{"id":6993,"date":"2021-10-22T14:30:45","date_gmt":"2021-10-22T09:00:45","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6993"},"modified":"2021-10-25T09:22:24","modified_gmt":"2021-10-25T03:52:24","slug":"suppose-a-population-a-has-100-observation-101-102-200-and-another-population-b-has-100-observations-151-152-250-if-v_a-and-v_b-represent-the-variance-of-the-two-popula","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/suppose-a-population-a-has-100-observation-101-102-200-and-another-population-b-has-100-observations-151-152-250-if-v_a-and-v_b-represent-the-variance-of-the-two-popula\/","title":{"rendered":"Suppose a population A has 100 observation 101, 102, ….. , 200 and another population B has 100 observations 151, 152, …. , 250. If \\(V_A\\) and \\(V_B\\) represent the variance of the two populations respectively, then \\(V_A\\over V_B\\) is"},"content":{"rendered":"
Since variance is independent of change of origin.<\/p>\n
Therefore, variance of observations 101, 102, …. , 200 is same as variance of 151, 152, ….. 250.<\/p>\n
\\(\\therefore\\)\u00a0 \\(V_A\\) = \\(V_B\\)<\/p>\n
\\(\\implies\\)\u00a0 \u00a0\\(V_A\\over V_B\\) = 1<\/p>\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