A machine dispensing a somewhat chunky material into packages
does not place exactly the same mass in each container. Studies
show that the actual mass dispensed each time is an approximately
normally distributed random variable with a standard deviation of
15.0 g, and a mean value which can be set at any desired value.
At what value must the mean be set to ensure that 90% of the
packages contain at least 650 g of material.

## Solution:

We are asked to find the value of the mean, μ, which will give a
10th percentile equal to 650 g.
Standard formulas from the theory of the normal distribution give
this as

10th percentile = μ-
z0.10 σ,

where -z0.10 is the
10th percentile of the standard
normal random variable (which we can determine to be -1.282, from
standard normal probability tables), and σ= 15.0 g, the
standard deviation.

Solving the above equation for the mean, μ, then gives:

μ=10th percentile +
z0.10σ

= 650 g + (1.282)(15.0 g)

= 669.23 g.

Thus, if the mean is set at 670 g, then at least 90% of the
containers filled by this machine will have 650 g or more
material in them.