Gareth56 Posted September 4, 2008 Share Posted September 4, 2008 If you're given an average value and the sample size can you determine from those values what the maximum & minimum values were that gave the average value? Thanks Link to comment Share on other sites More sharing options...
DrP Posted September 4, 2008 Share Posted September 4, 2008 I do not believe so. For example, the mean average is '4' from 5 samples. You could have 2, 2, 4, 6 and 6. You could have 1, 2, 4, 6 and 7. You could have 1, 1, 1, 3, 7 and 7. In each case the average if 4 and the number of values is 5, but there is no way of knowing what the min and max values are - they are different in each case. Link to comment Share on other sites More sharing options...
ajb Posted September 4, 2008 Share Posted September 4, 2008 As DrP, I can't see how you could do it, not without some (probability) distribution being given. Link to comment Share on other sites More sharing options...
chitrangda Posted September 4, 2008 Share Posted September 4, 2008 you need to know the dristribution for knowing the max nad min values. 1 Link to comment Share on other sites More sharing options...
Gareth56 Posted September 4, 2008 Author Share Posted September 4, 2008 Many thanks for your input. What it was was I heard an advertisment on the radio about a company blowing its own trumpet for check in times at an airport and was just wondering if they had given us the whole picture Averages can be so much removed from reality I suppose. Thanks again Link to comment Share on other sites More sharing options...
ajb Posted September 4, 2008 Share Posted September 4, 2008 Indeed one has to be very careful with averages. Usually the distribution is more interesting and fundamental. Link to comment Share on other sites More sharing options...
Kyrisch Posted September 7, 2008 Share Posted September 7, 2008 Well, given certain restraints like "the number cannot be negative", which is feasible in many real-world scenarios (number of siblings, number of cars, and other "survey questions") it can be done. Say the average is 8 of ten samples. Since the numbers have to be positive, the minimum is automatically zero. So, we have the equation [math]\frac{0 +0+0+0+0+0+0+0+0+x}{10}=8[/math] or [math]\frac{x}{10}=8[/math]. The maximum would then be 80. Link to comment Share on other sites More sharing options...
the tree Posted September 8, 2008 Share Posted September 8, 2008 Say the average is 8 of ten samples. Since the numbers have to be positive, the minimum is automatically zero.I'm fairly sure that the OP meant the min/max of either the population or the sample, not the min/max possible. Although you can often make reasonable assumptions about the distribution (i.e. that it's Normal or Possion or whatever depending on the situation) you'd need to know both the mean and the variance to come up with some reasonable guesses about min/max values which you may as well declare to be the boundaries of a confidence interval. Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now