library(DATA606)

3.2 What percent of a standard normal distribution is found in each region? Be sure to draw a graph.

# a. Z > -1.13
normalPlot(bounds = c(-1.13, Inf))

# b. Z < 0.18
normalPlot(mean = 0, sd = 1, bounds = c(-Inf,0.18))

# c. Z > 8
normalPlot(mean = 0, sd = 1, bounds = c(8,Inf))

# d. |Z| < 0.5
normalPlot(mean = 0, sd = 1, bounds = c(-0.5,0.5))

3.4 Triathlon times, Part I.

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: The finishing times of the Men, Ages 30 - 34 group has a mean of 4313 seconds with a standard deviation of 583 seconds. The finishing times of the Women, Ages 25 - 29 group has a mean of 5261 seconds with a standard deviation of 807 seconds. The distributions of finishing times for both groups are approximately Normal.

Remember: a better performance corresponds to a faster finish.

  1. Write down the short-hand for these two normal distributions.

    Men (30-34): \(N(\mu = 4313, \sigma = 583)\) Women (25-29): \(N(\mu = 5261, \sigma = 807)\)

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

Z_leo <- (4948 - 4313) / 583;Z_leo
## [1] 1.089194
Z_mary <- (5513 - 5261) /807;Z_mary
## [1] 0.3122677

  1. Did Leo or Mary rank better in their respective groups? Explain your reasoning. Mary rank better in her group than Leo did in his group since Mary’s Z score is closer to the mean

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

p_leo <- (1-pnorm((4948 - 4313) / 583)) * 100; p_leo
## [1] 13.80342
  1. What percent of the triathletes did Mary finish faster than in her group?
p_mary <- (1-pnorm((5513 - 5261) /807)) * 100; p_mary
## [1] 37.74186
  1. If the distributions of finishing times are not nearly normal, would your answers to parts (b) - (e) change? Explain your reasoning. Yes the answers may change because calculations/percentages would be different

3.18 Heights of female college students.

Below are heights of 25 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. Based on the calculation below, the data have 68-96-100%, which approximate the rule
f_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)

# 68% or 1 standard deviation from the mean?
sd_1 <- sum(f_heights > (61.52 - 4.58) & f_heights < (61.52 + 4.58)) / length(f_heights) * 100; sd_1
## [1] 68
# 95% or 2 standard deviations from the mean?
sd_2 <- sum(f_heights > (61.52 - (2* 4.58)) & f_heights < (61.52 + (2* 4.58))) / length(f_heights) * 100; sd_2
## [1] 96
# 99% or 3 standard deviations from the mean?
sd_3 <- sum(f_heights > (61.52 - (3* 4.58)) & f_heights < (61.52 + (3* 4.58))) / length(f_heights) * 100; sd_3
## [1] 100
  1. Do these data appear to follow a normal distribution? Explain your reasoning using the graphs provided below. Though not perfect, based on histogram, bell curve, and probability plot, it looks like the distribution can be considered nearly normal
hist(f_heights, probability = TRUE, ylim = c(0, 0.10))
x <- min(f_heights):max(f_heights)
y <- dnorm(x = x, mean = 61.52, sd = 4.58)
lines(x = x, y = y, col = "blue")

qqnorm(f_heights)
qqline(f_heights)

3.22 Defective rate.

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?
(.98^9)*(.02)
## [1] 0.01667496
  1. What is the probability that the machine produces no defective transistors in a batch of 100?
(.98^100)
## [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/expected value 
1/.02
## [1] 50
#standard deviation
sqrt((1-.02)/(0.02^2))
## [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?
# mean/expected value 
1/.05
## [1] 20
#standard deviation
sqrt((1-.05)/(0.05^2))
## [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? The effect of higher probability decreases the wait time of the event

3.38 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.
(choose(3,2)) * (0.51)^2 * (.49)
## [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.
# combinations: bbg, bgb, gbb
(.51 * .51 * .49) + (.51 * .49 * .51) + (.49 * .51 * .51)
## [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 (b) approach entails deriving the different combinations, and then adding their probabilities. In the case of 3 boys from 8 kids, it’s adding 56 different combinations!

3.42 Serving in volleyball.

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?
(choose(9,2)) * (0.15)^3 * (.85)^7
## [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? 0.15

  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? In (b), the calculation pertains to the probability of a single independent event having 2 possible outcome of either success or failure; while the scenario in (a) is more involve in the probability of specific outcome of success in sequence, looking for the player’s 3rd successful serve in the 10th serve