{"id":6999,"date":"2021-10-22T14:37:42","date_gmt":"2021-10-22T09:07:42","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6999"},"modified":"2021-10-25T09:21:49","modified_gmt":"2021-10-25T03:51:49","slug":"let-x_1-x_2-x_n-be-n-observations-such-that-sumx_i2-400-and-sumx_i-80-then-a-possible-value-of-among-the-following-is","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/let-x_1-x_2-x_n-be-n-observations-such-that-sumx_i2-400-and-sumx_i-80-then-a-possible-value-of-among-the-following-is\/","title":{"rendered":"Let \\(x_1\\), \\(x_2\\), ….. , \\(x_n\\), be n observations such that \\(\\sum{x_i}^2\\) = 400 and \\(\\sum{x_i}\\) = 80. Then, a possible value of among the following is"},"content":{"rendered":"

Question :<\/h2>\n

Let \\(x_1\\), \\(x_2\\), ….. , \\(x_n\\), be n observations such that \\(\\sum{x_i}^2\\) = 400 and \\(\\sum{x_i}\\) = 80. Then, a possible value of among the following is<\/p>\n

(a) 12<\/p>\n

(b) 9<\/p>\n

(c) 18<\/p>\n

(d) 15<\/p>\n

Solution :<\/h2>\n

Given \u00a0\\(\\sum{x_i}^2\\) = 400 and \\(\\sum{x_i}\\) = 80<\/p>\n

\\(\\because\\) \\(\\sigma^2\\) \\(\\ge\\) 0<\/p>\n

\\(\\therefore\\) \u00a0\\(\\sum{x_i}^2\\over n\\) – \\(({\\sum{x_i}\\over n})^2\\) \\(\\ge\\) 0<\/p>\n

\\(\\implies\\) \\(400\\over n\\) – \\(6400\\over n^2\\) \\(\\ge\\) 0<\/p>\n

\\(\\therefore\\)\u00a0 \u00a0n \\(\\ge\\) 16<\/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

A scientist is weighing each of 30 fishes. Their mean weight worked out is 30 g and a standard deviation of 2 g. Later, it was found that the measuring scale was misaligned and always under reported every fish weight by 2 g. The correct mean and standard deviation in gram of fishes are respectively.<\/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 mean of the numbers a, b, 8, 5, 10 is 6 and the variance is 6.80. Then, find the values of a and b?<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

Question : Let \\(x_1\\), \\(x_2\\), ….. , \\(x_n\\), be n observations such that \\(\\sum{x_i}^2\\) = 400 and \\(\\sum{x_i}\\) = 80. Then, a possible value of among the following is (a) 12 (b) 9 (c) 18 (d) 15 Solution : Given \u00a0\\(\\sum{x_i}^2\\) = 400 and \\(\\sum{x_i}\\) = 80 \\(\\because\\) \\(\\sigma^2\\) \\(\\ge\\) 0 \\(\\therefore\\) \u00a0\\(\\sum{x_i}^2\\over n\\) – …<\/p>\n

Let \\(x_1\\), \\(x_2\\), ….. , \\(x_n\\), be n observations such that \\(\\sum{x_i}^2\\) = 400 and \\(\\sum{x_i}\\) = 80. Then, a possible value of among the following is<\/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":"\nLet \\(x_1\\), \\(x_2\\), ..... , \\(x_n\\), be n observations such that \\(\\sum{x_i}^2\\) = 400 and \\(\\sum{x_i}\\) = 80. 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Then, a possible value of among the following is (a) 12 (b) 9 (c) 18 (d) 15 Solution : Given \u00a0(sum{x_i}^2) = 400 and (sum{x_i}) = 80 (because) (sigma^2) (ge) 0 (therefore) \u00a0(sum{x_i}^2over n) – … Let (x_1), (x_2), ….. , (x_n), be n observations such that (sum{x_i}^2) = 400 and (sum{x_i}) = 80. 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