
the normal dist
Hey guys,
Cereal dispensing machine problem
Suppose the standard deviation is not known but can be fixed at certain levels by adjusting the cereal dispensing machine. What is the largest std. dev. that will allow the actual value of cereal dispensed to fall within 1 ounce of the mean with prob of at least .95?
The mean is 11.67 oz. per box of cereal.
This involves Zscores I'm pretty sure...

Are you looking for something like this?
$\displaystyle P(\mu 2\sigma <Z < \mu+ 2\sigma) = 0.95$
$\displaystyle P( 11.67 2\sigma < Z< 11.67+ 2\sigma) = 0.95$
Solve for $\displaystyle \sigma $

Thanks for the response. I guess I'm confused as to how to solve this. Do I calculate the Zscore and set each equn. equal to that numer and solve for sigma?

Using symmetry
$\displaystyle P( 0 < Z< 11.67+ 2\sigma) = \frac{0.95}{2}$
$\displaystyle P( Z< 11.67+ 2\sigma) P(Z<0) = 0.475$
Now apply the inverse Z?