fairylight Posted September 15, 2004 Share Posted September 15, 2004 OK im prob being really stupid, but i cant for the life of me work out how to do this problem. can someone help??? A random variable X takes the integer value x with probability P(X=x) defined by P(X=x) = kx^2, x = 1, 2, 3 P(X=x) = k(7-x)^2 x = 4,5,6 P(X=x) = 0 otherwise Find the value of k Link to comment Share on other sites More sharing options...
MandrakeRoot Posted September 15, 2004 Share Posted September 15, 2004 Sum of probabilities should be one. Since you have only 6 possible values for x the equation you have to solve is : k + 4k + 9k + k(7 - 4)^2 + k(7 - 5)^2 + k(7 - 6)^2 = 1 which would be 28k = 1, making k = 1/28 Mandrake Link to comment Share on other sites More sharing options...
Dave Posted September 15, 2004 Share Posted September 15, 2004 Beat me to it Works for continuous as well as discrete stuff, except you have to use integration to find the probability density. Link to comment Share on other sites More sharing options...
MandrakeRoot Posted September 16, 2004 Share Posted September 16, 2004 Or discrete with infinite (but countable) values, needing infinite sums. Mandrake Link to comment Share on other sites More sharing options...
Guest lucylooloo Posted September 25, 2004 Share Posted September 25, 2004 Thank you sir may I I am dire need of some assistance in understanding predicting probabilities, I am just lost with most of this Link to comment Share on other sites More sharing options...
Dave Posted September 25, 2004 Share Posted September 25, 2004 What do you find difficult about them? Link to comment Share on other sites More sharing options...
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