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. \(Z < -1.35\)

  2. \(Z > 1.48\)

  3. \(-0.4 < Z < 1.5\)

  4. \(|Z| > 2\)

  5. \(Z < -1.35\)

Answer

## [1] "Percentage of normal distrbution is   8.85"

  1. \(Z > 1.48\)

Answer

paste("Percentage of normal distrbution is  ",round(pnorm(-1.48, mean=0, sd=1),digits=4)*100)
## [1] "Percentage of normal distrbution is   6.94"
normalPlot(mean = 0, sd = 1, bounds=c(1.48, 4), tails=FALSE )

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

Answer

paste("Percentage of normal distrbution is  ",round((pnorm(1.5, mean=0, sd=1) - pnorm(-0.4, mean=0, sd=1))  ,digits=4)*100)
## [1] "Percentage of normal distrbution is   58.86"
normalPlot(mean = 0, sd = 1, bounds=c(-0.4, 1.5), tails=FALSE )

  1. \(|Z| > 2\)

Answer

paste("Percentage of normal distrbution is  ",round((pnorm(2, mean=0, sd=1) - pnorm(-2, mean=0, sd=1))  ,digits=4)*100)
## [1] "Percentage of normal distrbution is   95.45"
normalPlot(mean = 0, sd = 1, bounds=c(-2, 2), tails=FALSE )

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.

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

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

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

  5. What percent of the triathletes did Mary finish faster than in her group?

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

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

Answer

For Leo

\(N(\mu=4313, \sigma=583)\)

For Mary

\(N(\mu=5261, \sigma=807)\)

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

Answer

paste("Z score of Leo is ",round((4948-4313)/583,digits=2))
## [1] "Z score of Leo is  1.09"
paste("Z score of Mary is ",round((5513-5261)/807,digits=2))
## [1] "Z score of Mary is  0.31"

The Z scores tell that Leo is 1.09 sd away from the average player and Mary is 0.31 sd away

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

Answer

paste("Leo played worst than ", round(pnorm(4948, mean=4313, sd=583), digits=4)*100, "% of the players ")
## [1] "Leo played worst than  86.2 % of the players "
paste("Mary played worst than ", round(pnorm(5513, mean=5261, sd=807), digits=4)*100, "% of the players ")
## [1] "Mary played worst than  62.26 % of the players "

Mary ranked better in her group when compared to Leo

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

Answer

paste("Leo finished faster than ", 100-round(pnorm(4948, mean=4313, sd=583), digits=4)*100,"% triathletes")
## [1] "Leo finished faster than  13.8 % triathletes"
  1. What percent of the triathletes did Mary finish faster than in her group?

Answer

paste("Mary finished faster than ", 100-round(pnorm(5513, mean=5261, sd=807), digits=4)*100,"% triathletes")
## [1] "Mary finished faster than  37.74 % triathletes"
  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 distribution wasn’t normal then the answers to c,d,e would change because the calculations are done assuming the distribution is normal.

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

  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.

Answer

Check for first 68% rule

pnorm(66.1,mean=61.52, sd=4.58) - pnorm(56.94, mean=61.52, sd=4.58)
## [1] 0.6826895

68% rule holds good

Check for 95% rule

pnorm(70.68, mean=61.52 , sd=4.58) - pnorm(52.36, mean=61.52, sd=4.58)
## [1] 0.9544997

95% rule hold good

Lastly, check for 99.7% rule

pnorm(75.26, mean=61.52, sd=4.58) - pnorm(47.78, mean=61.52, sd=4.58)
## [1] 0.9973002

99.7% rule hold good too.

Hence, the distribution follow 68-95-99.7 rule

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

Answer

The distribution follows the 68-95-99.7 rule as per our above verification, this states that the distribution is normal. Also from the qq plot we see it draws a straight line with skews at 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?

  2. What is the probability that the machine produces no defective transistors in a batch of 100?

  3. On average, how many transistors would you expect to be produced before the first with a defect? What is the standard deviation?

  4. 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?

  5. 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?

  6. What is the probability that the 10th transistor produced is the first with a defect?

Answer

# p(nth success)= p(failures)^n-1*p(success)
#failure probability q=0.02
#success probability p=0.98
paste("The probability that the 10th transistor produced is the first with a defect is ",round((.98^9*.02),digit=4)*100,"%")
## [1] "The probability that the 10th transistor produced is the first with a defect is  1.67 %"
  1. What is the probability that the machine produces no defective transistors in a batch of 100?

Answer

paste("Probability that the machine produces no defective transistors in a batch of 100", round(.98^100,digits=4)*100,"%" )
## [1] "Probability that the machine produces no defective transistors in a batch of 100 13.26 %"
  1. On average, how many transistors would you expect to be produced before the first with a defect? What is the standard deviation?

Answer

#expected value =1/p
paste("First defect is after ", 1/.02," trials" )
## [1] "First defect is after  50  trials"
#sd = sqrt(1-p/p*p)
paste("SD is ",round(sqrt((1-.02)/(.02*.02)), digits=2))
## [1] "SD is  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?

Answer

paste("Transistors produced before the first with a defect is ",1/0.05)
## [1] "Transistors produced before the first with a defect is  20"
paste("SD is ",sqrt((1-0.05)/(.05*.05)))
## [1] "SD is  19.4935886896179"
  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

We see from (c) and (d) that higher the probability, the chances of getting a defect is more quicker and the standard deviation is also small. Increasing the probability decreases the wait time until success.

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.

  2. 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.

  3. 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).

  4. Use the binomial model to calculate the probability that two of them will be boys.

Answer

p<- 0.51
dbinom(2,3,p)
## [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.

Answer

Possibilities -> boy-boy-girl, boy-girl-boy, girl-boy-boy

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

Answer of a) and b) match!!!

  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).

Answer

n!/k!(n-k)!, applying this formula we get 56 possibilities for this case. This makes the b) approach more tedious than approach a)

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?

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

  3. 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?

  4. What is the probability that on the 10th try she will make her 3rd successful serve?

Answer

p<- .15
n<- 10
k<-3
P<- choose(n-1, k-1)*p^k *(1-p)^(n-k)
paste("The probability that on the 10th try she will make her 3rd successful serve is",round(P*100,digits = 2),"%")
## [1] "The probability that on the 10th try she will make her 3rd successful serve is 3.9 %"
  1. Suppose she has made two successful serves in nine attempts. What is the probability that her 10th serve will be successful?

Answer

The probability that her 10th serve will be successful is 15% as each serve is independent.

  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

  1. discusses about an independent serve where case a) is joint probability