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.

#A
pnorm(-1.35)
## [1] 0.08850799
#B
1- pnorm(1.48)
## [1] 0.06943662
#C
pnorm(1.5) - pnorm(-0.4)
## [1] 0.5886145
#D
1-pnorm(2) + pnorm(-2)
## [1] 0.04550026
  1. \(Z < -1.35\)

8.85%

  1. \(Z > 1.48\)

6.94%

  1. \(-0.4 < Z < 1.5\)

58.86%

  1. \(|Z| > 2\)

4.55%

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.

Men, Ages 30 - 34: N(mean = 4313 seconds, sd = 583 seconds) Women, Ages 25 - 29: N(mean = 5261 seconds, sd = 807 seconds)

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

Leo has a z score of 1.089 and Mary has a z score of 0.312. This indicates that both were slower than the average of their groups, and they are a multiple of their z scores away from their means.

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

Mary did better respectively, as her z score is lower than Leo’s.

  1. What percent of the triathletes did Leo finish faster than in his group?
1 - pnorm(1.089194)
## [1] 0.1380342

Leo finished faster than 13.80% in his group.

  1. What percent of the triathletes did Mary finish faster than in her group?
1 - pnorm(0.3122677)
## [1] 0.3774185

Mary finished faster than 37.74% 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.

The answers to parts b-e are dependant on the assumption that the distributions are normal. If they are not normal, things like the z-scores used above will no longer be relevant.

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.
mean <- 61.52
sd <- 4.58
sds <- sd * c(1:3)

#This finds the proportion of heights within 1, 2, and 3 stdevs, respectively
pnorm(mean + sds, mean = mean, sd = sd) - pnorm(mean - sds, mean = mean, sd = sd)
## [1] 0.6826895 0.9544997 0.9973002

It does approximately follow the 68-95-99.7% rule

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

The data does approximately follow a normal distribution. The histogram appears to have an approximate bell curve, and the dots mostly fall on the line in the qq-plot

# 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.

#Setup
p = 0.02
  1. What is the probability that the 10th transistor produced is the first with a defect?
(1-p)^9 * p
## [1] 0.01667496
  1. What is the probability that the machine produces no defective transistors in a batch of 100?
(1-p)^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 = 1/p
sd = ((1-p)/p^2)^0.5
mean
## [1] 50
sd
## [1] 49.49747

mean = 50 sd = 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?
p2 <- 0.05
mean = 1/p2
sd = ((1-p2)/p2^2)^0.5
mean
## [1] 20
sd
## [1] 19.49359

mean = 20 sd = 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?

The time until success is proportional to the increase of the likelyhood. In the problems above, the likelyhood of a defect increased 2.5X, and the mean and sd before a defect shrank 2.5X in accordance

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.
boyp = 0.51
boys = 2
girlp = 0.49
girls = 1
n = 3


prob = (factorial(n)/(factorial(n - boys)*factorial(boys)))*(boyp^boys)*(girlp^girls)
prob
## [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
  2. but using the addition rule for disjoint outcomes. Confirm that your answers from parts (a) and (b) match.
boyp * boyp * girlp +
  boyp * girlp * boyp +
  girlp * boyp * boyp
## [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 in part b would be more tedious becuase it would involve writing out every combination of 3 boys among 8 kids, which is significantly more combinations than possible in part b.

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?
successp = 0.15
successes = 2
failp = 1-successp
fails = 7
n = 9


prob = (factorial(n)/(factorial(fails)*factorial(successes)))*(successp^successes)*(failp^fails) * 0.15
prob
## [1] 0.03895012

Finds probability of 2 successes within the first 9 tries, and multiplies by the chance of another success (on the 10th try)

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

0.15, assuming the events are independent. The first 9 attempts have no bearing on the 10th.

  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 questions are different. Part a asks for the probability of 2 successes in 9 and then another success; part b asks for the probability of one success, which just happens to be on the 10th try.