{"id":7010,"date":"2021-10-22T14:46:01","date_gmt":"2021-10-22T09:16:01","guid":{"rendered":"https:\/\/mathemerize.com\/?p=7010"},"modified":"2021-10-25T09:37:04","modified_gmt":"2021-10-25T04:07:04","slug":"the-mean-and-the-variance-of-a-binomial-distribution-are-4-and-2-respectively-then-the-probability-of-2-success-is","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/the-mean-and-the-variance-of-a-binomial-distribution-are-4-and-2-respectively-then-the-probability-of-2-success-is\/","title":{"rendered":"The mean and the variance of a binomial distribution are 4 and 2, respectively. Then, the probability of 2 success is"},"content":{"rendered":"

Solution :<\/h2>\n

Given that, mean = 4 \\(\\implies\\) np = 4<\/p>\n

And Variance = 2 \\(\\implies\\) npq = 2 \\(\\implies\\) 4q = 2<\/p>\n

\\(\\implies\\)\u00a0 q = \\(1\\over 2\\)<\/p>\n

\\(\\therefore\\)\u00a0 \u00a0p = 1 – q = 1 – \\(1\\over 2\\) = \\(1\\over 2\\)<\/p>\n

Also, n = 8<\/p>\n

Probability of 2 successes = P(X = 2) = \\(^8C_2\\)\\(p^2\\)\\(q^6\\)<\/p>\n

= \\(8!\\over {2!\\times 6!}\\) \\(\\times\\) \\(({1\\over 2})^2\\) \\(\\times\\) \\(({1\\over 2})^2\\)<\/p>\n

= \\(28\\over 256\\)<\/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

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Suppose a population A has 100 observation 101, 102, \u2026.. , 200 and another population B has 100 observations 151, 152, \u2026. , 250. If \\(V_A\\) and \\(V_B\\) represent the variance of the two populations respectively, then \\(V_A\\over V_B\\) is<\/a><\/p>\n

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Solution : Given that, mean = 4 \\(\\implies\\) np = 4 And Variance = 2 \\(\\implies\\) npq = 2 \\(\\implies\\) 4q = 2 \\(\\implies\\)\u00a0 q = \\(1\\over 2\\) \\(\\therefore\\)\u00a0 \u00a0p = 1 – q = 1 – \\(1\\over 2\\) = \\(1\\over 2\\) Also, n = 8 Probability of 2 successes = P(X = 2) = …<\/p>\n

The mean and the variance of a binomial distribution are 4 and 2, respectively. Then, the probability of 2 success 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":"\nThe mean and the variance of a binomial distribution are 4 and 2, respectively. 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