Let the p.m.f. of M be defined by f(x)=x/8,x=1,3,4.

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The probability mass function (p.m.f.) of the random variable M is defined by f(x) = x/8 for x = 1, 3, 4. The probabilities are calculated as P(M=1) = 1/8, P(M=3) = 3/8, and P(M=4) = 4/8, which sum to 1, confirming a valid distribution. The mean of M, denoted as E(M), is computed using the definition of expected value, but it is clarified that this distribution is not binomial.

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Let the p.m.f. of M be defined by f(x)=x/8,x=1,3,4. What is the mean of M?
Is this the same an E(M) of a binomial?
 
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I guess we have P(M=1)=1/8, P(M=3)=3/8 and P(M=4) = 4/8. These add to 1.
So the mean is ... what? Do the computation from the definition.
Yes, this is another name for E(M), but it is not binomial.
 

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