3.2) Area under the curve, Part II Mean=0 and SD=1 a) Z>-1.13

x<-pnorm(-1.13, mean=0, sd=1)
y<-1-x
y
## [1] 0.8707619
mean=0
sd=1
x<-seq(-3.5,3.5,length=100)*sd + mean
y<-dnorm(x,mean,sd)
plot(x, y, type="l")
polygon(c( x[x>=-1.13], -1.13 ),  c(y[x>=-1.13],0 ), col="red")

  1. Z>1.48
x<-pnorm(0.18, mean=0, sd=1)
x
## [1] 0.5714237
mean=0
sd=1
x<-seq(-3.5,3.5,length=100)*sd + mean
y<-dnorm(x,mean,sd)
plot(x, y, type="l")
polygon(c( x[x<=0.18], 0.18 ),  c(y[x<=0.18],0 ), col="red")

  1. Z>8 Pretty much zero
x<-pnorm(8, mean=0, sd=1)
y<-1-x
y
## [1] 6.661338e-16
mean=0
sd=1
x<-seq(-3.5,3.5,length=100)*sd + mean
y<-dnorm(x,mean,sd)
plot(x, y, type="l")

  1. |Z|>0.5
x<-pnorm(0.5, mean=0, sd=1)
y<-pnorm(-0.5, mean=0, sd=1)
z<-x-y
z
## [1] 0.3829249
cord.x <- c(-0.5,seq(-0.5,0.5,0.01),0.5) 
 cord.y <- c(0,dnorm(seq(-0.5,0.5,0.01)),0) 
 curve(dnorm(x,0,1),xlim=c(-3,3),main='Standard Normal') 
 polygon(cord.x,cord.y,col='red')

3.4) Triathalon Times, Part I (a) Write down the short-hand for these two normal distributions. X is the finishing time for males. X~N(mean=4314, sd=583)

Y is the finishing time for females Y~N(mean=5261, sd=807)

  1. What are the Z-scores for Leo’s and Mary’s finishing times? What do these Z-scores tell you?
#Leo
Z<-(4948-4313)/583
Z
## [1] 1.089194
#Mary
Z<-(5513-5261)/807
Z
## [1] 0.3122677

  1. Did Leo or Mary rank better in their respective groups? Explain your reasoning. Leo ranks better than Mary since he has a greater Z score.

  2. What percent of the triathletes did Leo finish faster than in his group?

x<-pnorm(1.0892, mean=0, sd=1)*100
x
## [1] 86.19672
  1. What percent of the triathletes did Mary finish faster than in her group?
x<-pnorm(0.3123, mean=0, sd=1)*100
x
## [1] 62.25937
  1. If the distributions of finishing times are not nearly normal, would your answers to parts
      1. change? Explain your reasoning. If finishing times are not normal, there would be no change for the Z scores. This is due to the fact that Z scores do not depend on what the distribution of the finishing times. We are unable to answer parts d and e because normaility is required for percentiles.

3.18) Heights of College Students (a) The mean height is 61.52 inches with a standard deviation of 4.58 inches. Use this information to determine if the heights approximately follow the 68-95-99.7% Rule. (61.62-4.58, 61.52+4.58)=(56.94, 66.1) At 68% (61.52-24.58, 61.52+24.58)=(52.36, 70.68) At 95% (61.52-33.58, 61.52+34.58)=(47.78, 75.26) We can conclude that the data follows the 68-95-99 rule

  1. Do these data appear to follow a normal distribution? Explain your reasoning using the graphs provided below Although there is evidence of some outliers, the data follows an approxmiate normal distribution in both plots.

3.22) Defective Rate (a) What is the probability that the 10th transistor produced is the first with a defect?

P=(1-.30)^9 *(.30)
P
## [1] 0.01210608
  1. What is the probability that the machine produces no defective transistors in a batch of 100?
P2=(1-0.02)^100
P2
## [1] 0.1326196
  1. On average, how many transistors would you expect to be produced before the first with a defect? What is the standard deviation?
ub=1/(0.02)
ub
## [1] 50
sdb<-sqrt((1-0.02)/(0.02)^2)
sdb
## [1] 49.49747

We expect 50 transistors and have a SD of 49.49747

  1. Another machine that also produces transistors has a 5% defective rate where each transistor is produced independent of the others. On average how many transistors would you expect to be produced with this machine before the first with a defect? What is the standard deviation?
ud<-1/(0.05)
ud
## [1] 20
sdd<-sqrt((1-0.05)/(0.05)^2)
sdd
## [1] 19.49359
  1. Based on your answers to parts (c) and (d), how does increasing the probability of an event affect the mean and standard deviation of the wait time until success? The higher the probability, then the mean and the standard deviation of the waiting time will decrease.

3.38) Male Children (a) Use the binomial model to calculate the probability that two of them will be boys.

probabilityOfBoy <- 0.51 
k <- 2 
n <- 3

factorialOfn <- factorial(n)
factorialOfk <- factorial(k)
factorialOfnminusk <- factorial(n-k)

probabilityOf2boysOf3 <- ( factorialOfn / (factorialOfk * factorialOfnminusk) ) * probabilityOfBoy^k * (1-probabilityOfBoy)^(n-k)
probabilityOf2boysOf3
## [1] 0.382347
  1. Write out all possible orderings of 3 children, 2 of whom are boys. Use these scenarios to calculate the same probability from part (a) by using the addition rule for disjoint outcomes. Confirm that your answers from parts (a) and (b) match. B=Boy, G=Girl GBG BGB BBG
probabilityOf2boysOf3_B <- ((1-probabilityOfBoy) * probabilityOfBoy * probabilityOfBoy) * 3
probabilityOf2boysOf3_B
## [1] 0.382347
  1. If we wanted to calculate the probability that a couple who plans to have 8 kids will have 3 boys, briefly describe why the approach from part (b) would be more tedious than the approach from part (a). Approach b would be more tedious because the number of possible permutations increases. In part a, there were only 3 possible orderings of two boys and a girl but with 8 children, it is not longer practical to list them all out.

3.42) Serving in Volley Ball other. (a) What is the probability that on the 10th try she will make her 3rd success

n <- 10    
k <- 3     
p <- 0.15  
tenth <- factorial(n - 1) / (factorial(k-1) * (factorial(n - k))) * p^k * (1-p)^(n-k)
tenth
## [1] 0.03895012

  1. Suppose she has made two successful serves in nine attempts. What is the probability that her 10th serve will be successful? She has a 0.15 probability of successful serve

  2. Even though parts (a) and (b) discuss the same scenario, the probabilities you calculated should be different. Can you explain the reason for this discrepancy? The probabilities are different because we are not given information regarding successes as we are in part b.