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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. \(Z < -1.35\)
require(fastGraph)
## Loading required package: fastGraph
pnorm(-1.35,mean=0,sd=1)
## [1] 0.08850799
shadeDist(-1.35)

  1. \(Z > 1.48\)
pnorm(1.48,mean=0,sd=1,,lower.tail=FALSE)
## [1] 0.06943662
shadeDist(1.48,lower.tail = FALSE)

  1. \(-0.4 < Z < 1.5\)
pnorm(c(-0.4,1.5),mean=0,sd=1)
## [1] 0.3445783 0.9331928
#Answer: The middle area is 0.9331928-0.3445783=0.5886145
shadeDist(c(-0.4,1.5),lower.tail = FALSE)

  1. \(|Z| > 2\)
pnorm(c(-2,2),0,1)
## [1] 0.02275013 0.97724987
# Answer:The area is 1-0.97724987+0.02275013=0.04550026

shadeDist(c(-2,2),lower.tail = TRUE)

Triathlon times, Part I a) Write down the short-hand for these two normal distributions.

Answer:

\(N(\mu=4313, \sigma=583)\) for man \(N(\mu=5261, \sigma=807)\) for Woman

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

Answer:

Leo’s Z-scores: (4948-4313)/583=1.089194 Mary’s Z-scores: (5513-5261)/807=0.3122677

Leo is 1.089 standard deviation above the mean and Mary is 0.312 standard deviation of the mean, which mean Leo perform better than Mary.

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

Answer: They all do better than the average because their z score is positive, but Leo’s z score is greater than Mary, so he do better than Mary.

  1. What percent of the triathletes did Leo finish faster than in his group?
pnorm(4948,4313,583,lower.tail = FALSE)
## [1] 0.1380342
  1. What percent of the triathletes did Mary finish faster than in her group?
pnorm(5513,5261,807,lower.tail = FALSE)
## [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.

Answer:

If the data are not nearly normal, the Z score of Leo and Mary will not change, but the answer of their probability will change because z score cannot use to calculate their probability. The result will be inaccurate if the data is not normal distribution.

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.
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)
summary(heights)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   54.00   58.00   61.00   61.52   64.00   73.00
sd(heights)
## [1] 4.583667
#Probability for falling within 1 standard deviation
pnorm(61.52+4.58,mean=61.52,sd=4.58)
## [1] 0.8413447
#Probability for falling within 2 standard deviation
pnorm(61.52+2*4.58,mean=61.52,sd=4.58)
## [1] 0.9772499
#Probability for falling within 3 standard deviation
pnorm(61.52+3*4.58,mean=61.52,sd=4.58)
## [1] 0.9986501
hist(heights)

qqnorm(heights)
qqline(heights)

qqnormsim(heights)

Answer: The probability of the first, second, and third standard deviation is 84.13%, 97.72% and 99.87%, respectively. It doesn’t follow the 68%, 95% and 99.7% rule.

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

Answer: The histogram and QQ-plot shows the data follow a normal distribution because those points seems to follow the line, except some deviation on both ends.

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?

p=0.98^9*0.02=0.01667496

p1<-0.98^9*0.02
p1
## [1] 0.01667496
  1. What is the probability that the machine produces no defective transistors in a batch of 100?
0.98^100
## [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?
#expect to be produced before the first with a defect
1/0.02
## [1] 50
#standard deviation
sqrt((1-0.02)/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?
#how many transistors would you expect to be produced with this machine before the first with a defect
1/0.05
## [1] 20
#standard deviation
sqrt((1-0.05)/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?

Answer: Increasing the probability of event will decreases 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.
factorial(3)/(factorial(2)*factorial(1))*0.51^2*(1-0.51)
## [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.

We have three chirldren A, B and C, the combination of two of them are boys as follows

BBG<-0.51*0.51*0.49
BGB<-0.51*0.49*0.51
GBB<-0.49*0.51*0.51

sum(BBG,BGB,GBB)
## [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).
factorial(8)/(factorial(3)*factorial(5))
## [1] 56

Answer: It is very tedious because we have 56 combinations to caculate the posiblility.

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?
choose(9,2)*0.15^3*0.85^7
## [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?

Answer: Everyone serrves is independtly, so the probability of her 10th serve will be sucessful is 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?

Answer: Part (a) calculate the third successful serve base on the 10th try, so it have fix number of trials, it is negative binomial distribution. Part (c) only discuss the third serve, which is independent.