gm533 week 2 lecture march 2012

Week 2 LectureGM 533 Applied Managerial StatisticsNot 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?


Chocolate 17 19 36

Vanilla 5 7 12

Strawberry 7 4 11

Sum 29 30 59

29 of the total 59 students are girls so the probability is29/59 or the decimal form (0.492)


•What is the probability the student is a girl given that the student prefers vanilla?


Chocolate 17 19 36

Vanilla 5 7 12

Strawberry 7 4 11

Sum 29 30 59

It was “given that” the studentprefers vanilla, so we are ONLY DEALING with the 12 students who prefer vanilla, so the probability is 5/12 or the decimal form (0.417)


•What is the probability the student prefers vanilla given the student is a girl?


Chocolate 17 19 36

Vanilla 5 7 12

Strawberry 7 4 11

Sum 29 30 59

It was “given that” the studentis a girl, so we are ONLY DEALING with the 29 students who are girls, so the probability is5/29 or the decimal form (0.172)NOTE THE DIFFERENCE FROM THEPREVIOUS EXAMPLE


•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?


•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


•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?


•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 otherparameters or guidance.


•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.


•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?


•This screams “Binomial”•You can either work these in Minitab or

Excel▫I will show you both


•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 oftrials• Enter 0.98 for Event probability• Select C1 (x) for Inputcolumn and C2 (P(x)) for Optional storageMake sure the radial Button is set to “Probability


•Click OK and you will now have see the probabilities under the P(x) column

•For example, the P(14) = 0.226093


•For our problem, we wanted to know the probability of “at least 13” so we will add the P(13) + P(14) + P(15)


•So P(x≥13) = 0.032299 + 0.226093 + 0.738569

which is0.996961

• 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


•1 – 0.003094 = 0.996961

•(The same result we got)


•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

0.9970


•You can find the binomial template and other cool statistical templates at

http://highered.mcgraw-hill.com/sites/0070620164/student_view0/excel_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.)





•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


•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” sothe probabilitywould be0.0896


•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 halfwould be at least10 of the 20questions.The answer wouldbe 0.0139Not too hot eh?


•I will post a link to these charts 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.

http://www.facebook.com/statcave

gm533 week 2 lecture march 2012

Education