library(tidyverse)
## -- Attaching packages --------------------------------------- tidyverse 1.3.1 --
## v ggplot2 3.3.5     v purrr   0.3.4
## v tibble  3.1.4     v dplyr   1.0.7
## v tidyr   1.1.3     v stringr 1.4.0
## v readr   2.0.1     v forcats 0.5.1
## -- Conflicts ------------------------------------------ tidyverse_conflicts() --
## x dplyr::filter() masks stats::filter()
## x dplyr::lag()    masks stats::lag()
library(openintro)
## Loading required package: airports
## Loading required package: cherryblossom
## Loading required package: usdata
library(dplyr)
#library(DATA606)
library(ggplot2)

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. function for graph

normal_area <- function(mean = 0, sd = 1, lb, ub, acolor = "lightgray", ...) {
    x <- seq(mean - 3 * sd, mean + 3 * sd, length = 100) 
    
    if (missing(lb)) {
       lb <- min(x)
    }
    if (missing(ub)) {
        ub <- max(x)
    }

    x2 <- seq(lb, ub, length = 100)    
    plot(x, dnorm(x, mean, sd), type = "n", ylab = "")
   
    y <- dnorm(x2, mean, sd)
    polygon(c(lb, x2, ub), c(0, y, 0), col = acolor)
    lines(x, dnorm(x, mean, sd), type = "l", ...)
}
  1. \(Z < -1.35\)
pnorm(-1.35)
## [1] 0.08850799
normal_area(mean = 0, sd = 1,  ub = -1.35, lwd = 2)

  1. \(Z > 1.48\)
1-pnorm(1.48)
## [1] 0.06943662
normal_area(mean = 0, sd = 1,  lb=1.48, lwd = 2)

  1. \(-0.4 < Z < 1.5\)
pnorm(1.5)-pnorm(-.4)
## [1] 0.5886145
normal_area(mean = 0, sd = 1, lb = -.4, ub = 1.5, lwd = 2)

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

## integer(0)

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.

DISCUSSION: Men Distribution–> N(4313, 583) Women Distribution–> N(5261,807)

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

They both finish below the mean (negative zscores). Mary is closer to the mean (smaller Z score)

  1. Did Leo or Mary rank better in their respective groups? Explain your reasoning.
x<-seq(-5,5,.01)
dens<-dnorm(x,0,1)
plot(x,dens,type='l')
abline(v=-.31)
abline(v=-1.08)

DISCUSSION:

Because Mary has a higher z score (-.31 ) than Leo (-1.08), Mary would rank higher.

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

Leo finishes faster than 13.8 percent of participants.

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

Mary finishes faster than in the .37 percent of runners.

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

Yes, all of the answers, zscores, rank, percentages would change based on the distribution. In the problem, we are instructed to base our calculations on a normal distribution.

For example, if the distribution was highly right skewed, the shape and therefore the area under the curve changes quite dramatically from the normal distribution and bell shaped curve.

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.
x<-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)
hist(x)

m.x<-mean(x)
s.x<-sd(x)


#Check if follows 68-95-99.7 rule, 68 below....
within1sd <- x[x >= m.x - s.x & x <= m.x + s.x]
length(within1sd) / length(x)
## [1] 0.68
#95 below.....
within2sd <- x[x >= m.x - 2*s.x & x <= m.x + 2*s.x]
length(within2sd) / length(x)
## [1] 0.96
#99.7 below.....
within3sd <- x[x >= m.x - 3*s.x & x <= m.x + 3*s.x]
length(within3sd) / length(x)
## [1] 1

The heights do approximately follow the 68-95-99.7 rule with .68, .96, 1 respectively.

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

The histogram depicts a bell-shaped symmetric curve which could be approximately normal distribution.

The qq plot also depicts the points falling on the straight line, also indicating a normal distribution.

# Use the DATA606::qqnormsim function, actually openintro - need to detach DATA606
df<-as.data.frame(x)
qqnormsim(sample=x,data=df)

From this comparison of 8 qq to actual data, it is reasonable to assume a normal distribution.

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?
#GGGGGGGGGD
((.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?
#Geometric distribution mean=1/p
1/.02
## [1] 50
# so 49


#sd sqrt((1-p)/p^2)
((1-.02)/((.02)^2))^.5
## [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?
#meand
1/.05
## [1] 20
#sd sqrt((1-p)/p^2)
((1-.05)/((.05)^2))^.5
## [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?

DISCUSSION: The increase in probability in the geometric distribution decrease the value of the mean and standard deviation.

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)*.49*.51^2
## [1] 0.382347
dbinom(2,3,.51)
## [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.

DISCUSSION: BBG

.51*.51*.49
## [1] 0.127449

BGB

.51*.49*51
## [1] 12.7449

GBB

.49*.51*.51
## [1] 0.127449

so P(BBG)+P(BGB)+P(GBB)

(.51*.51*.49)+(.51*.49*.51)+(.49*.51*.51)
## [1] 0.382347

DISCUSSION:

The answers from a & b match .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).
choose(8,3)
## [1] 56

DISCUSSION: There are 56 different combinations that of 8 children, 3 are boys.

To manually write and capture the 56 combinations is extremely tedious could easily result in making mistakes or omissions.

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?

DISCUSSION: This a negative binomial distribution p=.15,n=10,k=3

choose(9,2)*((.15)^3)*((.85)^7)
## [1] 0.03895012
dnbinom(7,3,.15)
## [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?

DISCUSSION:

p=.15. The trials are independent, so the on any particular event p=.15.

P(10th trial success|2 successes in 9)=P(A|B) P(A|B)=P(A \[intersect\] B)/P(B)

(9 2) (.15^3) * (.85 ^ 7) /(9 2) (.15^2) * (.85 ^ 7)==.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?

The difference is because it is not the same scenario. In the first case, we are calculating the entire sequence.

In the second case, we are calculating the probability of a trial, given something has already occurred.