a high speed dispensing machin fills boxes of laundry detergent with an ajustable mean of M ounces and a fixed

Question

a high speed dispensing machin fills boxes of laundry detergent with an ajustable mean of M ounces and a fixed standard deviation of .2 ounces. The amount dispensed is approximately normally distributed. what choice should be made of the mean M so that at least 99.5% of all boxes labelled 40 ounces will actually contain at least 40 ounces?

Answer 1

The nice thing about a normal distribution is that everything you need to know is contained in just two numbers: the mean and the standard deviation.

In the normal distribution, the fraction of the population inside N standard deviations is given by:

erf(n/sqrt(2))

where erf is the errof function. Here is a table of some values:

1 => 0.682689492137
2 => 0.954499736104
3 => 0.997300203937
4 => 0.999936657516
5 => 0.999999426697
6 => 0.999999998027

This means that 68.3% of the population is within 1 standard deviation of the mean, whatever it is, leaving the remaining 21.7% evenly split - half less than the median but more than one standard deviation away and half greater than the median.

In this case, you want at least 99.5% to be greater than 40 oz. If 40 oz. represents N standard deviations away from the mean, then you need 99% to be within N standard deviation (leaving 0.5% on the low side and 0.5% on the high side).

So first you need to figure out N from a table of values of erf (or a statistical calculator). Try:

http://en.wikipedia.org/wiki/Normal_distribution

Now that you know N, multiply it by the standard deviation to get the numerical distance from the mean.

Add this to the numerical value for the boundary to get the numerical value for the mean.