3.2

par(mfrow=c(2,2))
normalPlot(0,1, bounds=c(-1.3,5))
normalPlot(0,1, bounds=c(-5,.18))
normalPlot(0,1, bounds=c(8,100))
normalPlot(0,1, bounds=c(-.5,.5))

3.4
a) men: N(4313,583) women: N(5261,807)
b) The scores tell you where they rank amongst their respective groups

LeoZ <- (4948-4313)/583
MaryZ <- (5513-5261)/807
print(LeoZ)

## [1] 1.089194

print(MaryZ)

## [1] 0.3122677

3. Mary ranked better in her respective group because, while they took longer than the mean, Mary was much closer to the mean of her group - and thus was in a higher percentile.
   
4. Leo finished faster than: 1 - .8621 = 0.138 = 13.8%
5. Mary finished faster than: 1 - .6217 = 0.378 = 37.8%
6. If finish times were not nearly normal, then the zscores would not change. The answers for c-e would change because the density and percentiles would be different.

3.18

heights <- c(54, 55, 56, 56, 57, 58, 58, 59, 60, 60, 60, 61, 61, 62, 62, 63, 63, 63, 64, 65, 65, 67, 67, 69, 73)
qqnormsim(heights)

1. Yes, the data appers to follow the 68-95-99.7 rule
   
2. The data appears to follow the normal distribution as the QQplot shows that the datapoints align fairly linear

3.22

#a)  Probability of 10th transistor is first with defect
.98^9*.02^1

## [1] 0.01667496

#b)  Probability of no defective transistors in batch of 100
.98^100

## [1] 0.1326196

#c) On avg, how many produced before a defect?
1/.02

## [1] 50

#d)  Another machine has 5% rate, how many before defect?
1/.05

## [1] 20

#d)  std. dev?
sqrt(.95/.05^2)

## [1] 19.49359

5. The changes make the mean of the wait-time go down such that a defect will occur comparatively much sooner. The standard deviation similarly decreases.

3.38

n<-3
k<-2
p<-.51
#a)  two will be boys
factorial(n)/factorial(k)*factorial(n-k) * p^k * (1-p)^(n-k)

## [1] 0.382347

2. bbg = (.51)(.51)(.49) = 0.127449
   bgb = (.51)(.49)(.51) = 0.127449
   gbb = (.49)(.51)(.51) = 0.127449
   0.127449 + 0.127449 + 0.127449 = 0.382347

3. The binomial approach will be much more tedious because you will have to calculate and sum the proabilities of each possible combination meeting the criteria, with each combination requiring you to multiply 8 probabilities.

3.42

p<-.15
n<- 10
k<- 3
factorial(n-1)/(factorial(k-1)*factorial((n-1)-(k-1))) * p^k * (1-p)^(n-k)

## [1] 0.03895012

2. since it is independent, the 10th serve has a probability of .15
3. They are different because part a is asking for the probability of a specific combination of successful serves, part b is asking the probability of the 10th serve being successful. Since all serves are independent, the probability of any individual serve is .15.