{"id":6875,"date":"2021-10-21T21:35:59","date_gmt":"2021-10-21T16:05:59","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6875"},"modified":"2021-10-25T09:35:31","modified_gmt":"2021-10-25T04:05:31","slug":"if-mean-of-the-series-x_1-x2-x_n-is-barx-then-the-mean-of-the-series-x_i-2i-i-1-2-n-will-be","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/if-mean-of-the-series-x_1-x2-x_n-is-barx-then-the-mean-of-the-series-x_i-2i-i-1-2-n-will-be\/","title":{"rendered":"If mean of the series \\(x_1\\), \\(x^2\\), ….. , \\(x_n\\) is \\(\\bar{x}\\), then the mean of the series \\(x_i\\) + 2i, i = 1, 2, ……, n will be"},"content":{"rendered":"

Solution :<\/h2>\n

As given \\(\\bar{x}\\) = \\(x_1 + x_2 + …. + x_n\\over n\\)<\/p>\n

If the mean of the series \\(x_i\\) + 2i, i = 1, 2, ….., n be \\(\\bar{X}\\), then<\/p>\n

\\(\\bar{X}\\) = \\((x_1+2) + (x_2+2.2) + (x_3+2.3) + …. + (x_n + 2.n)\\over n\\)<\/p>\n

= \\(x_1 + x_2 + …. + x_n\\over n\\) + \\(2(1+2+3+….+n)\\over n\\)<\/p>\n

= \\(\\bar{x}\\) + \\(2n(n+1)\\over 2n\\)<\/p>\n

= \\(\\bar{x}\\) + n + 1.<\/p>\n

\n

Similar Questions<\/h3>\n

The mean and variance of a random variable X having a binomial distribution are 4 and 2 respectively, then P(X = 1) is<\/a><\/p>\n

The median of a set of 9 distinct observations is 20.5. If each of the largest 4 observation of the set is increased by 2, then the median of the new is<\/a><\/p>\n

The mean and the variance of a binomial distribution are 4 and 2, respectively. Then, the probability of 2 success is<\/a><\/p>\n

In a series of 2n observations, half of them equals a and remaining half equal -a. If the standard deviation of the observation is 2, then |a| equal to<\/a><\/p>\n

The average marks of boys in a class is 52 and that of girls is 42. The average marks of boys and girls combined is 50. The percentage of boys in the class is<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

Solution : As given \\(\\bar{x}\\) = \\(x_1 + x_2 + …. + x_n\\over n\\) If the mean of the series \\(x_i\\) + 2i, i = 1, 2, ….., n be \\(\\bar{X}\\), then \\(\\bar{X}\\) = \\((x_1+2) + (x_2+2.2) + (x_3+2.3) + …. + (x_n + 2.n)\\over n\\) = \\(x_1 + x_2 + …. + x_n\\over n\\) …<\/p>\n

If mean of the series \\(x_1\\), \\(x^2\\), ….. , \\(x_n\\) is \\(\\bar{x}\\), then the mean of the series \\(x_i\\) + 2i, i = 1, 2, ……, n will be<\/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":[43,58],"tags":[],"yoast_head":"\nIf mean of the series \\(x_1\\), \\(x^2\\), ..... , \\(x_n\\) is \\(\\bar{x}\\), then the mean of the series \\(x_i\\) + 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i = 1, 2, ……, n will be Read More »\" \/>\n<meta property=\"og:url\" content=\"https:\/\/mathemerize.com\/if-mean-of-the-series-x_1-x2-x_n-is-barx-then-the-mean-of-the-series-x_i-2i-i-1-2-n-will-be\/\" \/>\n<meta property=\"og:site_name\" content=\"Mathemerize\" \/>\n<meta property=\"article:published_time\" content=\"2021-10-21T16:05:59+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2021-10-25T04:05:31+00:00\" \/>\n<meta name=\"author\" content=\"mathemerize\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"mathemerize\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"1 minute\" \/>\n<script type=\"application\/ld+json\" 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