I know the probability of an outcome in a sample, how to apply to population? I admittedly am horrible at statistics. I know that the probability of an individual who has saved for retirement has saved less then 50,000 USD is 16.4% according to my sample data, but only 69% of workers say they have saved for retirement. So what is the probability for a random person selected from the entire population to have saved less than 50,000 USD?
So that means 31% of the population admits to not saving anything (which is less then 50,000 USD), so the probability of a random person selected from the entire population having saved less than 50,000 USD for retirement is actually greater than 16.4% - but I don't know how to calculate what it is more precisely.
This is homework, but I am so lost....
Any help is appreciated.
 A: Hint: Suppose you had 1000 workers.  
Then you might think about $690=1000\times 0.69$ of them had saved something for retirement and $310=1000-690$ had saved nothing.  
Of those, about 16.4% or $690 \times 0.164$ had saved something for retirement but less than $50,000. What is that as a percentage of 1000?
And how many had saved less than $50,000 including those who had save nothing?  What is that as a percentage of 1000?  
A: Looking at both your posts, you probably want to start here: Bayes theorem and work through the examples half-way down carefully.
Edit: This looks like a reasonable beginners' tutorial.
A: you may have to check the sample data more closely, but I think the 16.4% is a "conditional" percentage.  That is, 16.4% of the sampled units who said they saved for retirement have saved less than $50,000$.  If we write this as a fraction, it may help your understanding:
$$\frac{\text{Number of sampled units saving >0 and <50000}}{\text{Number of sampled units saving >0}}\times 100\text{%}=16.4\text{%}$$
(The denominator here may or may not include the sampled units with exactly $0$ savings, which is why you need to check the question).  And your 69% would be obtained by:
$$\frac{\text{Number of sampled units saving >0}}{\text{Number of sampled units saving >0 or exactly 0}}\times 100\text{%}=69\text{%}$$
I think this is the difficulty you are having.  Note that the numerator of the second probability is the same as the denominator of the first probability*, so if you multiply them, they cancel.  To further complete the question, is easiest done via a "decision process".  We must first:


*

*Decide whether the person admits to saving something or nothing.

*If we decide they admit saving nothing, then they also save less than $50,000$.  If we decided that they admit saving something, that they would further say that they saved less than $50,000$


Let me know if this helps.
*note: technically, they are frequencies, not probabilities, but this makes no difference to your problem here.
