HW4 Grading Key (Biostat) Assigned 9/20; due 9/27; quick grading as follows: A score of 5 will represent HW done neatly with diligence. A score of 4 will represent a "sloppy but completed work." A score of 3 will represent "incomplete work." A score of 2 will represent late work. I hope I do not see any 1s and 0s. 

4.3 February birthdays See online key ["Key to odd exercises link" is always active]
4.4 Survival following treatment
(A) The estimated probability is the relative frequency in the sample. This is equal to 475 � 601 = .7903.
(B) The estimated probability of not surviving is the complement of surviving. This is equal to 1 - .7903 = .2097.
(C) These estimated probabilities are based on observations (" experience"). Note that this type of estimation is not as reliable as probabilities based on logic. They are, however, are a practical way to get rough estimates of probabilities. 
4.5 Random sampling a small population, N = 26 See online key
4.9 Lifetime breast cancer incidence See online key
4.10 Smoking on campus  Let X represent the number of smokers in the sample X~b(2, .2)
(A) Probability mass function (pmf)
Pr(X = 0) = (2C0)(.20)(.82-0) = (1)(1)(.6400) = .6400
Pr(X = 1) = (2C1)(.21)(.82-1) = (2)(.2)(.8) = .3200
Pr(X = 2) = (2C2)(.22)(.82-2) = (1)(.04)(1) = .0400 
(B) Pr(X = 2) = .04, as calculated above. This means that we would seldom see 2 smokers in a sample .04 (i.e., 4%) of the time if the underlying assumptions are true (assumptions: independent trial, each with p = .2).
4.17 Standard Normal drill See online key
4.18 Standard Normal drill 2
(A)
Pr(Z < -1.42) = .0778
(B) Pr(Z < 1.25) = .8944
(C)  Pr(-1.42 < Z < 1.25) .8944 - .0778 = .8166 [Draw curve on board]
4.20A Z percentiles z.10 = -1.28  Notes: (a) Draw curve to demonstrate relation between cumulative probability and z score (b) It is helpful to develop the language to discuss standard Normal distributions: In words, "the 10th percentile on a standard Normal curve is negative one point two eight.") 
4.21A Hospital stay duration See online key
4.22 Weight of Alzheimer brains   Let X represent the weight of Alzheimer brains: X~N(1076.80, 105.76)
(A) Pr(X < 900) = ?
Standardize a value of 900: z = (900 - 1076.8) / 105.76 = -1.67. This means that this value 900 is 1.67 standard deviations below the population mean.
Draw curve and shade relevant area: Pr(X < 900) = Pr(Z < -1.67) 
Look up cumulative probability in Z table: Pr(Z < -1.67)  = .0475 
(B) Pr(X > 1200) = ?
Standardize a value of 1200: z = (1200 - 1076.8) / 105.76 = 1.16. This means that the value 1200 is 1.16 standard deviations above the population mean.
Draw curve and shade relevant area: Pr(X > 1200) = Pr(Z > 1.16) = .1230 [Note that this is a right tail probability, not a cumulative probability: Pr(Z > 1.16) = 1 - Pr(Z < 1.16) = 1 - .8770 = .1230 by law of complements.