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Exercise 7.1 · Q15
Q.

A random variable X has the following probability distribution :

X01234567
P (X)0k2k2k3kk²2k²7k² + k

Determine : (i) k (ii) P (X < 3) (iii) P ( X > 4)

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Summing all probabilities to 1: 0+k+2k+2k+3k+k2+2k2+(7k2+k)=10+k+2k+2k+3k+k^2+2k^2+(7k^2+k)=1, i.e. 9k+10k2=19k+10k^2=1, so 10k2+9k−1=010k^2+9k-1=0. By the quadratic formula, k=−9±12120=−9±1120k=\dfrac{-9\pm\sqrt{121}}{20}=\dfrac{-9\pm11}{20}, giving k=0.1k=0.1 or k=−1k=-1; rejecting the negative root, k=0.1k=0.1. …

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