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A random variable X has the following probability distribution:

x1234
p(x)2k4k3kk

The value of E(X) is:

Solution

✅ Correct Option: 4

Sum: 2k+4k+3k+k=10k=12k + 4k + 3k + k = 10k = 1, so k=0.1k = 0.1. Then E(X)=1(0.2)+2(0.4)+3(0.3)+4(0.1)=0.2+0.8+0.9+0.4=2.3E(X) = 1(0.2) + 2(0.4) + 3(0.3) + 4(0.1) = 0.2 + 0.8 + 0.9 + 0.4 = 2.3.

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