid":{"rendered":"https:\/\/mathemerize.com\/?p=4057"},"modified":"2022-02-26T15:29:37","modified_gmt":"2022-02-26T09:59:37","slug":"mean-by-step-deviation-method","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/mean-by-step-deviation-method\/","title":{"rendered":"Mean By Step Deviation Method and By Short Method"},"content":{"rendered":"

Here you will learn how to solve mean by using step deviation method and by short method and properties of mean.<\/p>\n

Let’s begin –<\/p>\n

Mean By Short Method<\/h2>\n

If the value of \\(x_i\\) are large, then calculation of A.M. by using mean<\/a><\/span> 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

\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\n\n
Class Interval<\/td>\n0-50<\/td>\n50-100<\/td>\n100-150<\/td>\n150-200<\/td>\n200-250<\/td>\n250-300<\/td>\n<\/tr>\n
\\(f_i\\)<\/td>\n17<\/td>\n35<\/td>\n43<\/td>\n40<\/td>\n21<\/td>\n24<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Solution : <\/span>Let assumed mean a = 175<\/p>\n\n\n\n\n\n\n\n\n\n\n
Class Interval<\/td>\nmid value \\((x_i)\\)<\/td>\n\\(d_i\\) = \\(x_i – 175\\)<\/td>\nfrequency \\(f_i\\)<\/td>\n\\(f_id_i\\)<\/td>\n<\/tr>\n
0-50<\/td>\n25<\/td>\n-150<\/td>\n17<\/td>\n-2550<\/td>\n<\/tr>\n
50-100<\/td>\n75<\/td>\n-100<\/td>\n35<\/td>\n-3500<\/td>\n<\/tr>\n
100-150<\/td>\n125<\/td>\n-50<\/td>\n43<\/td>\n-2150<\/td>\n<\/tr>\n
150-200<\/td>\n175<\/td>\n0<\/td>\n40<\/td>\n0<\/td>\n<\/tr>\n
200-250<\/td>\n225<\/td>\n50<\/td>\n21<\/td>\n1050<\/td>\n<\/tr>\n
250-300<\/td>\n275<\/td>\n100<\/td>\n24<\/td>\n2400<\/td>\n<\/tr>\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>\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   \\(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 \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
\\(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

Properties of Arithmetic Mean<\/strong><\/h4>\n

(i)  Sum of deviations of variate from their A.M. is always zero i.e. \\(\\sum\\)(\\(x_1 – \\bar{x}\\)) = 0, \\(\\sum\\)\\(f_i\\)(\\(x_1 – \\bar{x}\\)) = 0<\/p>\n

(ii)  Sum of square of deviations of variate from their A.M. is minimum i.e. \\(\\sum\\)(\\(x_1 – \\bar{x})^2\\) is minimum.<\/p>\n

(iii)  A.M. is independent of change of assumed mean i.e. it is not effected by any change in assumed mean.<\/p>\n

<\/p>\n\n\n

\n
Next – What are Measures of Dispersion Statistics ?<\/a><\/div>\n<\/div>\n\n\n\n
\n
Previous – What is the Formula for Mean Median and Mode<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Here you will learn how to solve mean by using step deviation method and by short method and properties of mean. Let’s begin – Mean By Short Method 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 …<\/p>\n

Mean By Step Deviation Method and By Short Method<\/span> Read More »<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-global-header-display":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":""},"categories":[22],"tags":[599,600,601],"yoast_head":"\nMean By Step Deviation Method and By Short Method - Mathemerize<\/title>\n<meta name=\"description\" content=\"In this post you will learn how to solve mean by using step deviation method and by short method and properties of mean.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mathemerize.com\/mean-by-step-deviation-method\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Mean By Step Deviation Method and By Short Method - 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