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. \(Z < -1.35\)
percA <- (pnorm(q = -1.35, mean = 0, sd = 1))*100
ggplot(NULL, aes(c(-3,3))) +
  geom_area(stat = "function", fun = dnorm, fill = "#00998a", xlim = c(-3, -1.35)) +
  geom_area(stat = "function", fun = dnorm, fill = "grey80", xlim = c(-1.35, 3))

The probability of \(Z < -1.35\) is \(8.8507991\)%.

  1. \(Z > 1.48\)
percB <- (1 - pnorm(q = 1.48, mean = 0, sd = 1))*100
ggplot(NULL, aes(c(-3,3))) +
  geom_area(stat = "function", fun = dnorm, fill = "#00998a", xlim = c(1.48, 3)) +
  geom_area(stat = "function", fun = dnorm, fill = "grey80", xlim = c(-3, 1.48))

The probability of \(Z > 1.48\) is \(6.9436623\)%.

  1. \(-0.4 < Z < 1.5\)
percC <- (pnorm(q = 1.5, mean = 0, sd = 1) - pnorm(-0.4, 0, 1)) * 100
ggplot(NULL, aes(c(-3,3))) +
  geom_area(stat = "function", fun = dnorm, fill = "#00998a", xlim = c(-.4, 1.5)) +
  geom_area(stat = "function", fun = dnorm, fill = "grey80", xlim = c(-3,-.4))+
  geom_area(stat = "function", fun = dnorm, fill = "grey80", xlim = c(1.5:3))

The probability of \(-0.4 < Z < 1.5\) is \(58.861454\)%.

  1. \(|Z| > 2\)
percD <- (pnorm(-2, 0, 1)*2) * 100
ggplot(NULL, aes(c(-3,3))) +
  geom_area(stat = "function", fun = dnorm, fill = "#00998a", xlim = c(-3,-2)) +
  geom_area(stat = "function", fun = dnorm, fill = "#00998a", xlim = c(2,3))+
  geom_area(stat = "function", fun = dnorm, fill = "grey80", xlim = c(-2,2))

The probability of \(|Z| > 2\) is \(4.5500264\)%.

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.

Leo - N(mean = 4313, sd =583) Mary - N(mean = 5261, sd =807)

  1. What are the Z-scores for Leo’s and Mary’s finishing times? What do these Z-scores tell you?
Leo <- (4948-4313) / 583
Mary <- (5513-5261) / 807

The Z-score for Leo was \(1.0891938\), this tells us her deviation from the mean and that she ran longer than average. The Z- score for Mary was \(0.3122677\), this tells us her deviation from the mean and that she ran longer than average.

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

Mary ranked better than Leo in her respective group because both scores landed above the mean however Mary’s was closer towards the mean than was Leo.

  1. What percent of the triathletes did Leo finish faster than in his group?
Leo <- (1 - pnorm(Leo, 0, 1))*100

Leo ran faster than \(13.8034211\)% of people.

  1. What percent of the triathletes did Mary finish faster than in her group?
Mary <- (1 - pnorm(Mary, 0, 1))*100

Mary ran faster than \(37.7418559\)% of people.

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

Answers (b) and (c) would not change (b) is a factor of standard deviation and mean so unless they were to change our answer for (b) will be constant. For (c), the question is relating the two runners. It is possible that a alter how they did compared to one another. For instance if a runner runs a very very good time they will reduce the mean and increase the standard deviation, reducing the z-score. However it is unlikely even with a few major outliers that Leo did better than Mary in respect to their groups.

Answers (d) and (e) will change because these percentiles assume a normal distribution. If the distribution is not normal than the percent of runners who did better or worse than Mary and Leo will change.

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.
## [1] 0.68
(sum(heights < 61.52 + (2*4.58)) - sum(heights < 61.52 - (2*4.58))) / length(heights)
## [1] 0.96
(sum(heights < 61.52 + (3*4.58)) - sum(heights < 61.52 - (3*4.58))) / length(heights)
## [1] 1

The heights do follow the 68-95-99.7 rule indicating the normal distribution however all values fall within three standard deviations but it still appears to be distributed normally.

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

In the histogram we notice a unimodal peak at the same location of the normal distribution line’s peak. We also notice both tails decreasing in counts relative to the line. There are no counts beyond three standard deviations which is statistically normal.In the QQ plot the points are all relatively close to the line indicating normal probabilities. The theoretical and sample quantiles match very well with the exception of the two points at the 2 and -2 quantile, but statistically speaking this is to be expected.

# Use the DATA606::qqnormsim function
qqnormsim(heights)

This QQ plots above show our data and eight simulated plots. The “Normal QQ Plot (Data)” actually looks to be the most normal out of the bunch.

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?
trans_error_10 <- (.98*.98*.98*.98*.98*.98*.98*.98*.98*.02) * 100

There is a \(1.6674955\)% chance of the 10th transistor having a defect.

  1. What is the probability that the machine produces no defective transistors in a batch of 100?
no_error <- (.98^100)

There is a \(`no_error`\)% chance of the 100 transistors not having a defect.

  1. On average, how many transistors would you expect to be produced before the first with a defect? What is the standard deviation?
mean2 <- 1 / .02
sd2 <- ( (1-.02) / (.02^2) )^(1/2)

The first defect on average will occur after \(50\) transistors and the standard deviation is \(49.4974747\).

  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?
mean5 <- 1 / .05
sd5 <- ( (1-.05) / (.05^2) )^(1/2)

The first defect on average will occur after \(20\) transistors and the standard deviation is \(19.4935887\).

  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?

By increasing the probability of an event, the average occurance of the event also increases. Adversely increasing the probability will reduce the standard deviation of the event taking place.

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.
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.

Out of three children you can have bbg, bgb, gbb.

(.51*.51*.49) + (.51*.49*.51) + (.49*.51*.51)
## [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).

Our solution from part (b) would be correct but tedious due to the fact that each outcome has to be written out. So to find this probability we would have to take into considertation 56 outcomes.

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?
dbinom(2, 9, .15) * .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?

.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?

They are different because all events are independent of one another. For scenario (a) we need 7 fails and 2 successes and a success on the last attempt. We find out the probability of hitting any two shots our of 9 and multiply by the 15% chance of making the last shot. In part (b) we know that no matter what previous results are she has a 15% chance of making the next shot.