Using Z-Scores to Calculate ProbabilityDate: 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/ |
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