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Consider a game in which the player loses $5 with probability 1/2, breaks even with probability...

Consider a game in which the player loses $5 with probability 1/2, breaks even with probability 1/3, and wins $10 with probability 1/6. Construct a probability distribution to represent this game. Also find the mean, variance and standard deviation.

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Answer #1

The probability distribution is summarized in the following table:

Winning (x) P[X=x]
-5 1/2
0 1/3
10 1/6

Now,

Mean = E[X] = \sumx.p(x)

= -5*1/2 + 0*1/3 + 10*1/6

= -5/2 + 5/3

= -5/6

Variance= E[X2] - (E[X])2

E[X2] = \sumx2.p(x)

= (-5)2*1/2 + 0*1/3 +102 *1/6

= 25/2 + 50/3

= 175/6

So, Variance = 175/6 -(-5/6)2

= 175/6 -25/36

= 1025/6

Standard deviation = vart = V1025/6

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