3.2 Area under the curve, Part II

a.) Z > -1.13
P(Z> -1.13 )=1-P(Z< -1.13 )

1-pnorm(-1.13,mean=0,sd=1)
## [1] 0.8707619
  1. Z < 0.18
pnorm(0.18)
## [1] 0.5714237
  1. Z > 8
    P(Z > 8 )=1-P(Z < 8 ) = 0
1-pnorm(8)
## [1] 6.661338e-16
  1. |Z| < 0.5
    P(|Z|<0.5)=P(Z< 0.5 ) -P(Z < ???0.5 )
pnorm(0.5) - pnorm(-0.5)
## [1] 0.3829249

3.4 Triathlon times, Part I

a.Write down the short-hand for these two normal distributions.
Women: N(??=5261,??=807) Men: N(??=4313,??=583)

  1. What are the Z-scores for Leo’s and Mary’s finishing times? What do these Z-scores tell you?
    Mary: Z=(5513-5261) / 807 =0.3122677 Leo: z=(4948-4314) / 583= 1.089194 A Z-score is a numerical measurement of a value’s relationship to the mean.Leo and Mary is 1.089 and 0.312 sd above mean respectively.

c.Did Leo or Mary rank better in their respective groups? Explain your reasoning.
z vaule of mary is lower than leo’s, so Mary did better in the competetion.

  1. What percent of the triathletes did Leo finish faster than in his group?
    we are looking for z_leo>1.089
    1-P(z_leo<1.089)
1-pnorm(1.089)
## [1] 0.1380769
  1. What percent of the triathletes did Mary finish faster than in her group?
    find Z_Mary>0.312 P(Z>0.312)=1???P(Z<0.312)
1-pnorm(0.312)
## [1] 0.3775203
  1. If the distributions of finishing times are not nearly normal, would your answers to parts
      1. change? Explain your reasoning.
        Yes, sinece we use z value to compare it is all base on normal distrubution.

3.18 Heights of female college students.

  1. 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.
    yes, it doese follow the rule.
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)
heights.df <- as.data.frame(heights)
heights.mean <- mean(heights.df$heights)
heights.sd <- sd(heights.df$heights)
percent.vec <- vector()
for (i in 1:3) {
  percent.vec[i] <- nrow(subset(heights.df, heights > heights.mean - i*heights.sd & heights < heights.mean + i*heights.sd))/nrow(heights.df)*100
  cat(i,"standard deviations =",percent.vec[i],"%\n")
}
## 1 standard deviations = 68 %
## 2 standard deviations = 96 %
## 3 standard deviations = 100 %
  1. Do these data appear to follow a normal distribution? Explain your reasoning using the graphs provided below.

Yes, most of point fall on the line and the data graph look alike with other sim graph.

qqnormSim(heights)

3.22 Defective rate.

a.What is the probability that the 10th transistor produced is the first with a defect?

dgeom(9,0.02)
## [1] 0.01667496

b.What is the probability that the machine produces no defective transistors in a batch of 100?

(1-0.02)^100
## [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?
    mean=50, sd=0.144
df_rate=0.02
mean =1/df_rate
mean
## [1] 50
sd = sqrt((df_rate)/(1-df_rate)^2)
sd
## [1] 0.1443075
  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?
    mean=20, sd=0.235
df_rated <- 0.05
mean_d <- 1/df_rated
sd_d <-sqrt((df_rated)/(1-df_rated)^2)
mean_d
## [1] 20
sd_d
## [1] 0.2353756
  1. Based on your answers to parts (c) and (d), how does increasing the probability of an event a???ect the mean and standard deviation of the wait time until success?
    From (c) and (d), we can see that increasing the defective rate decreases the mean and standard deviation.

3.38 Male children.

  1. Use the binomial model to calculate the probability that two of them will be boys.
p_boy = 0.51
dbinom(2, 3,p_boy)
## [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) but using the addition rule for disjoint outcomes. Confirm that your answers from parts (a) and (b) match.

There are three possible combinations for two boys and 1 girl: BBG, BGB, GBB.

  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).
    the probability will be 0.2098. Using part b will have to list all the combination options.
dbinom(3, 8,p_boy)
## [1] 0.2098355

3.42 Serving in volleyball.

a.(a) What is the probability that on the 10th try she will make her 3rd successful serve?

p<-0.15
dnbinom(10-3,3,0.15)
## [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?
    the probabitlity is still same 15%.

  2. Even though parts (a) and (b) discuss the same scenario, the probabilities you calculated should be di???erent. Can you explain the reason for this discrepancy?
    In part a, we were looking into a even of 10, and in part b is event of 1.It is different.