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Okay i got a story problem and here it is. S soft drink machine is regulated so it discharges an averaged of u ounces per cup. If the ounces of fill are normally distributed with standard deviation equal to .3 ounce give the setting for u so that the 8 ounce cups will overflow only 1% of the time.
Im not sure how to start. I figure i need to find the z score but im not sure what else to do? any ideas?
This is quite easy. If I get you right, what you are supposed to do is to find the average fill volume (lets call it E) causing overflow 1% of the time. (Not overflow 99% of the time)
As you mention, the first thing to do is to look up the z-value in a table of normal curve areas. Such tables normally tabulate the area covered by the curve between the average, x=E, and x=E+z. Since the area of the curve between -infinity and E is 0.50, you are supposed to find a z corresponding to an area of
A = 0.99 - 0.50 = 0.49
From the table, z=2.33.
Now what does this mean? This means that the risk that a random fill volume is more than 2.33 standard deviations above the average E is 1-0.99=0.01, or 1%.
And since the volume of the cup is V=8, you get the relation
E + 2.33sigma = V
Since the standard deviation is sigma=0.3, the answer is:
E = 8 - 2.33*0.3 = 7.3 ounce