Area under the curve, Part I. (4.1, p. 142) What percent of a standard normal distribution \(N(\mu=0, \sigma=1)\) is found in each region? Be sure to draw a graph.

  1. \(Z < -1.35\)
pnorm(-1.35)
## [1] 0.08850799
visualize.norm(stat=-1.35,mu=0,sd=1,section="lower")

  1. \(Z > 1.48\)
pnorm(1.48)
## [1] 0.9305634
visualize.norm(stat=1.48,mu=0,sd=1,section="upper")

  1. \(-0.4 < Z < 1.5\)
visualize.norm(stat=c(0.4,1.5),mu=0,sd=1,section="bounded")

  1. \(|Z| > 2\)
visualize.norm(stat=c(-2,2),mu=0,sd=1,section="bounded")

Triathlon times, Part I (4.4, p. 142) In triathlons, it is common for racers to be placed into age and gender groups. Friends Leo and Mary both completed the Hermosa Beach Triathlon, where Leo competed in the Men, Ages 30 - 34 group while Mary competed in the Women, Ages 25 - 29 group. Leo completed the race in 1:22:28 (4948 seconds), while Mary completed the race in 1:31:53 (5513 seconds). Obviously Leo finished faster, but they are curious about how they did within their respective groups. Can you help them? Here is some information on the performance of their groups:

Remember: a better performance corresponds to a faster finish.

  1. Write down the short-hand for these two normal distributions.
#Ages of men 30-40: 
m_mean <- 4313
m_sd <- 583 

#Ages of women 30-34: 
w_mean <- 5261 
w_sd <- 807
  1. What are the Z-scores for Leo’s and Mary’s finishing times? What do these Z-scores tell you?
Zscores_Leo <- (4948 - 4313) / 583
Zscores_Leo
## [1] 1.089194
Zscores_Mary <- (5513 - 5261) / 807
Zscores_Mary
## [1] 0.3122677

Leo’s result is 1.09 standard deviations from the mean while Mary’s results is 0.31 standard deviations from the mean.

  1. Did Leo or Mary rank better in their respective groups? Explain your reasoning.

Solution: Mary ranks better since her result is closer to the median than Leo’s.The Z-score for Mary’s time is 0.31 which means she finished 0.31 standard deviations above the mean for her age group.

  1. What percent of the triathletes did Leo finish faster than in his group?
pnorm(Zscores_Leo)
## [1] 0.8619658

Solution: Leo finished faster then 86% of the other competitors in his group.

  1. What percent of the triathletes did Mary finish faster than in her group?
pnorm(Zscores_Mary)
## [1] 0.6225814

Solution: Mary finished faster then 62% of the other competitors in her group.

  1. If the distributions of finishing times are not nearly normal, would your answers to parts (b) - (e) change? Explain your reasoning.

Solution: Z-scores for non normal distributions are relevant for analysis in a multiple group. We will not be able to compare two people from different groups based on Z-scores of non normal distributions. We cannot use the normal probability table to calculate the probabililties and percentiles without a normal model for parts (d) through (e).

Heights of female college students Below are heights of 25 female college students.

\[ \stackrel{1}{54}, \stackrel{2}{55}, \stackrel{3}{56}, \stackrel{4}{56}, \stackrel{5}{57}, \stackrel{6}{58}, \stackrel{7}{58}, \stackrel{8}{59}, \stackrel{9}{60}, \stackrel{10}{60}, \stackrel{11}{60}, \stackrel{12}{61}, \stackrel{13}{61}, \stackrel{14}{62}, \stackrel{15}{62}, \stackrel{16}{63}, \stackrel{17}{63}, \stackrel{18}{63}, \stackrel{19}{64}, \stackrel{20}{65}, \stackrel{21}{65}, \stackrel{22}{67}, \stackrel{23}{67}, \stackrel{24}{69}, \stackrel{25}{73} \]

  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.
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
##  [1] 54 55 56 56 57 58 58 59 60 60 60 61 61 62 62 63 63 63 64 65 65 67 67
## [24] 69 73
mean_heights <- 61.52
sd_heights <- 4.58
1-2*pnorm(mean_heights+sd_heights,mean=mean_heights,sd=sd_heights,lower=FALSE)
## [1] 0.6826895
1-2*pnorm(mean_heights+2*sd_heights,mean=mean_heights,sd=sd_heights,lower=FALSE)
## [1] 0.9544997
1-2*pnorm(mean_heights+3*sd_heights,mean=mean_heights,sd=sd_heights,lower=FALSE)
## [1] 0.9973002

Solution: The heights of the students seems to follow the 68-95-99.7% Rule since they appear to be normally distributed.

  1. Do these data appear to follow a normal distribution? Explain your reasoning using the graphs provided below.

The data appear to follow a normal distribution from the histogram but in the normal probability plot, the data follows a straight line with some deviations.

# Use the DATA606::qqnormsim function

Defective rate. (4.14, p. 148) A machine that produces a special type of transistor (a component of computers) has a 2% defective rate. The production is considered a random process where each transistor is independent of the others.

  1. What is the probability that the 10th transistor produced is the first with a defect?
p <- .02
n <- 10
P <- (1-p)^(n-1)*p
p
## [1] 0.02

Solution: The probability that the 10th transistor produced is the first with a defect is 0.02.

  1. What is the probability that the machine produces no defective transistors in a batch of 100?
round((1-p)^100, 3)
## [1] 0.133

There is 0.133 chance that the machine produces no defective transistors in a batch of 100.

  1. On average, how many transistors would you expect to be produced before the first with a defect? What is the standard deviation?
Ex_Val <- 1/p
Ex_Val
## [1] 50
sd <- sqrt((1-p)/(p^2))
sd
## [1] 49.49747

On average that 50 transistors would be produced before the first with a defect with a standard deviation of 49.5.

  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?
p <- .05
Ex_Val <- 1/p
Ex_Val
## [1] 20
sd <- sqrt((1-p)/(p^2))
sd
## [1] 19.49359

On average 20 transistors would be produced before the first with a defect on this machine with a standard deviation of 19.5.

  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?

Solution: Increasing the probability of an event makes both the expected value and the standard deviation to decrease.

Male children. While it is often assumed that the probabilities of having a boy or a girl are the same, the actual probability of having a boy is slightly higher at 0.51. Suppose a couple plans to have 3 kids.

  1. Use the binomial model to calculate the probability that two of them will be boys.
#getting two boys 

p = .51
P = round(dbinom(2,3,p),3)
P
## [1] 0.382

Solution: There is a 0.382 chance that exactly two of the kids will be boys.

  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.
P <- .51*.51*.49 + .51*.49*.51 + .49*.51*.51
P
## [1] 0.382347

p(a Boy) = 0.51 p(a Birl) = 0.49

There’re 3 possibilities: 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 approach from part (b) would be more tedious because it will be harder to list all the cases with 3 boys and 5 girls than cases with 2 boys and a girl.

Serving in volleyball. (4.30, p. 162) A not-so-skilled volleyball player has a 15% chance of making the serve, which involves hitting the ball so it passes over the net on a trajectory such that it will land in the opposing team’s court. Suppose that her serves are independent of each other.

  1. What is the probability that on the 10th try she will make her 3rd successful serve?
p = .15
P = dbinom(2,9,p)
round(P * .15,3)
## [1] 0.039

Solution: There a 0.039 chance that she will make her 3rd successful serve on the 10th trial.

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

Solution: With two successful serve in nine attempts, the probability that the 10th serve will be successful is the probability of making one successful serve, which is .15 or 15%.

  1. 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?

Solution: Part (b) calculates the probability that two of the previous nine attempts are successesful, while part (a) calculates the probability that the 10th attempt is a successful with the previou nine attempts successful.