# the normal dist

• Oct 27th 2010, 04:39 PM
sfspitfire23
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 Z-scores I'm pretty sure...
• Oct 27th 2010, 04:57 PM
pickslides
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 $\displaystyle \sigma$
• Oct 27th 2010, 07:12 PM
sfspitfire23
Thanks for the response. I guess I'm confused as to how to solve this. Do I calculate the Z-score and set each equn. equal to that numer and solve for sigma?
• Oct 27th 2010, 07:18 PM
pickslides
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?