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. \(|Z| > 2\)

We know the mean is 0 and standard deviation is 1 we will assume the number of observations to be 1000 lets see what we get from our simulated results

normalPlot(bounds=(c(-1.35,Inf)))

so for (a)

1-0.911
## [1] 0.089

for (b)

normalPlot(bounds=(c(1.48,Inf)))

c

normalPlot(bounds=(c(-0.4,1.5)))

d

normalPlot(bounds=(c(2,Inf)))

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.

Answer (a)

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

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

Answer (b)

Z Score For Leo

Z_Leo <- (4948-4313)/583
Z_Leo
## [1] 1.089194

Z Score for Mary

Z_Mary <- (5513-5261)/807
Z_Mary
## [1] 0.3122677

According to Leo’s Z score his time was 1.09 standard deviations above the mean. Mary’s Z score suggest that her time was 0.31 standard deviations above the mean.

Answer (c)

Looks like Leo had a better Z score and was 1.08 standard deviations above the mean so he was higher in his group.

Answer (d)

pnorm(1.09)
## [1] 0.8621434
print("Leo Finished faster than 86.21% people in his group")
## [1] "Leo Finished faster than 86.21% people in his group"

Answer (e)

pnorm(0.31)
## [1] 0.6217195
print("Mary Finished faster than 62.17% people in his group")
## [1] "Mary Finished faster than 62.17% people in his group"

Answer(d)

The Z scores would stay the same. However the percentiles would change if the distribution was not normal. So the answers to parts d and e would be different if the distributions of finishing times were not nearly 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.

# Use the DATA606::qqnormsim function

Answer (a)

pnorm(61.52+4.58,mean=61.52,sd=4.58)
## [1] 0.8413447
pnorm(61.52+2*4.58,mean=61.52,sd=4.58)
## [1] 0.9772499
pnorm(61.52+3*4.58,mean=61.52,sd=4.58)
## [1] 0.9986501

From the above results we can suggest that it does not follow the 68-95-99% rule

Answer (b) 

female_height = 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(female_height)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   54.00   58.00   61.00   61.52   64.00   73.00
hist(female_height,xlab = "Height of Female College Student in cm", main = "Histogram For Female Height")

qqnormsim(female_height)

It seems thpough as there are a few outliers the distribution is nearly normal.

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?

Answer (a)

I will be using the pgeom function to determine the probability of the 10th Transistor produced is defective

pgeom(10-1,0.02)
## [1] 0.1829272

Answer (b)

The probability that the manchine produces no defective transistors in a batch of 100 we can use the pgeom function to calculate

1-pgeom(100,0.02)
## [1] 0.1299672

Answer (c) 

mean<-1/0.02

std<- sqrt((1-0.02)/((0.02)*(0.02)))


mean
## [1] 50
std
## [1] 49.49747

Answer (d)

mean<-1/0.05

std<- sqrt((1-0.05)/((0.05)*(0.05)))


mean
## [1] 20
std
## [1] 19.49359

Answer (e)

Based on the answer to c and d if we increase the probability the value of the mean and the standard deviation will decrease. So, in the case when the probability was lower the mean was 50 but when probability increased the mean decreased. Similarly for the standard deviation first when probability was lower it was 49.497 but when probability increased the value for Standard deviation became 19.493.

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

Answer: P(having a boy)=0.51 , n=3 and k=2 we will use dbinom function and plug in the values

dbinom(2,3,0.51)
## [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.Confirm that your answers from parts (a) and (b) match.

Answer:

BBG BGB GBB

probability = (0.51^2)*0.49*3
probability
## [1] 0.382347

The values in the answer (a) and (b) are the same.

  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

I think that the apporach from Part b where you have to use the addition rule for disjoint it will be more tedious because of the number of scenerios and it will just be hard to draw all the combinations.

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?

Answer (a)

Lets use the choose function to calculate this.

paste ("P(making 3rd successful serve on 10th try)=",choose(9,2)*0.15^3*0.85^7)
## [1] "P(making 3rd successful serve on 10th try)= 0.0389501162261719"

Answer (b)

The answer is 0.15 since each serve is independant so the probability of the 10th attempt will be same as the previous 9.

Answer (c):

calculated should be different. Can you explain the reason for this discrepancy?

It seems as though parts a and b are quite similar in terms of having success in the 10th serve. However, if we read the question carefully we can understand that they are quite different in part a the question is referring to the number of successes however in part b the question is asking about what is the probability that she makes a successful serve on 10th try. Since, all events are indepenedant a succesfull server on 10th try would be same as any other try.