Hello,

I need assistance with the attached problem. I've also attached an example for reference. Please assist by letting me know what should go in the empty boxes as seen in the example.

Thanks for your help.

? QUESTION Fill in the P (X=x) values to give a legitimate probability distribution for the discrete random variable X, whose possible values are - 2, 2, 4, 5, and 6. Value x of X P(X = x) - 2 0.27 2 0.10 4 0.29 EXPLANATION The values that the discrete random variable X may take are -2, 2, 4, 5, and 6. Thus, it must be that we have the following. P(X= -2) +P(X=2) +P(X=4) +P(X=5) +P(X=6)=1 In the above, OSP(X=x) $1 for x= -2, 2, 4, 5, 6. From the information given, we have this. 0.27 + 0.10 + P(X=4) + P(X=5) +0.29 =1 That gives us the following. P (X=4) + P(X=5)=1- (0.27 + 0.10 + 0.29)=0.34 To complete the probability distribution, then, we must choose P (X"=4) and P (X"= 5) to be any numbers from 0 to 1 such that this holds. P(X=4) + P(X=5)=0.34 For example, we could choose P (X= 4) = 0.12 and P (X=5)=0.22. ANSWER Here is one possible answer. Value x of X P(X = x) -2 0.27 2 0.10 4 0.12 5 0.22 6 0.29Fill in the P (X=x) values to give a legitimate probability distribution for the discrete random variable X, whose possible values are -2, 0, 4, 5, and 6. Value x of X P( X = x) -2 0.22 O 4 5 0.25 6 0.29 X 5