Find the mean of the probability distribution of a random variable X which if š(š) =
š„+1
20
for X= 1, 2, 3, 4, and 5.
EX=āx=15xā x+120=120(1ā 2+2ā 3+3ā 4+4ā 5+5ā 6)=3.5EX=\sum_{x=1}^5{x\cdot \frac{x+1}{20}}=\frac{1}{20}\left( 1\cdot 2+2\cdot 3+3\cdot 4+4\cdot 5+5\cdot 6 \right) =3.5EX=āx=15āxā 20x+1ā=201ā(1ā 2+2ā 3+3ā 4+4ā 5+5ā 6)=3.5
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