The Professor's Response

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A student e-mailed me this question:
"My teacher gave us the below problem and I am not able to figure it out. Can you help me.

If X is a random variable that can take on the values 1, 2, 3, or 4, find a probability distribution for X with a mean = 3

Thanks for your help

Alex "

See the professor's answer below.

Hi Alex,

This problem does not have a unique solution because there are four unknowns and only two equations possible. However, that doesn't mean we cannot solve the problem for a solution. It just means there will be many solutions possible.

Here is what we know:

 X P(X) X*P(X) 1 a a 2 b 2b 3 c 3c 4 d 4d Totals 1.00 3.00 = µ

We know the probabilities must sum to 1.00, and the sum of X*P(X) must equal 3.00. This forms two equations.

a+b+c+d = 1 and a+2b+3c+4d = 3.

Now, you can just plug in any probability for a and b you want (make sure the values are between 0 and 1). For example, if I make a = 0.10 and b = 0.20, the first equation becomes: 0.10 + 0.20 + c + d = 1.00.  This simplifies to c + d = 0.70.

Then the second equation becomes: 0.10 + 2(0.20) + 3c + 4d = 3.00. This gives us 3c + 4d = 2.50.

Now, you can use substitution to solve: c = 0.70 - d.  plug this expression for c into the second equation to get::

3(0.70 - d) + 4d = 2.50

Solve for d

2.10 -3d + 4d =2.50

2.10 + d = 2.50

d = 2.50 - 2.10

d = 0.40

If d = 0.40, we know c must be 0.30. (because the sum of a, b, c, and d must be one).

Here is the final table:

 X P(X) X*P(X) 1 0.10 0.10 2 0.20 0.40 3 0.30 0.90 4 0.40 1.60 totals 1.00 3.00

I hope that helps,

Professor McGuckian