# Welcome to PMBA0608: Economics/Statistics Foundation Fall 2006 Session 7: September 30.

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Welcome to PMBA0608: Economics/Statistics Foundation Fall 2006 Session 7: September 30

How was the exam?  Any questions of exam questions?

Next Class  October 18  I will be in Eastern campus Local students can meet with me at 7:00 pm (half hour before class)  Study Chapters 3 and 4 of Stat book  Study Chapter 5 of Econ book  Send me your questions

Assignment 3 (Due on or before October 14) 1.Application 3.17, Page 110 of Stat 2.Application 3.19, Page 110 of Stat 3.Application 3.27, Page 115 of Stat 4.Exercise 3.31, Page 123 of Stat 5.Application 3.33, Page 123 of Stat

Stat Book: Section 3.4 Let’s spin once P (1 given that red has occurred) = ? P (1 given that red has occurred) = ½= 0.5 P (1\red)= P (1 and red) / P (red) P (1\red)= 0.25 / 0.5 = 0.5 This is called conditional probability

Let’s practice  In Europe, 88% of all households have a television. 51% of all households have a television and a VCR. What is the probability that a household has a VCR given that it has a television? 1.173% 2.58% 3.42% 4.None of the above.

Where did 58% come from?  P (TV) = 0.88  P (TV & VCR) = 0.51  P (VCR\TV) = 0.51/0.88  P (VCR\TV) =0.58 or 58%

Let’s spin twice P (1 on second spin given that red has occurred on the first spin) = ? P (1 on second spin given that red has occurred on the first spin) = ¼ = 0.25 1 on the second spin is independent from red on the first spin so P ( 1 on second spin\red has occurred on the first spin ) = P (1)

Let’s play cards now  A card is chosen at random from a deck of 52 cards. It is then replaced and a second card is chosen. What is the probability of choosing a jack the second time given that we chose an eight the first time?  P(jack\8) = ?  P (jack\8) = P (jack)= 4/52

Unions versus Intersections  P (AB) = intersection of A and B = Probability of A and B happening simultaneously  P (AB) = P (A) P (B\A)

OK ready to practice?  100 individuals  50 male and 20 of them smoke  P (male & smoker) = ?  P (male & smoker) = P (male)* P (smoker/male)  P (male & smoker) = ½ * 2/5 = 2/10 =0.2 or 20%

Unions versus Intersections  P (AUB) = union = Probability of A or B or both  P (AUB) = P (A) + P (B) – P (AB)

And…. Let’s practice  100 individuals  50 male and 20 of them smoke  50 female and 10 of them smoke  P (male or smoker) =?  P (male or smoker) = P (male) + P (smoker) – P (male and smoker)  P (male or smoker) = 0.5 + 0.3 - 0.2 = 0.6 or 60%

What is P(AUB) now?  A and B are mutually exclusive  P (AUB) = P (A) + P (B)

Please work on this problem with a partner  1% of women at age forty who participate in routine screening have breast cancer. 80% of women with breast cancer will get positive mammographies. 9.6% of women without breast cancer will also get positive mammographies. A woman in this age group had a positive mammography in a routine screening. What is the probability that she actually has breast cancer?

Here is the problem  P (cancer) = 0.01  P (positive \cancer) = 0.8  P (positive \no cancer) =0.096  P (cancer\positive) = ?  If we use conditional probability P (cancer\positive) = P (cancer and positive)/P (positive)

Let’s draw a map. Cancer No Cancer P =0.99 P = 0.01 Positive, P = 0.8 Negative, P =0.2 Positive, P=0.096 Negative, P = 0.904  P (cancer and positive) = 0.01 * 0.8 = 0.008  P (positive) = P (cancer and positive) + P (no cancer and positive) = 0.008 + 0.095=0.103

Now let’s plug this into the conditional probability formula  P (cancer\positive) = P (cancer and positive)/P (positive)  P (cancer\positive) = 0.008/0.103  P (cancer\positive) =0.078  This is Bayes’ rule  If your mamo result is positive, there is only 7.8% chance that you have breast cancer Not bad ha?

Are you confused? Let’s put it differently  10,000 women  Group 1: 100 women with breast cancer. (1%) 80% (80) positive 20% (20) negative  Group 2: 9,900 women without breast cancer. 9.6% (almost 950)positive mamo The rest (8950) negative mamo

So, we have 4 groups of women  Group A: 80 women with breast cancer, and a positive mammography.  Group B: 20 women with breast cancer, and a negative mammography.  Group C: 950 women without breast cancer, and a positive mammography.  Group D: 8,950 women without breast cancer, and a negative mammography.

What is P (cancer\positive)?  P (cancer\positive)= Number of women with cancers/ total number of women with positive tests  P (cancer\positive)= 80/ (80+950) = 0.078 or 7.8%

Baye’s Rule P (cancer\ positive) = p (positive\cancer)*p (cancer) divided by p (positive\cancer)*p (cancer) + p (positive\no cancer)*p (no cancer) P (cancer\positive) = (0.8 * 0.01)/ (0.8 * 0.01) + (0.096* 0.99) P (cancer\positive)= 0.008/(0.008 +0.095) = 0.078

What is a random variable?  Value of it depends on the outcome of an experiment  Example The rate of return on the portfolio of your stocks is a random variable Let’s call that r Is r discrete or continuous? It is continuous

Let’s think of a discrete random variable  Let’s suppose that there are only 3 possible outcomes for the return on your stock portfolio: \$0, \$100, and \$150 Now your return R is a discrete variables  Now suppose that there is 50% chance that you make \$100 and 25% chance that you make \$150. What are the chances that you make \$0? 25% R is a discrete random variable  1≥P (R) ≥0  ΣP (R) = 1

Question: What is your expected return RP (R) \$00.25 \$1000.5 \$1500.25 E(R) = μ = ΣR * P (R) E(R) = (0 * 0.25) + (100 * 0.5) + (150* 0.25) E(R) =\$87.5 Note: don’t call this average Average is for certain outcomes Expected value is for uncertain outcomes

Will you always make \$87.5?  No  If you repeat this investment an infinite number of times on average you will make \$87.5  But each time you may make less or more  So there is a distribution of returns

Variance and standard deviation of distribution of returns  Variance σ 2 = Σ (R – μ) 2 P (R) What is it in our problem?  Standard deviation = square root of variance

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