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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? Look up 1.75 in the tables and we get… ZX = 0.4 = 1

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[4] z

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

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[13] z ::::::::::: :::::::::::

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[14] ( b) What percentage of cans have a dye concentration < 4.4 mL? ZX Look up in the tables and we get… The area above Z = (and by symmetry the area below Z = 2.25) equals 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? % 4.01% 94.77%

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

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[17] z ::::::::::: ::::::::::: Perhaps, Z = ???

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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 isnt too bad

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

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[20] (e) What value for ? = % 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 - )/ = ( )/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 Perhaps, z = ???

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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 = (6) = A reorder point of 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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