Answer:
The standard deviation of your weight over a day is of 1.1547 pounds.
Step-by-step explanation:
Uniform probability distribution:
An uniform distribution has two bounds, a and b, and the standard deviation is:
[tex]S = \sqrt{\frac{(b-a)^2}{12}}[/tex]
Assume that the changes in water weight is uniformly distributed between minus two and plus two pounds in a day.
This means that [tex]a = -2, b = 2[/tex]
What is the standard deviation of your weight over a day?
[tex]S = \sqrt{\frac{(2 - (-2))^2}{12}} = \sqrt{\frac{4^2}{12}} = \sqrt{\frac{16}{12}} = 1.1547[/tex]
The standard deviation of your weight over a day is of 1.1547 pounds.
