ValuexofX PXx 3 024 0 018 3 013 5 6 Fill in the value

ValuexofX

P=Xx

?3 = 0.24

0 = 0.18

3 = 0.13

5 = ?

6 = ?

Fill in the \"Fill values in the table below to give a legitimate probability distribution for the discrete random variable \"\" , whose possible values are \"\" ,\"\" ,\"\" ,\"\" , and \"\".

Solution

0.24+0.18+0.13 = 0.55

As I explained before, the sum of all probabilities must be 1, and probabilities lie in range [0,1], so we can distributed remaining 0.45 in 2 variables any way we like, for example:

P(X=5) = 0.2, and P(X=6) = 0.25

ValuexofX P=Xx ?3 = 0.24 0 = 0.18 3 = 0.13 5 = ? 6 = ? Fill in the values in the table below to give a legitimate probability distribution for the discrete rand

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