mba 522 Normal Example 1

mba 522 Normal Example 1 - words, it is 0.3). That is...

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According to the Bureau of Labor Statistics, the average weekly pay for a U.S. production worker was $441.84 ( The World Almanac, 2000 ). Assume that available data indicate that production worker wages were normally distributed with a standard deviation of $90. How much did a production worker have to earn to be in the top 20% of wage earners? Graphic solution to the problem by Dr. J X (weekly Earnings) $441.84 ? Z 0 0.84 Question is asking us to find the weekly earnings corresponding to top 20% (i.e., probability is 0.2). Since salary axis increases from left to right, the top 20% would be this right tail of curve. This area is 30% of the area under the curve (in other
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Unformatted text preview: words, it is 0.3). That is because half of the curve is 0.5, and we know the right tail is 0.2. So, this area must be 0.3 This area is 0.5 of the total area under the curve due to normal curve symmetry. Total of these 2 areas equals 80% of area under the curve 1 Z=0.84 was used to answer the question. It is read from the table by finding within the table the number that comes closest to 0.8. Search in the table to find 0.7995. From that point go to the left and to the top margins to locate Z=0.84. Now use 0.84 as Z in the Z formula to find X which is the answer to the question ($517.44) 0.84 = (X 441.84) / 90. Solve for X = 517.44...
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