{"id":6877,"date":"2021-10-21T21:38:20","date_gmt":"2021-10-21T16:08:20","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6877"},"modified":"2021-10-25T09:34:29","modified_gmt":"2021-10-25T04:04:29","slug":"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","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/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\/","title":{"rendered":"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"},"content":{"rendered":"

Solution :<\/h2>\n

As given \\(\\bar{x}\\) = 4, n = 5 and \\({\\sigma}^2\\) = 5.2. If the remaining observations are \\(x_1\\), \\(x_2\\) then<\/p>\n

\\({\\sigma}^2\\) = \\(\\sum{(x_i – \\bar{x})}^2\\over n\\) = 5.2<\/p>\n

\\(\\implies\\) \\({(x_1-4)}^2 + {(x_2-4)}^2 + {(1-4)}^2 + {2-4)}^2 + {(6-4)}^2\\over 5\\) = 5.2<\/p>\n

\\(\\implies\\) \\({(x_1-4)}^2 + {(x_2-4)}^2\\) = 9\u00a0 …..(1)<\/p>\n

Also \\(\\bar{x}\\) = 4 \\(\\implies\\) \\(x_1 + x_2 + 1 + 2 + 6\\over 5\\) = 4 \\(\\implies\\) \\(x_1 + x_2\\) = 11\u00a0 ….(2)<\/p>\n

from eq.(1), (2)\u00a0 \\(x_1\\), \\(x_2\\) = 4, 7<\/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

If the mean deviation about the median of numbers a, 2a, \u2026. , 50a is 50, then |a| is equal to<\/a><\/p>\n

All the students of a class performed poorly in mathematics. The teacher decided to give grace marks of 10 to each of the students. Which statistical measure will not change even after the grace marks were given ?<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

Solution : As given \\(\\bar{x}\\) = 4, n = 5 and \\({\\sigma}^2\\) = 5.2. If the remaining observations are \\(x_1\\), \\(x_2\\) then \\({\\sigma}^2\\) = \\(\\sum{(x_i – \\bar{x})}^2\\over n\\) = 5.2 \\(\\implies\\) \\({(x_1-4)}^2 + {(x_2-4)}^2 + {(1-4)}^2 + {2-4)}^2 + {(6-4)}^2\\over 5\\) = 5.2 \\(\\implies\\) \\({(x_1-4)}^2 + {(x_2-4)}^2\\) = 9\u00a0 …..(1) Also \\(\\bar{x}\\) = 4 \\(\\implies\\) …<\/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<\/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":"\nThe mean and variance of 5 observations of an experiment are 4 and 5.2 respectively. 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If the remaining observations are (x_1), (x_2) then ({sigma}^2) = (sum{(x_i – bar{x})}^2over n) = 5.2 (implies) ({(x_1-4)}^2 + {(x_2-4)}^2 + {(1-4)}^2 + {2-4)}^2 + {(6-4)}^2over 5) = 5.2 (implies) ({(x_1-4)}^2 + {(x_2-4)}^2) = 9\u00a0 …..(1) Also (bar{x}) = 4 (implies) … The mean and variance of 5 observations of an experiment are 4 and 5.2 respectively. 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