1. ## lost:(

A soft-drink machine can be regulated so that it discharges an average of x ounces per cup. If the ounces of fill are normally distributed with standard deviation 0.3 ounce, give the setting for x so that 8-ounce cups will overflow only 1% of the time.

from the normal distribution table, the probability of X where X is greater than x is always greater than 0.5. how do i find a value x such that the probability of it overflowing is only 1%?

2. Originally Posted by alexandrabel90
A soft-drink machine can be regulated so that it discharges an average of x ounces per cup. If the ounces of fill are normally distributed with standard deviation 0.3 ounce, give the setting for x so that 8-ounce cups will overflow only 1% of the time.

from the normal distribution table, the probability of X where X is greater than x is always greater than 0.5. how do i find a value x such that the probability of it overflowing is only 1%?
Hi alexandrabel90,

The part in blue is the probability that $Z\le X$ when $X\ge x$

for $Z=\frac{X-x}{0.3}$

In this case you need to work with the region in the tails of graph of the normal curve.

The 0.5% to the right of 99.5% added to the first 0.5%
constitutes the 1% region in a 2-tailed situation,
or the first 1% or last 1% (9% to 100%) constitutes the 1% region
in a 1-tail situation.