Graded Exercises

3.2 Area under the curve, Part II.

What percent of a standard normal distribution N(µ = 0, = 1) is found in each region? Be sure to draw a graph.

  1. Z>1.13
Area under the curve = 1 - 0.8708 = 0.13
x <- seq(-4,4, length=1000)
y <- dnorm(x, 0, 1)
plot(x, y, type="n", main = "Z>1.13")
lines(x, y)
i <- x >= 0.13 & x <= 4
polygon(c(0.13,x[i],4), c(0,y[i],0), col="red")

  1. Z<0.18
Area under the curve = 0.5714
x <- seq(-4,4, length=1000)
y <- dnorm(x, 0, 1)
plot(x, y, type="n", main = "Z<0.18")
lines(x, y)
i <- x >= -4 & x <= 0.57
polygon(c(-4,x[i],0.57), c(0,y[i],0), col="red")

  1. Z>8
Area under the curve = 8 sd from the mean is nearly 0
  1. |Z| < 0.5
Area under the curve = 0.6915-0.3085 = 0.383
x <- seq(-4,4, length=1000)
y <- dnorm(x, 0, 1)
plot(x, y, type="n", main = "|Z| < 0.5")
lines(x, y)
i <- x >= -0.38 & x <= 0.38
polygon(c(-0.38,x[i],0.38), c(0,y[i],0), col="red")

3.4 Triathlon times, Part I.

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 ???nished 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:

. The ???nishing times of the Men, Ages 30 - 34 group has a mean of 4313 seconds with a standard deviation of 583 seconds. . The ???nishing times of the Women, Ages 25 - 29 group has a mean of 5261 seconds with a standard deviation of 807 seconds. . The distributions of ???nishing times for both groups are approximately Normal.

Remember: a better performance corresponds to a faster ???nish.

  1. Write down the short-hand for these two normal distributions.
Men: N(µ = 4313, ?? = 583); Women: N(µ = 5261, ?? = 807)
  1. What are the Z-scores for Leo’s and Mary’s ???nishing times? What do these Z-scores tell you?
Zleo = (4948-4313)/583 = 1.089; Zmary = (5513-5261)/807 = 0.312.
The Z-scores show that Leo performed 1.089 SD’s above the mean, while Mary was only 0.312 SD’s above.
  1. Did Leo or Mary rank better in their respective groups? Explain your reasoning.
Given Leo’s and Mary’s Z-scores (1.089 & 0.312 respectively), Leo performed worse than Mary within their respective competition groups (making more time than the average)
  1. What percent of the triathletes did Leo ???nish faster than in his group?
1 - 0.8621 = 0.137 -> 13.7% (13th percentile)
  1. What percent of the triathletes did Mary ???nish faster than in her group?
1 - 0.6217 = 0.378 -> 37.8% (37th percentile)
  1. If the distributions of ???nishing times are not nearly normal, would your answers to parts (b) - (e) change? Explain your reasoning.
Yes, the Z scores calculation formula and the normal probability table assume a normal distribution

3.18 Heights of female college students.

Below are heights of 25 female college students. 1 54, 2 55, 3 56, 4 56, 5 57, 6 58, 7 58, 8 59, 9 60, 10 60, 11 60, 12 61, 13 61, 14 62, 15 62, 16 63, 17 63, 18 63, 19 64, 20 65, 21 65, 22 67, 23 67, 24 69, 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.
Z1SD -> Area between 56.9 and 66.1. Heights from #4 to #21 fall within this area, meaning 18 out of 25 = 18/25 = 72%
Z2SD -> Area between 52.3 and 70.68. Heights from #1 to #24 fall within this area, meaning 24 out of 25 = 24/25 = 96%
Z3SD -> Area between 47.7 and 75.26. Heights from #1 to #25 fall within this area, meaning 25 out of 25 = 25/25 = 100%
  1. Do these data appear to follow a normal distribution? Explain your reasoning using the graphs provided below.
library("png")
pp <- readPNG("CollegeFHgts.png")
plot.new() 
rasterImage(pp,0,0,1,1)

Yes, they appear to follow a normal distribution, very close approximation to the 68-95-99.7% rule: 72-96-100%.
The histogram resembles a normal distribution line and the normal probability plot is very close to a straight line with a stepwise pattern (due to the discrete data points)

3.22 Defective rate.

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 ???rst with a defect?
P(9 non-defective, 10th defective) = (0.989)0.02 = 0.1764 (in this case defective means success)
  1. What is the probability that the machine produces no defective transistors in a batch of 100?
P(100 non-defective, 101th defective) = 1-P(101th defective) = 1-(0.98)^101 = 0.87
  1. On average, how many transistors would you expect to be produced before the ???rst with a defect? What is the standard deviation?
P(defective) = 0.02, E(X) = 1/0.02 = 50 transistors; sd = sqrt((0.98)/(0.02^2)) = 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 ???rst with a defect? What is the standard deviation?
P(defective) = 0.05, E(X) = 1/0.05 = 20 transistors; sd = sqrt((0.95)/(0.05^2)) = 19.5
  1. Based on your answers to parts (c) and (d), how does increasing the probability of an event a???ect the mean and standard deviation of the wait time until success?
Increasing the probability greatly affects the mean and sd, in this case increasing from 0.02 to 0.05 the defective rate, decreased the waiting time until next defect by more than half (from 50 to 20). Geometric dictributions vary exponentially according to the changes in probabilities (higher probability -> more common the event)

3.38 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.
P(2 boys of total 3 kids) = choose(3,2)(0.51^2)(0.49^1) = 0.38
  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. Con???rm that your answers from parts (a) and (b) match.
P(1st=Boy, 2nd=Boy, 3rd=Girl) = (0.51)^2*(0.49)^1 ##### P(1st=Boy, 2nd=Girl, 3rd=Boy) = (0.51)2(0.49)^1 ##### P(1st=Girl, 2nd=Boy, 3rd=Boy) = (0.51)2(0.49)^1 ##### Therefore: 3 possible orderings times the same probabilites = 3(0.51)2(0.49)1 = 0.38
  1. If we wanted to calculate the probability that a couple who plans to have 8 kids will have 3 boys, brie???y describe why the approach from part (b) would be more tedious than the approach from part (a).
Approach a): P(3 boys of total 8 kids) = choose(8,3)(0.51^3)(0.49^5) = 0.21
Approach b): Compared to a), this approach would be much more tedious as choose(8,3) = 56, meaning 56 possible senarios to outline

3.42 Serving in volleyball.

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?
P(3 successful serves on 10th try) = choose(9,2)(0.15^3)(0.85^7) = 0.039
  1. Suppose she has made two successful serves in nine attempts. What is the probability that her 10th serve will be successful?
P(success on 10th try) = 0.15 (all serve tries are independent, so the probability is the same as the described success rate)
  1. Even though parts (a) and (b) discuss the same scenario, the probabilities you calculated should be di???erent. Can you explain the reason for this discrepancy?
On part a) the probabilities are calculated as a negative binomial distribution, where multiple combinations (scenarios) are possible in order to get to 2 successful serves prior to the last one - this affects the overall probability result. On part b), the probability calculation is simpler as no other combinations (scenarios) need to be considered, only the success rate of the last try