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MA4104 Business Statistics Spring 2008, Lecture 03

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

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MA4104 Business Statistics Spring 2008, Lecture 03

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  1. MA4104 Business StatisticsSpring 2008, Lecture 03 Examples Class: Using the Normal Tables ( and some computer alternatives )

  2. A machine used to regulate the amount of dye dispensed can be set so that it discharges an average of m = 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 m so that only 1% of cans of paint will be unacceptable. (e) Keeping m fixed at 5.3 mL, what reduction in s is required so that just 1% of cans of paint will be unacceptable.

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

  4. 5.3 6.0 (a) What percentage of cans have a dye concentration > 6.0 mL? s = 0.4 s = 1 X Z 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%

  5. 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”.

  6. Even easier in the stats package R

  7. 4.4 -2.25 (b) What percentage of cans have a dye concentration < 4.4 mL? s = 0.4 s = 1 X Z 5.3 0 Look up +2.25 in the tables and we get…

  8. 4.4 -2.25 (b) What percentage of cans have a dye concentration < 4.4 mL? s = 0.4 s = 1 X Z 5.3 0 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%

  9. 1.22% 4.01% 94.77% (c) What percentage of cans have a dye concentration between 4.4 mL and 6.0 mL? 5.3 5.3 4.4 6.0 6.0 4.4

  10. (d) What value for m? s = 0.4 s = 1 1% m 0 6.0

  11. Perhaps, Z = 2.325 ???

  12. There is an EXCEL function NORMINV that does this exactly: The exact value is 2.326, so our approximation of 2.325 isn’t too bad 

  13. (d) What value for m? s = 0.4 s = 1 1% m 0 6.0 2.325

  14. (e) What value for s? s s = 1 1% 5.3 0 6.0 2.325

  15. Example: Kev’s Garage Kev’s 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) .

  16. Area = .2033 Area = .7967 .83 Example: Kev’s Garage • Standard Normal Distribution z = (X -  )/ = (20 - 15)/6 = .83 Z 0

  17. Example: Kev’s Garage Extra If the manager of Kev’s 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 z 0

  18. Perhaps, z = 1.645 ???

  19. Example: Kev’s Garage Extra If the manager of Kev’s 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 z 0 z = 1.645

  20. Example: Kev’s Garage Extra 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 Kev’s should set the reorder point at 25 gallons to keep the probability under .05

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