The Professor's Response
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