Graded Homework. P. 163, #29 P. 170, #35, 37. Graded Homework, cont. P. 163, #29 U. Of Pennsylvania 1,033 admitted early (E) 854 rejected outright (R) 964 deferred (D) Typically 18% of deferred early admission are admitted in the regular admission process (173.5)
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P. 163, #29
P. 170, #35, 37
P(E) = 1033/2851 = 0.362
P(R) = 854/2851 = 0.300
P(D) = 964/2851 = 0.338
Yes, a student cannot be both admitted and deferred, so P(E∩D)=0
1033/2375 = 0.435
1033/2851 + (964/2851)(.18) = 0.423
P. 170, 35
P. 170, 35
P(Manager|Female) = 0.17/0.46 = 0.37
P(Precision production|Male) = 0.11/0.54 = 0.20
No, P(Manager|Female) = 0.37, P(Manager) = 0.34
P. 170, 37
P(PC) = .37P(Y) = .14
P(Y|PC) = .19P(O|PC) = .81
P(PC|Y) = P(Y ∩ PC)/P(Y) = [P(PC)P(Y|PC)]/P(Y) = [(.37)(.19)]/(.14) = 0.5
P(PC|O) = P(O ∩ PC)/P(O) = [P(PC)P(O|PC)]/P(O) = [(.37)(.81)]/(.86) = 0.35
People under 24 years old are more likely to use credit cards.
Yes, otherwise they can’t establish a credit history and the companies want customers who will make heavy use of the cards. They could put strict limits on the maximum balance for the card.
P(A ∩ B) = P(B)P(A|B)
P(A ∩ B) = P(A)P(B|A)
If events A and B are independent then P(A|B) = P(A)P(B). In this special case the multiplication rule reduces from:
P(A ∩ B) = P(B)P(A|B)
P(A ∩ B) = P(B)P(A)
Assume we take the four aces out of a deck of cards and we draw twice with replacement:
Are A and B statistically independent in this case?
If we sample without replacement the outcomes will not be statistically independent.
However, if we are drawing from a large population the change in probability will be so small we can treat the draws as being statistically independent.
A technique used to modify a probability given additional information.
Assume that 10% of the population has a disease. Assume there is a test to see if someone has the disease but it is not very accurate.
Assume we want to calculate the probability that someone has the disease if the test says they have the disease.
A1 = Has the disease
B = Test says the patient has the disease
P(A1) = .1P(A2) = .9
P(B|A1) = .8P(B|A2) = .3
Assume that 40% of a company’s parts are produced in Boston and 60% are produced in Chicago. Also assume that 20% of the parts produced in Boston are defective, and 10% of the parts produced in Chicago are bad.
A randomly chosen part is defective. Use Bayes Theorem to find the probability the part came from Boston.
A1 = Boston
B = Part is defective
P(A1) = .4P(A2) = .6
P(B|A1) = .2P(B|A2) = .1
Given k steps (or rounds) in an experiment and ni possible outcomes at step i, the total number of possible outcomes is:
N! = (N)(N-1)(N-2)…(2)(1)
5! = (5)(4)(3)(2)(1) = 120
What is the value of 4! ?
(4)(3)(2)(1) = 24
What is the value of 5!/3! ?
[(5)(4)(3)(2)(1)]/[(3)(2)(1)] = (5)(4) = 20
By definition 0! = 1
The number of permutations of N objects taken n at a time:
Assume a broker is going to pick 3 stocks from a pool of 10 stocks. Also assume he will invest 60% of his money in one stock, 30% in another, and 10% in another. How many portfolios can be constructed?
The number of combinations of N objects taken n at a time:
A bank is constructing a bond based on mortgages. It is going to base the bond from four mortgages, it has ten mortgages to choose from. How many ways can the bond be structured?
Three employees will be chosen from an office of 8 workers for a committee to evaluate a new production technique. How many possible committees could be formed?
Assume a club has 5 members and they are going to elect a president, treasurer, and secretary. How many ways can the offices be filled?
A magazine is going to recommend two of ten products to its readers. It will identify the rankings of the two products that are selected. How many potential rankings are there?
P. 151, 1, 3 + redo 3 assuming order is important (counting rules)
P. 169, 31 (Multiplication rule)
P. 177, 43 (Bayes’ theorem)