So the question is this: A machine fills bottles with soda. The volume with which the bottles are filled is almost normally distributed. The standard deviation is 3.5 ml and the mean value is controlled by the setting on the machine. What should we set the machine to expect at least 977 out of 1000 bottles to contain at least 500ml?
In the answer they they juse a digital tool to guess. So my question is, if it woulb be possible to do this in the form of an equation.
looking up the function for normal distrubution I found this:
$f(x) = \frac{1}{{\sigma}\sqrt{2\pi}} \cdot \exp \left[-\frac{1}{2} \left(\frac{x-u}{\sigma}\right)^2 \right] $
I then put up this equation
$\frac{\int_{500}^{\infty} f(x)dx}{\int_{-\infty}^{\infty} f(x)dx} = 0.997$
using wolfram to get the primal function I get this: $ \frac{1}{2}\operatorname{erf}\left(\frac{x-500}{3\sqrt{2}}\right) + \text{constant} $ where 500 is the mean and 3 is the standard deviation
expanding the equation to be solvable
$\frac{\left( \frac{1}{2}\operatorname{erf}( \frac{x+500}{3.5 \cdot \sqrt{2}})\right)-\left( \frac{1}{2}\operatorname{erf}( \frac{x+5000}{3.5 \cdot \sqrt{2}})\right)}{\left(\frac{1}{2}\operatorname{erf}( \frac{x- 5000}{3.5 \cdot \sqrt{2}})\right)-\left(\frac{1}{2}\operatorname{erf}( \frac{x+5000}{3.5 \cdot \sqrt{2}})\right)} = 0.997 $
I changed infinity to a large number since it just returned false with infinty, the result I get is 5000 or whatever I change the upper intergral to. I am not very experienced with intergrals so if anyone got another way to do this please let me know.
The answer btw is 507.