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Week 2 Lecture
GM 533
Applied Managerial Statistics
Not to be posted or stolen, etc.
without my permission. Students
can download a copy for their own
personal use.
Week 2 GM 533 Lecture
• Third Graders at a local elementary school were
  asked their favorite ice cream of 3 choices.
• The table below summarized their choices.

              Ice Cream    Girls   Boys
              Chocolate       17      19
              Vanilla          5      7
              Strawberry       7      4
Week 2 GM 533 Lecture
• Before answering any questions, let’s get our
  totals.
        Ice Cream    Girls       Boys    Sum

        Chocolate        17         19     36

        Vanilla              5      7      12

        Strawberry           7      4      11

        Sum              29         30     59
Week 2 GM 533 Lecture
• What is the probability we randomly choose a
  girl?

Ice Cream    Girls       Boys    Sum    29 of the total 59 students are
                                        girls so the probability is
Chocolate        17         19     36   29/59 or the decimal form (0.492)
Vanilla              5      7      12

Strawberry           7      4      11

Sum              29         30     59
Week 2 GM 533 Lecture
• What is the probability the student is a girl given
  that the student prefers vanilla?
Ice Cream    Girls       Boys    Sum
                                        It was “given that” the student
Chocolate        17         19     36   prefers vanilla, so we are ONLY
                                        DEALING with the 12 students
Vanilla              5      7      12   who prefer vanilla, so the
                                        probability is 5/12
Strawberry           7      4      11   or the decimal form (0.417)
Sum              29         30     59
Week 2 GM 533 Lecture
• What is the probability the student prefers
  vanilla given the student is a girl?
Ice Cream    Girls       Boys    Sum
                                        It was “given that” the student
Chocolate        17         19     36   is a girl, so we are ONLY
                                        DEALING with the 29 students
Vanilla              5      7      12   who are girls, so the probability is
                                        5/29 or the decimal form (0.172)
Strawberry           7      4      11   NOTE THE DIFFERENCE
                                        FROM THE
Sum              29         30     59   PREVIOUS EXAMPLE
Week 2 GM 533 Lecture
• Bob is playing two games with his family. In one
  game he has a 30% chance of winning, in the
  other, he has a 80% chance of winning. What is
  the probability Bob will lose both games?
Week 2 GM 533 Lecture
• He would have to lose the first game AND the
  second game. The probability of winning was
  given, thus we will use the complement as the
  probability of losing. Remember he has to lose
  both (so we multiply).

      (1 - 0.30)(1 - 0.80) = (0.7)(0.2) = 0.14
Week 2 GM 533 Lecture
• The probability that a component will work is
  95%. If we choose three of these components at
  random, what is the probability they will all
  work?
Week 2 GM 533 Lecture
• This would be

     (0.95)^3 or (0.95)(0.95)(0.95) = 0.857

• The probability that none will work would be

   (0.05)^3 or (0.05)(0.05)(0.05) = 0.000125
Week 2 GM 533 Lecture
• A new product is being manufactured and it has
  been determined the following probability
  distribution holds for its profitability.

           Profit     Probability of Scenario

         ($100,000)            0.25

          $50,000              0.55

         $250,000              0.20
Week 2 GM 533 Lecture
  • The company asks your opinion on the overall
    profitability. You could consider the - $100K as
    the cost to produce the new product.

(-100000)(0.25) + (50,000)(0.55) + (250,000)(0.20) = $52,500

It seems we will realize a profit of $52,500, we might need other
parameters or guidance.
Week 2 GM 533 Lecture
• A staffing company estimates that 90% of their
  placements last at least 4 weeks. Looking at a
  random sample of 17 placements, calculate the
  mean number of placements that will stay at
  least 4 weeks.
Week 2 GM 533 Lecture
• A staffing company estimates that 90% of their
  placements last at least 4 weeks. Looking at a
  random sample of 17 placements, calculate the
  mean number of placements that will stay at
  least 4 weeks.

               np = 17(0.9) = 15.3
Week 2 GM 533 Lecture
• Your product works 98% of the time when it hits
  the market. If you randomly call 15 of your new
  customers, what is the probability that at least 13
  of your new customers will have had a good first
  experience with your new product?
Week 2 GM 533 Lecture
• This screams “Binomial”
• You can either work these in Minitab or Excel
 ▫ I will show you both
Week 2 GM 533 Lecture
• In Minitab
 ▫ Enter “x” in gray box
   below C1 and P(x) in gray
   box below C2
 ▫ Fill in cells under “x” with
   numbers 0 through 15
Week 2 GM 533 Lecture
• Go to Calc >> Probability Distributions >> Binomial
• Enter 15 for number of
trials
• Enter 0.98 for Event
probability
• Select C1 (x) for Input
column and C2 (P(x)) for
Optional storage
Make sure the radial
Button is set to “Probability
Week 2 GM 533 Lecture
• Click OK and you
  will now have see
  the probabilities
  under the P(x)
  column
• For example, the
  P(14) = 0.226093
Week 2 GM 533 Lecture
• For our problem,
  we wanted to
  know the
  probability of “at
  least 13” so we
  will add the P(13)
  + P(14) + P(15)
Week 2 GM 533 Lecture
• So P(x≥13) =
  0.032299 +
  0.226093 +
  0.738569
which is
0.996961
• An alternate way is to
  use “Cumulative
  probability” and enter
  “12” for the Input
  constant… This gives
  you the probability of
  12 or less, we could
  then subtract the
  result from 1 to get
  the probability of 13
  or more
Week 2 GM 533 Lecture
• 1 – 0.003094 =
  0.996961
• (The same result we
  got)
Week 2 GM 533 Lecture
• Don’t be afraid, these are easier than what you
  think
Week 2 GM 533 Lecture
• On these type problems, I actually use an Excel
  template that I will be happy to share with you
Week 2 GM 533 Lecture
• You can find the binomial template and other
  cool statistical templates at
http://highered.mcgraw-
  hill.com/sites/0070620164/student_view0/exc
  el_templates.html

Just download the file to your computer and go to
 the “Review” tab at the top of Excel and select
 “Unprotect Sheet” (this allows you to use your
 own data, etc.)
Week 2 GM 533 Lecture
• On the following problems, I am going to use the
  Excel Template
• It is a matter of preference for me
• You will need to ask your instructor
Week 2 GM 533 Lecture
• Paula’s Pizza has a 90% chance of delivering
  their pizzas in under 37 minutes. Out of 11
  deliveries, what is the probability that fewer than
  9 pizzas will be delivered within 37 minutes?
Week 2 GM 533 Lecture
• Using Binomial Template


“Fewer than 9”
would be the
same as
“At most 8” so
the probability
would be
0.0896
Week 2 GM 533 Lecture
• A quiz has 20 multiple choice questions with
  four possible answers for each question. If only
  one of the answers is correct and a student
  guesses on all questions, what is the probability
  that the student will get at least half of the
  questions correct?
Week 2 GM 533 Lecture
• Use Binomial Template


At least half
would be at least
10 of the 20
questions.
The answer would
be 0.0139
Not too hot eh?
Week 2 GM 533 Lecture
• I post other helpful information on my Statcave
  site at http://www.facebook.com/statcave
• You DO NOT have to be a Facebook
  person to see this
• It is convenient for me to post there and it is just
  for fun. It is NOT required to go to there.

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Gm533 week 2 lecture sep 2012

  • 1. Week 2 Lecture GM 533 Applied Managerial Statistics Not to be posted or stolen, etc. without my permission. Students can download a copy for their own personal use.
  • 2. Week 2 GM 533 Lecture • Third Graders at a local elementary school were asked their favorite ice cream of 3 choices. • The table below summarized their choices. Ice Cream Girls Boys Chocolate 17 19 Vanilla 5 7 Strawberry 7 4
  • 3. Week 2 GM 533 Lecture • Before answering any questions, let’s get our totals. Ice Cream Girls Boys Sum Chocolate 17 19 36 Vanilla 5 7 12 Strawberry 7 4 11 Sum 29 30 59
  • 4. Week 2 GM 533 Lecture • What is the probability we randomly choose a girl? Ice Cream Girls Boys Sum 29 of the total 59 students are girls so the probability is Chocolate 17 19 36 29/59 or the decimal form (0.492) Vanilla 5 7 12 Strawberry 7 4 11 Sum 29 30 59
  • 5. Week 2 GM 533 Lecture • What is the probability the student is a girl given that the student prefers vanilla? Ice Cream Girls Boys Sum It was “given that” the student Chocolate 17 19 36 prefers vanilla, so we are ONLY DEALING with the 12 students Vanilla 5 7 12 who prefer vanilla, so the probability is 5/12 Strawberry 7 4 11 or the decimal form (0.417) Sum 29 30 59
  • 6. Week 2 GM 533 Lecture • What is the probability the student prefers vanilla given the student is a girl? Ice Cream Girls Boys Sum It was “given that” the student Chocolate 17 19 36 is a girl, so we are ONLY DEALING with the 29 students Vanilla 5 7 12 who are girls, so the probability is 5/29 or the decimal form (0.172) Strawberry 7 4 11 NOTE THE DIFFERENCE FROM THE Sum 29 30 59 PREVIOUS EXAMPLE
  • 7. Week 2 GM 533 Lecture • Bob is playing two games with his family. In one game he has a 30% chance of winning, in the other, he has a 80% chance of winning. What is the probability Bob will lose both games?
  • 8. Week 2 GM 533 Lecture • He would have to lose the first game AND the second game. The probability of winning was given, thus we will use the complement as the probability of losing. Remember he has to lose both (so we multiply). (1 - 0.30)(1 - 0.80) = (0.7)(0.2) = 0.14
  • 9. Week 2 GM 533 Lecture • The probability that a component will work is 95%. If we choose three of these components at random, what is the probability they will all work?
  • 10. Week 2 GM 533 Lecture • This would be (0.95)^3 or (0.95)(0.95)(0.95) = 0.857 • The probability that none will work would be (0.05)^3 or (0.05)(0.05)(0.05) = 0.000125
  • 11. Week 2 GM 533 Lecture • A new product is being manufactured and it has been determined the following probability distribution holds for its profitability. Profit Probability of Scenario ($100,000) 0.25 $50,000 0.55 $250,000 0.20
  • 12. Week 2 GM 533 Lecture • The company asks your opinion on the overall profitability. You could consider the - $100K as the cost to produce the new product. (-100000)(0.25) + (50,000)(0.55) + (250,000)(0.20) = $52,500 It seems we will realize a profit of $52,500, we might need other parameters or guidance.
  • 13. Week 2 GM 533 Lecture • A staffing company estimates that 90% of their placements last at least 4 weeks. Looking at a random sample of 17 placements, calculate the mean number of placements that will stay at least 4 weeks.
  • 14. Week 2 GM 533 Lecture • A staffing company estimates that 90% of their placements last at least 4 weeks. Looking at a random sample of 17 placements, calculate the mean number of placements that will stay at least 4 weeks. np = 17(0.9) = 15.3
  • 15. Week 2 GM 533 Lecture • Your product works 98% of the time when it hits the market. If you randomly call 15 of your new customers, what is the probability that at least 13 of your new customers will have had a good first experience with your new product?
  • 16. Week 2 GM 533 Lecture • This screams “Binomial” • You can either work these in Minitab or Excel ▫ I will show you both
  • 17. Week 2 GM 533 Lecture • In Minitab ▫ Enter “x” in gray box below C1 and P(x) in gray box below C2 ▫ Fill in cells under “x” with numbers 0 through 15
  • 18. Week 2 GM 533 Lecture • Go to Calc >> Probability Distributions >> Binomial • Enter 15 for number of trials • Enter 0.98 for Event probability • Select C1 (x) for Input column and C2 (P(x)) for Optional storage Make sure the radial Button is set to “Probability
  • 19. Week 2 GM 533 Lecture • Click OK and you will now have see the probabilities under the P(x) column • For example, the P(14) = 0.226093
  • 20. Week 2 GM 533 Lecture • For our problem, we wanted to know the probability of “at least 13” so we will add the P(13) + P(14) + P(15)
  • 21. Week 2 GM 533 Lecture • So P(x≥13) = 0.032299 + 0.226093 + 0.738569 which is 0.996961
  • 22. • An alternate way is to use “Cumulative probability” and enter “12” for the Input constant… This gives you the probability of 12 or less, we could then subtract the result from 1 to get the probability of 13 or more
  • 23. Week 2 GM 533 Lecture • 1 – 0.003094 = 0.996961 • (The same result we got)
  • 24. Week 2 GM 533 Lecture • Don’t be afraid, these are easier than what you think
  • 25. Week 2 GM 533 Lecture • On these type problems, I actually use an Excel template that I will be happy to share with you
  • 26. Week 2 GM 533 Lecture • You can find the binomial template and other cool statistical templates at http://highered.mcgraw- hill.com/sites/0070620164/student_view0/exc el_templates.html Just download the file to your computer and go to the “Review” tab at the top of Excel and select “Unprotect Sheet” (this allows you to use your own data, etc.)
  • 27. Week 2 GM 533 Lecture • On the following problems, I am going to use the Excel Template • It is a matter of preference for me • You will need to ask your instructor
  • 28. Week 2 GM 533 Lecture • Paula’s Pizza has a 90% chance of delivering their pizzas in under 37 minutes. Out of 11 deliveries, what is the probability that fewer than 9 pizzas will be delivered within 37 minutes?
  • 29. Week 2 GM 533 Lecture • Using Binomial Template “Fewer than 9” would be the same as “At most 8” so the probability would be 0.0896
  • 30. Week 2 GM 533 Lecture • A quiz has 20 multiple choice questions with four possible answers for each question. If only one of the answers is correct and a student guesses on all questions, what is the probability that the student will get at least half of the questions correct?
  • 31. Week 2 GM 533 Lecture • Use Binomial Template At least half would be at least 10 of the 20 questions. The answer would be 0.0139 Not too hot eh?
  • 32. Week 2 GM 533 Lecture • I post other helpful information on my Statcave site at http://www.facebook.com/statcave • You DO NOT have to be a Facebook person to see this • It is convenient for me to post there and it is just for fun. It is NOT required to go to there.