Question

An adult can lose or gain two pounds of water in the course of a day. Assume that the changes in water weight are uniformly distributed between minus two and plus two pounds in a day. What is the standard deviation of your weight over a day?

Answer 1

Answer: 1.15 pounds

Step-by-step explanation:

For uniform distribution.

The standard deviation is  :

$\sigma=\sqrt{\dfrac{(b-a)^2}{12}}$

, where a = Lower limit of interval [a,b].

b = Upper limit of interval [a,b].

Given : The changes in water weight are uniformly distributed between minus two and plus two pounds in a day.

i.e. Interval =  [-2 , +2]

Here , a= -2 and b= 2

Then, the standard deviation is  :

$\sigma=\sqrt{\dfrac{(2-(-2))^2}{12}}$

$\sigma=\sqrt{\dfrac{(2+2)^2}{12}}$

$\sigma=\sqrt{\dfrac{16}{12}}=\sqrt{1.3333}=1.15468610453\approx1.15$

Hence, the standard deviation of your weight over a day =  1.15 pounds