Question 294458
The time required to make 1100 gallons of synthetic rubber at a plant in South America in a recent year was normally distributed with a mean of 16 hours and a standard deviation of 3 hours. What is the probability that it will take more than 19 hours to make 1100 gallons of synthetic rubber?
<pre><font size = 4 color = "indigo"><b>
[Notice that the "1100 gallons" has nothing to do with the problem]

We calculate the z-score for 19

{{{z=(x-mu)/sigma}}}

{{{z=(19-16)/3}}}

{{{z=3/3=1}}}

We draw the normal curve and shade under the curve
to the right of z=1:
 
{{{drawing(400,200,-5,5,-.5,1.5, graph(400,200,-5,5,-.5,1.5, exp(-x^2/2)), 
 
line(1.1,0,1.1,exp(-1.1^2/2)),line(1,0,1,exp(-1/2)),
line(1.2,0,1.2,exp(-1.2^2/2)),
line(1.3,0,1.3,exp(-1.3^2/2)),
line(1.4,0,1.4,exp(-1.4^2/2)),
line(1.5,0,1.5,exp(-1.5^2/2)),
line(1.6,0,1.6,exp(-1.6^2/2)),
line(1.7,0,1.7,exp(-1.7^2/2)),
line(1.8,0,1.8,exp(-1.8^2/2)),
line(1.9,0,1.9,exp(-1.9^2/2)),
line(2.0,0,2.0,exp(-2.0^2/2))
line(2.1,0,2.1,exp(-2.1^2/2)),
line(2.2,0,2.2,exp(-2.2^2/2)),
line(2.3,0,2.3,exp(-2.3^2/2)),
line(2.4,0,2.4,exp(-2.4^2/2)),
line(2.5,0,2.5,exp(-2.5^2/2)),
line(2.6,0,2.6,exp(-2.6^2/2)),
line(2.7,0,2.7,exp(-2.7^2/2)),locate(4.8,-.01,z),locate(4.8,.2,z)
  
)}}}
 
The percentage of the area that is shaded is what we are after.

Now you've probably learned that 34.1% of the area under
the normal curve lies between z=0 and z=1 and that 50% of 
all of the area under the normal curve lies to the right 
of z=0.  So subtracting the 34.1% of the area between
z=0 and z=1 from 50% gives the percent of shaded area,
which is the probability that it will take more that 19
hours.  50%-34.1% = 15.9% is the desired probability, or,
expressed as a decimal, 0.159.

Edwin</pre>