Fill in the P(X = x) values in the table below to give a legitimate probability

Question

Fill in the P(X = x) values in the table below to give a legitimate probability

distribution for the discrete random variable X , whose possible values are -6 , 3 ,

4 , 5 , and 6.

The values that the discrete random variable X may take are -6 , 3 , 4 , 5 , and 6.

Thus, it must be that

P( X = -6) + P( X = 3) + P( X = 4) + P( X = 5 ) + P( X = 6) = I ,

with O < P( X = x) < 1 for x = -6 , 3 , 4 , 5 , 6.

From the information given, we have

0.18 + P(X = 3)+0.18 +0.21 + P(X = 6) = 1,

which means that

P(X= 3)+P(X= 6) = 1-(0.18+0.18+0.21) = 0.43.

To complete the probability distribution, then, we must choose P(X = 3) and

P(X = 6) to be any numbers from Oto 1 such that

P(X = 3)+ P(X = 6) = 0.43.

For example, we could choose P(X = 3) = 0.37 and P(X = 6) = 0.06.

One correct answer is:

My question is, how can we get the answer of 0.37 and 0.06? I went over the explanition many times, but I couldn't find out the way of calcuating the answer! Could you please tell in details how to get the answer?

Thank you

Explanation / Answer

I dont think that you have to calculate particular solution for the problem. There are infinite posssible solutions in that case.

The required solution will be

P(X = 3)+ P(X = 6) = 0.43.

And that's enough.

So as long as P(X = 3)+ P(X = 6) = 0.43 you will have a probability mass function.

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