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[1] MA4104 Business Statistics Spring 2008, Lecture 03 Examples Class: Using the Normal Tables ( and some computer alternatives )

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[2] A machine used to regulate the amount of dye dispensed can be set so that it discharges an average of = 5.3 mL of dye per can of paint. The amount of dye discharged is known to have a normal distribution with a standard deviation of 0.4 mL. (a) What percentage of paint cans have a dye concentration greater than 6.0 mL? (b) What percentage of paint cans have a dye concentration less than 4.4 mL? (c) What percentage of paint can have a dye concentration of between 4.4 mL and 6.0 mL? (d) If more than 6 mL of dye is discharged when making a certain shade of blue, the shade is unacceptable. Determine the setting for so that only 1% of cans of paint will be unacceptable. (e) Keeping fixed at 5.3 mL, what reduction in is required so that just 1% of cans of paint will be unacceptable.

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[3] ( a) What percentage of cans have a dye concentration > 6.0 mL? 0 1.75 Look up 1.75 in the tables and we get… 6.0 5.3 ZX = 0.4 = 1

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[4] z.00.01.02.03.04.05.06.07.08.09 0.0 0.50000.50400.50800.51200.51600.51990.52390.52790.53190.5359 0.1 0.53980.54380.54780.55170.55570.55960.56360.56750.57140.5753 0.2 0.57930.58320.58710.59100.59480.59870.60260.60640.61030.6141 0.3 0.61790.62170.62550.62930.63310.63680.64060.64430.64800.6517 0.4 0.65540.65910.66280.66640.67000.67360.67720.68080.68440.6879 0.5 0.69150.69500.69850.70190.70540.70880.71230.71570.71900.7224 0.6 0.72570.72910.73240.73570.73890.74220.74540.74860.75170.7549 0.7 0.75800.76110.76420.76730.77040.77340.77640.77940.78230.7852 0.8 0.78810.79100.79390.79670.79950.80230.80510.80780.81060.8133 0.9 0.81590.81860.82120.82380.82640.82890.83150.83400.83650.8389 1.0 0.84130.84380.84610.84850.85080.85310.85540.85770.85990.8621 1.1 0.86430.86650.86860.87080.87290.87490.87700.87900.88100.8830 1.2 0.88490.88690.88880.89070.89250.89440.89620.89800.89970.9015 1.3 0.90320.90490.90660.90820.90990.91150.91310.91470.91620.9177 1.4 0.91920.92070.92220.92360.92510.92650.92790.92920.93060.9319 1.5 0.93320.93450.93570.93700.93820.93940.94060.94180.94290.9441 1.6 0.94520.94630.94740.94840.94950.95050.95150.95250.95350.9545 1.7 0.95540.95640.95730.95820.95910.95990.96080.96160.96250.9633 1.8 0.96410.96490.96560.96640.96710.96780.96860.96930.96990.9706 1.9 0.97130.97190.97260.97320.97380.97440.97500.97560.97610.9767 2.0 0.97720.97780.97830.97880.97930.97980.98030.98080.98120.9817

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[5] ( a) What percentage of cans have a dye concentration > 6.0 mL? 0 1.75 Look up 1.75 in the tables and we get… 0.9599 So that the area above Z = +1.75 equals 0.0401 or 4.01% 6.0 5.3 ZX = 0.4 = 1

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[6] Statistical tables: are a hangover from the early and middle parts of the last century; look old-fashioned, and make the subject of statistics look old-fashioned; are redundant if you know how to use some basic functions in EXCEL, or the stats package R; unfortunately, probably still needed for the purposes of teaching large classes.

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[11] Even easier in the stats package R

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[12] ( b) What percentage of cans have a dye concentration < 4.4 mL? -2.25 0 4.4 5.3 ZX = 0.4 = 1 Look up +2.25 in the tables and we get…

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[13] z.00.01.02.03.04.05.06.07.08.09 0.0 0.50000.50400.50800.51200.51600.51990.52390.52790.53190.5359 0.1 0.53980.54380.54780.55170.55570.55960.56360.56750.57140.5753 0.2 0.57930.58320.58710.59100.59480.59870.60260.60640.61030.6141 0.3 0.61790.62170.62550.62930.63310.63680.64060.64430.64800.6517 0.4 0.65540.65910.66280.66640.67000.67360.67720.68080.68440.6879 0.5 0.69150.69500.69850.70190.70540.70880.71230.71570.71900.7224 0.6 0.72570.72910.73240.73570.73890.74220.74540.74860.75170.7549 ::::::::::: ::::::::::: 2.0 0.97720.97780.97830.97880.97930.97980.98030.98080.98120.9817 2.1 0.98210.98260.98300.98340.98380.98420.98460.98500.98540.9857 2.2 0.98610.98640.98680.98710.98750.98780.98810.98840.98870.9890 2.3 0.98930.98960.98980.99010.99040.99060.99090.99110.99130.9916 2.4 0.99180.99200.99220.99250.99270.99290.99310.99320.99340.9936 2.5 0.99380.99400.99410.99430.99450.99460.99480.99490.99510.9952

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[14] ( b) What percentage of cans have a dye concentration < 4.4 mL? -2.25 0 4.4 5.3 ZX Look up +2.25 in the tables and we get… 0.9878 The area above Z = +2.25 (and by symmetry the area below Z = 2.25) equals 0.0122 or 1.22% = 0.4 = 1

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[15] ( c) What percentage of cans have a dye concentration between 4.4 mL and 6.0 mL? 4.4 5.3 6.0 4.46.0 1.22% 4.01% 94.77%

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[16] (d) What value for ? = 0.4 = 1 6.0 1% 0

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[17] z.00.01.02.03.04.05.06.07.08.09 0.0 0.50000.50400.50800.51200.51600.51990.52390.52790.53190.5359 0.1 0.53980.54380.54780.55170.55570.55960.56360.56750.57140.5753 0.2 0.57930.58320.58710.59100.59480.59870.60260.60640.61030.6141 0.3 0.61790.62170.62550.62930.63310.63680.64060.64430.64800.6517 0.4 0.65540.65910.66280.66640.67000.67360.67720.68080.68440.6879 0.5 0.69150.69500.69850.70190.70540.70880.71230.71570.71900.7224 0.6 0.72570.72910.73240.73570.73890.74220.74540.74860.75170.7549 ::::::::::: ::::::::::: 2.0 0.97720.97780.97830.97880.97930.97980.98030.98080.98120.9817 2.1 0.98210.98260.98300.98340.98380.98420.98460.98500.98540.9857 2.2 0.98610.98640.98680.98710.98750.98780.98810.98840.98870.9890 2.3 0.98930.98960.98980.99010.99040.99060.99090.99110.99130.9916 2.4 0.99180.99200.99220.99250.99270.99290.99310.99320.99340.9936 2.5 0.99380.99400.99410.99430.99450.99460.99480.99490.99510.9952 Perhaps, Z = 2.325 ???

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[18] There is an EXCEL function NORMINV that does this exactly: The exact value is 2.326, so our approximation of 2.325 isnt too bad

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[19] (d) What value for ? 2.325 = 0.4 = 1 6.0 1% 0

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[20] (e) What value for ? 2.325 0 = 1 6.0 1% 5.3

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[21] Example: Kevs Garage Kevs garage sells a popular multi-grade motor oil. When the stock of this oil drops to 20 gallons, a replenishment order is placed. The store manager is concerned that sales are being lost due to stock-outs while waiting for an order. It has been determined that lead-time demand is normally distributed with a mean of 15 gallons and a standard deviation of 6 gallons. The manager would like to know the probability of a stock-out, that is, the probability P(X > 20).

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[22] Standard Normal Distribution 0.83 Z Example: Kevs Garage Area =.7967 Area =.2033 z = (X - )/ = (20 - 15)/6 =.83

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[23] If the manager of Kevs Garage wants the probability of a stock-out to be no more than.05, what should the reorder point be? z represents the Z value cutting the tail area of.05 Area =.05 Area =.95 0 z Example: Kevs Garage Extra

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[24] z.00.01.02.03.04.05.06.07.08.09 0.0 0.50000.50400.50800.51200.51600.51990.52390.52790.53190.5359 0.1 0.53980.54380.54780.55170.55570.55960.56360.56750.57140.5753 0.2 0.57930.58320.58710.59100.59480.59870.60260.60640.61030.6141 0.3 0.61790.62170.62550.62930.63310.63680.64060.64430.64800.6517 0.4 0.65540.65910.66280.66640.67000.67360.67720.68080.68440.6879 0.5 0.69150.69500.69850.70190.70540.70880.71230.71570.71900.7224 0.6 0.72570.72910.73240.73570.73890.74220.74540.74860.75170.7549 0.7 0.75800.76110.76420.76730.77040.77340.77640.77940.78230.7852 0.8 0.78810.79100.79390.79670.79950.80230.80510.80780.81060.8133 0.9 0.81590.81860.82120.82380.82640.82890.83150.83400.83650.8389 1.0 0.84130.84380.84610.84850.85080.85310.85540.85770.85990.8621 1.1 0.86430.86650.86860.87080.87290.87490.87700.87900.88100.8830 1.2 0.88490.88690.88880.89070.89250.89440.89620.89800.89970.9015 1.3 0.90320.90490.90660.90820.90990.91150.91310.91470.91620.9177 1.4 0.91920.92070.92220.92360.92510.92650.92790.92920.93060.9319 1.5 0.93320.93450.93570.93700.93820.93940.94060.94180.94290.9441 1.6 0.94520.94630.94740.94840.94950.95050.95150.95250.95350.9545 1.7 0.95540.95640.95730.95820.95910.95990.96080.96160.96250.9633 1.8 0.96410.96490.96560.96640.96710.96780.96860.96930.96990.9706 1.9 0.97130.97190.97260.97320.97380.97440.97500.97560.97610.9767 2.0 0.97720.97780.97830.97880.97930.97980.98030.98080.98120.9817 Perhaps, z = 1.645 ???

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[25] If the manager of Kevs Garage wants the probability of a stockout to be no more than.05, what should the reorder point be? z represents the z value cutting the tail area of.05 Area =.05 Area =.95 0 z Example: Kevs Garage Extra z = 1.645

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[26] The corresponding value of X is given by x = + z = 15 + 1.645(6) = 24.87 A reorder point of 24.87 gallons will place the probability of a stock-out during lead-time at.05 Perhaps Kevs should set the reorder point at 25 gallons to keep the probability under.05 Example: Kevs Garage Extra

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