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(q = -1.35, mean = 0, sd = 1)
## [1] 0.08850799
normalPlot(mean= 0, sd = 1, bounds=c(-4,-1.35), tails = FALSE)

  1. \(Z > 1.48\)
1-pnorm(q = 1.48, mean = 0, sd = 1)
## [1] 0.06943662
normalPlot(mean= 0, sd = 1, bounds=c(1.48,4), tails = FALSE)

  1. \(-0.4 < Z < 1.5\)
pnorm(q = 1.5, mean = 0, sd = 1) - pnorm(q = -0.4, mean = 0, sd = 1)
## [1] 0.5886145
normalPlot(mean= 0, sd = 1, bounds=c(-0.4,1.5), tails = FALSE)

  1. \(|Z| > 2\)
(1- pnorm(q = 2, mean = 0, sd = 1)) + pnorm(q = -2, mean = 0, sd = 1)
## [1] 0.04550026

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.

Male: N(\(\mu \ \)=4313, \(\sigma \ \ \)=583)

Female: N(\(\mu \ \)=5261, \(\sigma \ \ \)=807)

  1. What are the Z-scores for Leo’s and Mary’s finishing times? What do these Z-scores tell you?

Leo has a z-score of 1.089 and Mary has a z-score of 0.312. This means that Leo finished the race slower than the average male participant in his division (1.089 standard deviations above the mean), and Mary finished the race slower than the average female participant in her division (0.312 standard deviations above the mean). 

leoTime <- 4948
maryTime <- 5513
maleMean <- 4313
maleSd <- 583
femaleMean <- 5261
femaleSd <- 807

(leoTime-maleMean)/maleSd
## [1] 1.089194
(maryTime-femaleMean)/femaleSd
## [1] 0.3122677
  1. Did Leo or Mary rank better in their respective groups? Explain your reasoning.

When we compare the 2 z-scores, we can see that Leo was more slow than the average than Mary was. This is because Leo was a higher deviation from the mean than Mary was.

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

Leo finished faster than about 14% of the participants in his group. 

1-pnorm(q = leoTime, mean = maleMean, sd = maleSd)
## [1] 0.1380342
  1. What percent of the triathletes did Mary finish faster than in her group?

Mary finished faster than about 38% of the participants in her group.

1-pnorm(q = maryTime, mean = femaleMean, sd = femaleSd)
## [1] 0.3774186
  1. If the distributions of finishing times are not nearly normal, would your answers to parts (b) - (e) change? Explain your reasoning.

Yes, these answers would change because we couldn’t use z scores to compare how well Mary and Leo did against their age group and against each other.

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.

Yes, the heights approximately follow the 68-95-99.7% Rule.

meanHeight <- 61.52
sdHeight <- 4.58

# 68% rule: 68% of the data falls within +/- 1 sd from the mean
pnorm(meanHeight + sdHeight, mean = meanHeight, sd = sdHeight) - pnorm(meanHeight - sdHeight, mean = meanHeight, sd = sdHeight)
## [1] 0.6826895
# 95% rule: 95% of the data falls within +/- 2 sd from the mean
pnorm(meanHeight + (2*sdHeight), mean = meanHeight, sd = sdHeight) - pnorm(meanHeight - (2*sdHeight), mean = meanHeight, sd = sdHeight)
## [1] 0.9544997
# 99.7% rule: 99.7% of the data falls within +/- 3 sd from the mean
pnorm(meanHeight + (3*sdHeight), mean = meanHeight, sd = sdHeight) - pnorm(meanHeight - (3*sdHeight), mean = meanHeight, sd = sdHeight)
## [1] 0.9973002
  1. Do these data appear to follow a normal distribution? Explain your reasoning using the graphs provided below.

Yes, this data appears to follow a normal distribution because (1) the actual data closely follows the normal density overlay and (2) aside from a few fluctuations at the tails, the data does not stray far from the normal probability plot. Additionally, looking at 8 additional normal probability plots, we can see that the height data looks very similar. 

qqnormsim(heights)

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?

There is a probability of 0.017 that the 10th transistor is the first with a defect.

pDefect <- 0.02
ntrials <- 10

((1 - pDefect)^ (ntrials-1))*pDefect
## [1] 0.01667496
  1. What is the probability that the machine produces no defective transistors in a batch of 100?

There is a probability of 0.133 that the machine does not produce any defective transistors in a batch of 100.

ntrials <- 100

(1-pDefect)^ntrials
## [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?

On average, it would take about 50 trials before a defect, with a standard deviation of 49.5. 

1/(pDefect)
## [1] 50
sqrt((1-pDefect)/(pDefect^2))
## [1] 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?

On average, it would take 20 trials before a defect, with a standard deviation of 19.5.

pDefect2 <- 0.05
1/(pDefect2)
## [1] 20
sqrt((1-pDefect2)/(pDefect2^2))
## [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?

As one increases the probability of a defect, the mean and standard deviation of the first occurance decreases. This makes sense because at each trial, there is a higher likelihood of the defect happening.

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.

The probability of having 2 boys is 0.38.

numChildren <- 3
numBoys <- 2
probBoy <- 0.51

dbinom(numBoys, size=numChildren, prob=probBoy) 
## [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.

Possibilities: BBG, BGB, GBB – all possibilities have the same probability because we need to multiply the probability of having a boy times the probability of having a boy times the probability of having a girl. However, since there are 3 ways of this occurrence, we need to add all three options together. (or multiply the probability of 2 boys by 3) 

prob2Boys <- probBoy*probBoy*(1-probBoy)
prob2Boys * 3
## [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).

The approach from part b would be more tedious because we need to figure out how many combinations of 3 boys within the 8 children there would be. In part A, we don’t have to figure that out at all - we can just plug in our numbers.

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?

The probability of making her 3rd successful serve on the 10th try is 0.039

numTrials <- 10
numSuccess <- 3
probSuccess <- 0.15

dnbinom(numTrials-numSuccess,numSuccess, probSuccess)
## [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 probability of her 10th serve being successful is 0.15 because the serves are independent of one another.

  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?

The reason for this difference is because in part a we are looking for a specific sequence of serves. We want to know that the 3rd successful attempt is on the 10th try, so the sequencing of the events makes the final probability dependent on the other serves. In part b, we want to know the probability of success on the 10th try, without any basis on the previous serves, so it is an independent event with a probability of 0.15.