Using Z-Scores to Calculate Probability

Date: 05/25/2008 at 16:56:42
From: Kate
Subject: Statistics

Trains carry bauxite ore from a mine in Canada to an aluminum 
processing plant in northern New York state in hopper cars.  Filling 
equipment is used to load ore into the hopper cars.  When functioning 
properly, the actual weights of ore loaded into each car by the 
filling equipment at the mine are approximately normally distributed 
with a mean of 70 tons and a standard deviation of 0.9 ton.  If the 
mean is greater than 70 tons, the loading mechanism is overfilling.

If the filling equipment is functioning properly, what is the 
probability that the weight of the ore in a randomly selected car will
be 70.7 tons or more?

Date: 05/27/2008 at 18:46:08
From: Doctor Grosch
Subject: Re: Statistics

Hi, Kate:

This is a problem in determining the Z-function.  Visualize a diagram 
of the normal, bell-shaped distribution.  Visualize also how the 
standard deviation fits in the diagram.  The area under the normal 
curve from + 1 SD to - 1 SD comprises about 2/3 of the total area of 
the curve.  That is, for this problem, the mean sits at 70 tons on 
the horizontal axis and the area under the normal curve from the 
points on the horizontal axis from 70-0.9 tons to 70+0.9 tons (69.1 
tons to 70.9 tons) contains about 2/3 of the area of the whole curve.

Now, the area under the curve from - infinity to the mean is half the 
total area and that from the mean to + infinity is the other half of 
the total area.  In this context area is the same as probability.  
The question is what the probability is of having 70.7 tons or more 
of ore in a randomly selected car.  Don't let the "randomly selected" 
bit spook you.  That's a necessary precondition for getting a 
reliable answer.  Don't be concerned about how the workers in the 
field make sure that the car is randomly selected.  Just accept, as a 
given, that they've managed to select the car randomly and go from 
there.

Your main task, in a problem like this, is to figure out the critical 
z-value and to make certain that you pick it correctly.  As you 
probably know, the formula for z follows:

       x - mu
  Z = --------
        sigma

where x is the point in question, namely 70.7 tons of ore, mu is the
mean value (in this case, 70 tons of ore), and sigma is the standard
deviation.

      70.7 - 70
  Z = ---------
         0.9

  Z = 0.7/0.9 = 0.778

That gives you the entry-value for the tabulated z-value in your 
table for the normal curve.  Look at the following link:

  Hyperstat Online: Normal Distribution
    http://davidmlane.com/hyperstat/z_table.html 

It's a normal-curve calculator.  All you have to do is to put your z-
value into the indicated space and it calculates the area or 
probability for you and gives you a picture of the situation on the 
normal curve.  The only difference from the problem-statement is 
that the curve is centered on zero as its mean, whereas your problem
centers the mean at 70.  You can disregard that distinction because
calculating the Z-value universalizes the normal curve and centers its
mean at zero.  

You are interested in the area of the curve ABOVE the critical 
Z-value, so put your figure, 0.778, in the indicated space (above) and
you see the area or probability: 0.218, to three significant figures,
which is usually adequate.  The calculator gives you six significant
figures but the last three are meaningless, for most purposes.

- Doctor Grosch, The Math Forum
  http://mathforum.org/dr.math/