{"id":7002,"date":"2021-10-22T14:40:28","date_gmt":"2021-10-22T09:10:28","guid":{"rendered":"https:\/\/mathemerize.com\/?p=7002"},"modified":"2021-10-25T09:19:23","modified_gmt":"2021-10-25T03:49:23","slug":"a-random-variable-x-has-poisson-distribution-with-mean-2-then-px-1-5-equal-to","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/a-random-variable-x-has-poisson-distribution-with-mean-2-then-px-1-5-equal-to\/","title":{"rendered":"A random variable X has poisson distribution with mean 2. Then, P(X > 1.5) equal to"},"content":{"rendered":"
Now, P(X > 1.5) = P(2) + P(3) + …… \\(\\infty\\)<\/p>\n
= 1 – [P(0) + P(1)]<\/p>\n
= 1 – \\((e^{-2} + {e^{-2}(2)\\over 1})\\) = 1 – \\(3\\over e^2\\)<\/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 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