Q. 84.7( 13 Votes )

A random variable

Answer :

Given: A random variable X with its probability distribution.

(i) As we know the sum of all the probabilities in a probability distribution of a random variable must be one.



Hence the sum of probabilities of given table:


0 + k + 2k + 2k + 3k + k2 + 2k2 + 7K2 + k = 1


10K2 + 9k = 1


10K2 + 9k – 1 = 0


(10K-1)(k + 1) = 0



It is known that probability of any observation must always be positive that it can’t be negative.


So


(ii) P(X < 3) = ?


P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)


= 0 + k + 2k


= 3k



(iii) P(X > 6) = ?


P(X > 6) = P(X = 7)


= 7K2 + k





(iv) P(0 < X < 3) = ?


P(0 < X < 3) = P(X = 1) + P(X = 2)


= k + 2k


= 3k



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