library(DATA606)
## 
## Welcome to CUNY DATA606 Statistics and Probability for Data Analytics 
## This package is designed to support this course. The text book used 
## is OpenIntro Statistics, 3rd Edition. You can read this by typing 
## vignette('os3') or visit www.OpenIntro.org. 
##  
## The getLabs() function will return a list of the labs available. 
##  
## The demo(package='DATA606') will list the demos that are available.

3.2 Area under the curve, Part II. What percent of a standard normal distribution N(µ = 0,= 1) is found in each region? Be sure to draw a graph.

  1. Z > -1.13
pnorm(1.13, mean= 0, sd= 1 ) 
## [1] 0.8707619
normalPlot(bounds = c(-1.13, Inf))

  1. Z < 0.18
pnorm(0.18, mean= 0, sd= 1)   
## [1] 0.5714237
normalPlot(bounds = c(-Inf, 0.18))

(c) Z > 8

pnorm(8, mean= 0, sd= 1)
## [1] 1
normalPlot(bounds = c(8, Inf))

  1. |Z| < 0.5
pnorm(0.5, mean=0, sd= 1)
## [1] 0.6914625
normalPlot(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 ???nished 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 ???nishing times of the Men, Ages 30 - 34 group has a mean of 4313 seconds with a standard deviation of 583 seconds. . The ???nishing times of the Women, Ages 25 - 29 group has a mean of 5261 seconds with a standard deviation of 807 seconds. . The distributions of ???nishing times for both groups are approximately Normal. Remember: a better performance corresponds to a faster ???nish.

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

$N(=4313, =583) $ Men Ages 30 - 34: N(µ =4313, o =583)

Women Ages 25-29: N(µ =5261, o =807)

  1. What are the Z-scores for Leo’s and Mary’s ???nishing times? What do these Z-scores tell you?
leo <- 4948
leo_uMale <- 4313
leo_sdMale <- 583
mary <- 5513
mary_ufemale <- 5261
mary_sdfemale <- 807


(leo - leo_uMale) / leo_sdMale
## [1] 1.089194
(mary - mary_ufemale)/mary_sdfemale
## [1] 0.3122677
  1. Did Leo or Mary rank better in their respective groups? Explain your reasoning.

Mary’s Z score was lower than Leo’s . Therefore she had performed better than leo performed in their respective groups. Mary is 0.31 away from the mean while Leo is 1.08 SD from the mean.

  1. What percent of the triathletes did Leo ???nish faster than in his group?
pnorm(leo, leo_uMale, leo_sdMale, lower.tail = FALSE)
## [1] 0.1380342
  1. What percent of the triathletes did Mary ???nish faster than in her group?
pnorm( mary, mary_ufemale, mary_sdfemale,lower.tail = FALSE)
## [1] 0.3774186
  1. If the distributions of ???nishing times are not nearly normal, would your answers to parts (b) - (e) change? Explain your reasoning.

Yes. In this problem the Z score was determined by a normal distribution. We couldn’t determine how they did in comparison to their respective groups.

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.
hts <- 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)
m <- mean(hts)
sd <- sd(hts)

hts_upper1 <-  m + sd
hts_upper1
## [1] 66.10367
hts_lower1 <-  m - sd
hts_lower1
## [1] 56.93633
hts_upper2 <-  m + (2*sd)
hts_upper2
## [1] 70.68733
hts_lower2 <-  m - (2*sd)
hts_lower2
## [1] 52.35267
hts_upper3 <-  m + (3*sd)
hts_upper3
## [1] 75.271
hts_lower3 <-  m - (3*sd)
hts_lower3
## [1] 47.769
length(hts[hts >= hts_lower1 & hts <= hts_upper1]) / length(hts) * 100
## [1] 68
length(hts[hts >= hts_lower2 & hts <= hts_upper2]) / length(hts) * 100
## [1] 96
length(hts[hts >= hts_lower3 & hts <= hts_upper3]) / length(hts) * 100
## [1] 100

In this problem the data which is shown followes approx. the 68-95-99.7 rule - 68-96-100

  1. Do these data appear to follow a normal distribution? Explain your reasoning using the graphs provided below.
hist(hts, probability = T)
x <- 40:100
y <- dnorm(x=x, m + sd)
lines(x=x,y=y, col= "red")

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 ???rst with a defect?
pf <- 0.02
ps <- 1 - pf
n <- 10

round(dgeom(n, ps), 2)
## [1] 0
  1. What is the probability that the machine produces no defective transistors in a batch of 100?
pf <- 0.02
ps <- 1- pf
n <- 100

round(ps^n, 2)
## [1] 0.13
  1. On average, how many transistors would you expect to be produced before the ???rst with a defect? What is the standard deviation?
pf <- 0.02
ex <- 1/pf
ex
## [1] 50
sd <- sqrt((1-pf)/(pf^2))
sd
## [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 ???rst with a defect? What is the standard deviation?
pf <- 0.05
ex <- 1/pf
ex
## [1] 20
sd <- sqrt((1-pf)/(pf^2))
sd
## [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?

When the probability of failure is bigger, the event is more common, meaning the expected number of trials before a success and the standard deviation of the waiting time are smaller.

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.
n <- 3
k <- 2
pboy1 <- 0.51
choose(n, k) * (1 - pboy1)^(n - k) * (pboy1)^k
## [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. Con???rm that your answers from parts (a) and (b) match.
children <- c(pboy1 * pboy1 * (1 - pboy1), pboy1 * (1 - pboy1) * pboy1, (1 - pboy1) * 
    pboy1 * pboy1)
sum(children)
## [1] 0.382347
  1. If we wanted to calculate the probability that a couple who plans to have 8 kids will have 3 boys, brie???y describe why the approach from part (b) would be more tedious than the approach from part (a).

Part B you’d have to write out the entire # of combinations vs. using combinations to solve for the answer much quicker

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?
p <- 0.15
n <- 10
k <- 3

choose(n-1, k-1) * (1-p)^(n-k) * p^k
## [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?

Each attempt is independent of the last, so 15%

  1. 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?

According to the text: In the negative binomial case, we examine how many trials it takes to observe a fixed number of successes and require that the last observation be a success.