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. \(|Z| > 2\)

##ANSWER: (a) From the normal probability tables, 8.85% of the normal curve lies to the left of the z-score (-1.35).

## [1] 0.08850799

##ANSWER: (b) From the normal probability tables, 6.94% of the normal curve lies to the right of the z-score (1.48).

# use the DATA606::normalPlot function
1 - pnorm(1.48, mean = 0, sd = 1)
## [1] 0.06943662
DATA606::normalPlot(mean = 0, sd = 1, bounds = c(1.48, 4), tails = FALSE)

##ANSWER: (c) From the normal probability tables, 58.9% of the normal curve lies between the z-scores -0.4 and 1.5.

pnorm(1.5, mean = 0, sd = 1) - pnorm(-0.4, mean = 0, sd = 1)
## [1] 0.5886145
DATA606::normalPlot(mean = 0, sd = 1, bounds = c(-0.4, 1.5), tails = FALSE)

##ANSWER: (d) From the normal probability tables, 4.55% of the normal curve lies below z < -2 and above z > 2.

(1 - pnorm(2, mean = 0, sd = 1)) + pnorm(-2, mean = 0, sd = 1)
## [1] 0.04550026
DATA606::normalPlot(mean = 0, sd = 1, bounds = c(-2, 2), tails = TRUE)

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.

##Answer: Men: N(mu = 4313, sd = 583) ## Women: N(mu = 5261, sd = 807)

  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-score = (4948 - 4313) / 583 = 1.09 ## Mary’s Z-score = (5513 - 5261) / 807 = 0.31

Leo’s finishing time was 1.09 standard deviations above the finishing times of men while Mary’s finishing time was 0.31 standard deviations above that of women.

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

##Answer: 86.2% of the runners were faster than Leo in his group while only 62.2% of the runners within Mary’s group were faster than her. We can also look at the z-scores, Leo’s finishing time was 1.09 standard deviations away from the mean, Mary’s finishing time was only 0.31 standard deviations away from the mean, hence the percentage of runners who finished faster than her is lower.

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

##Answer: 86.2% of runners finished faster than Leo, so Leo must have been faster than the remaining 1 - 0.862 = 13.8% of runners.

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

##Answer: 62.2% of runners finished faster than Mary, so Mary must have been faster than the remaining 1 - 0.622 = 37.8% of runners.

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

##Answer: Z-scores can be calculated for non-normal distributions, so anser (b) would remain the same. However, the other answers would change because we wouldn’t be able to use the normal probability tables to look up the percentages based on the z-scores since it’s not a normal or approximately 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.

##ANSWER: 68% of the 25 students (17 students) must fall with +-1sd of the mean, i.e. 56.9 < x < 66.1 which is found to be true looking at the values (See below everything else falls within 1sd of the mean). ## 95% of the 25 students (24 students) must fall with +-2sd of the mean, i.e. 52.4 < x < 70.7 which is found to be true looking at the values (i.e. the values 54, 55, 56, 56, 67, 67, 69 fall between 2sd and 3sd of the mean). ## 99.7% of the 25 students (all of them) must fall with +-3sd of the mean, i.e. 47.8 < x < 75.3 which is found to be true looking at the values (i.e. only 73 falls outside 2sd of the mean).

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

##ANSWER: Yes, these data follow a normal distribution as depicted by the histogram which is unimodal and symmetric. Also, we can see that the Q-Q plot has most of the points close to the straight line and the plot is linear. Hence, the data follow a normal distribution.

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

##ANSWER: Using the qqnormsim function, we compare the actual Q-Q plot to a few simulated plots. The actual plot is more normal than some of the simulated plots. ——————————————————————————–

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?

##ANSWER: Since each transistor is independent of the other, we can use the multiplication rule to calculate the probability. The 10th transistor being the first defective one means the previous 9 were non-defective. If the defect rate is 2%, then the non-defect rate is 98%.

##10th transistor defect probability = (P(Transistor without defect)^9) * P(Transistor with defect) = (0.98^9) * 0.02 = 0.016675

(0.98^9) * 0.02
## [1] 0.01667496
  1. What is the probability that the machine produces no defective transistors in a batch of 100?

##ANSWER: No defective transistors in a batch of 100 means 100 independent events of non-defective transistors, again using the multiplication rule, we get 0.13262

(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?

##ANSWER: This would be 1/0.02, 50 transistors before we encounter the first defective transistor. The sd is 49.5.

1/0.02
## [1] 50
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?

##ANSWER: This would be 1/0.05, 20 transistors before we encounter the first defective transistor. The sd is 19.5.

1/0.05
## [1] 20
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: If we increase the probability, the wait time till we encounter the first defective transistor is reduced.(i.e. in the examples above the wait time was reduced from 50 to 20 transistors). Also the sd decreases as well.

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.

##Answer: The probability that both of them will be boys is 38.23%.

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 outcomes. Confirm that your answers from parts (a) and (b) match.

##ANSWER: The same answer can also be derived manually as the sume of the probabilities of the three outcomes = BBG, BGB and GBB. The probability of two boys and a girl = 0.49 * 0.49 * 0.51 = 0.127449, there are 3 combinations hence 3 * 0.127449 = 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).

##ANSWER: it’s very tedious to come up with all of the combinations, some of which are given below. It’s simpler to use the formula, also below.

GGGGGBBB

GGGGBBBG

GGGBBBGG

GGBBBGGG

GBBBGGGG

BBBGGGGG

BGGGGGBB

BBGGGGGB

GBGGBBGG

GBBGGGBG

dbinom(3, 8, 0.51)
## [1] 0.2098355

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?

##ANSWER: The probability of doing so is 3.9%.

choose(9,2) * (.15)^2 * (1 - .15)^(9-2) * (.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?

##ANSWER: The probability of doing so is 0.15%, since she already has two successful serves.

  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: a is marginal probability of making the 3rd successful serve, b is the conditional probability of making the 3rd sucecssful serve.