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If the probability distribution of a random variable X is given as

xix_i0123
pip_i2k22k^2k2k^23k23k^2kk

Then the mean of X is

Solution

✅ Correct Option: 4

Sum of probabilities =1= 1: 6k2+k=16k^2 + k = 1, i.e. 6k2+k−1=06k^2 + k - 1 = 0. Solving: k=−1+512=13k = \frac{-1 + 5}{12} = \frac{1}{3} (positive root).

Mean =∑xipi=0+k2+6k2+3k=7k2+3k=79+1=169= \sum x_i p_i = 0 + k^2 + 6k^2 + 3k = 7k^2 + 3k = \frac{7}{9} + 1 = \frac{16}{9}.

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