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Course Documents

Chapter 1 - Intro

Chapter 2 - Methods for Describing Sets of Data

Chapter 3 - Probability

Chapter 4 - Discrete Random Variables

Chapter 5 - Normal Random Variables

Chapter 6 - Sampling Distributions

Chapter 7 - Confidence Intervals

Chapter 8 - Tests of Hypothesis: One Sample

Chapter 9 - Confidence Intervals and Hypothesis Tests: Two Samples

Sample Exam I: Chapters 1 & 2

Sample Exam II: Chapters 3 & 4

Sample Exam III: Chapters 5 & 6

Sample Exam IV: Chapters 7 & 8

Ask the Professor Forum

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 "

Posted to STATS 1 on Sunday, October 19, 2014   Replies: 3


Professor Mcguckian
10/20/2014
2:29 AM EST

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


.
10/20/2014
12:21 PM EST
Thank you so much for your help.
I never though about plugging values to solve the problem.

I am so happy for your website and for the details of your examples.

Sincerely

Alex

Professor Mcguckian
10/20/2014
4:22 PM EST

You're welcome Alex. I am glad the site has been helpful.

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