Sample mean and sample variance

  • Thread starter Thread starter Shackman
  • Start date Start date
  • Tags Tags
    Mean Variance
Shackman
Messages
22
Reaction score
2
Homework Statement
The sample mean and sample variance of five data values are, respectively, 104 and 4. If three of the data values are 102 100 and 105, what are the other two values?The attempt at a solution
My idea was to get two equations for the two unknowns, let's call them x and y. The equation from the sample mean was easy enough to find.

(102 + 100 + 105 + x + y) / 5 = 104
102 + 100 + 105 + x + y = 520
x + y = 213

The equation obtained from the sample variance is way more difficult because of the algebra. I start off ((102-104)^2 + (100-104)^2 + (105 - 100)^2 + (x-104)^2 + (y-104)^2)/4 = 4. As you can see this will get messy in a hurry, when you substitute for x or y using the sample mean equation it will be a fourth degree polynomial. Is there an easier way to do this?
 
Last edited:
  • Like
Likes JACOBINA and (deleted member)
Physics news on Phys.org
An alternative equation you can use is \sum_i x_i^2/n-(\bar x)^2 = 4.
 
  • Like
Likes Krittapas chaichaum
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top