{"id":6980,"date":"2021-10-22T02:28:51","date_gmt":"2021-10-21T20:58:51","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6980"},"modified":"2021-10-25T09:30:26","modified_gmt":"2021-10-25T04:00:26","slug":"find-the-variance-of-first-n-even-natural-numbers","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/find-the-variance-of-first-n-even-natural-numbers\/","title":{"rendered":"Find the variance of first n even natural numbers ?"},"content":{"rendered":"

Solution :<\/h2>\n

\\(\\therefore\\) Variance = [\\({1\\over n}\\sum{(x_i)^2}\\)] – \\((\\bar{x})^2\\)<\/p>\n

= \\(1\\over n\\)[\\(2^2 + 4^2 + ….. + (2n)^2\\)] – \\((n+1)^2\\)<\/p>\n

= \\(1\\over n\\)\\(2^2 [ 1^2 + 2^2 + ….. + n^2]\\) – \\((n+1)^2\\)<\/p>\n

= \\(4\\over n\\) \\(n(n + 1)(2n + 1)\\over 6\\) – \\((n+1)^2\\)<\/p>\n

= \u00a0\\((n + 1)[(2n + 1) – 3(n + 1)\\over 3\\)<\/p>\n

= \\((n + 1)(n – 1)\\over 3\\) = \\(n^2 – 1\\over 3\\)<\/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 mean and variance of 5 observations of an experiment are 4 and 5.2 respectively. If from these observations three are 1, 2 and 6, then remaining will be<\/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 variance of first 50 even natural numbers is<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

Solution : \\(\\therefore\\) Variance = [\\({1\\over n}\\sum{(x_i)^2}\\)] – \\((\\bar{x})^2\\) = \\(1\\over n\\)[\\(2^2 + 4^2 + ….. + (2n)^2\\)] – \\((n+1)^2\\) = \\(1\\over n\\)\\(2^2 [ 1^2 + 2^2 + ….. + n^2]\\) – \\((n+1)^2\\) = \\(4\\over n\\) \\(n(n + 1)(2n + 1)\\over 6\\) – \\((n+1)^2\\) = \u00a0\\((n + 1)[(2n + 1) – 3(n + 1)\\over 3\\) …<\/p>\n

Find the variance of first n even natural numbers ?<\/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":"\nFind the variance of first n even natural numbers ?<\/title>\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\/find-the-variance-of-first-n-even-natural-numbers\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Find the variance of first n even natural numbers ?\" \/>\n<meta property=\"og:description\" content=\"Solution : (therefore) Variance = [({1over n}sum{(x_i)^2})] – ((bar{x})^2) = (1over n)[(2^2 + 4^2 + ….. + (2n)^2)] – ((n+1)^2) = (1over n)(2^2 [ 1^2 + 2^2 + ….. + n^2]) – ((n+1)^2) = (4over n) (n(n + 1)(2n + 1)over 6) – ((n+1)^2) = \u00a0((n + 1)[(2n + 1) – 3(n + 1)over 3) … Find the variance of first n even natural numbers ? 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