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? Round your answer to 2 decimal places.

Answer 1

Answer:

i really don't care

Step-by-step explanation:

Answer 2

Answer:

Standard deviation of the weight over a day is 1.15

Step-by-step explanation:

Since it is given that water weight is uniformly distributed between -2 to 2.  

So applying formula for variance of uniformly distribution as follows,  

$variance=\dfrac{\left (b-a\right )^{2}}{12}$

From the given data, values of a and b is , $a=-2$, $b=2$..

Substituting the values,  

$variance=\dfrac{\left (2-\left (-2 \right )\right )^{2}}{12}$

Now, $-\left ( -2 \right )=2$

$variance=\dfrac{\left (2+2\right )^{2}}{12}$

$variance=\dfrac{\left (4\right )^{2}}{12}$

$variance=\dfrac{16}{12}$

Dividing the fraction by 4,

$variance=\dfrac{4}{3}$

So, the value of variance is $\dfrac{4}{3}$

The formula for standard deviation is given as,

$Standard\:deviation=\sqrt{variance}$

Substituting the value,

$Standard\:deviation=\sqrt{\dfrac{4}{3}}$

$Standard\:deviation=1.15470$

Rounding to 2 decimal places,

$Standard\:deviation=1.15$

So, the value of standard deviation is 1.15.