Thread: 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%?

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 when

for

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.