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 (b) Z<0.18 (c) Z>8 (d) |Z| < 0.5
normalPlot(bounds=c(-1.13,Inf),mean=0,sd=1)#87%

normalPlot(bounds=c(-Inf,.18),mean=0,sd=1)#57%

normalPlot(bounds=c(8,Inf),mean=0,sd=1) #0%

normalPlot(bounds=c(-.5,.5),mean=0,sd=1) #38%

3.4 Triathlon times, Part I.

  1. Write down the short-hand for these two normal distributions.
    Men = N(µ=4313,??=583) | Women = N(µ=5261,??=807)
  2. What are the Z-scores for Leo’s and Mary’s ???nishing times? What do these Z-scores tell you?
    (ObservedEvent - mean / SD)
LeoZ = (4948-4313)/583
LeoZ
## [1] 1.089194
MaryZ = (5513-5261)/807
MaryZ
## [1] 0.3122677

  1. Did Leo or Mary rank better in their respective groups? Explain your reasoning.
    Mary did better than Leo in relation to their respective groups. You can see this through Z score comparison. Whichever number is lower (since this is a race the goal is to have the lowest time) is a better performance.

  2. What percent of the triathletes did Leo ???nish faster than in his group?

  3. What percent of the triathletes did Mary ???nish faster than in her group?

pnorm(LeoZ,lower.tail = FALSE) # This is the % of Leo's group he beat
## [1] 0.1380342
pnorm(MaryZ,lower.tail=FALSE)# This is the % of Mary's group she beat
## [1] 0.3774186
  1. If the distributions of ???nishing times are not nearly normal, would your answers to parts (b) - (e) change? Explain your reasoning
    Answer B and C would not change as the Z score does not account for whether or not distributions are normal, its simply something you use if so.
    Answer E would change because normal probability measures rely on a normal distribution.

3.18 Heights of female college students.

  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.
    Using the 68-95-99 rule, and looking at the heights column. About 80% (20 out of 25) of the sample space is within 1 SD, 100% are within 3 SD (Only 1 item outside 2 SD) means yeah…eyeballing this, I can easily say this is within the normal distributions.
  2. Do these data appear to follow a normal distribution? Explain your reasoning using the graphs provided below.
heights = 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)
heightsMean = mean(heights)
heightsSD = sd(heights)
hist(heights,ylim=c(0,.13),probability=TRUE)
x = 45:80
y = dnorm(x=x,mean=heightsMean,sd = heightsSD)
lines(x=x,y=y,col='purple')

qqnormsim(heights)

Looks approximately normal to me, slightly right skewed, nothing too big. I’d call this a normal distribution, nothings perfect. If it were, that wouldn’t be normal :)

3.22 Defective rate

  1. What is the probability that the 10th transistor produced is the ???rst with a defect?
    Geometric distribution!
probFailure = .02
dgeom(9,prob = probFailure)
## [1] 0.01667496
  1. What is the probability that the machine produces no defective transistors in a batch of 100?
.98^100
## [1] 0.1326196
  1. On average, how many transistors would you expect to be produced before the ???rst with a defect? What is the standard deviation?
eV = 1/.02
eV
## [1] 50
  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?
    Var(X) = 1-p / p^2
sqrt((1-.05)/(.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? Increasing the probability of failure the event that it is a failure becomes more probable. So the SD of waiting time and sample space before failure become smaller.

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.
dbinom(2,3,0.51)
## [1] 0.382347
  1. Write out all possible orderings of 3 children, 2 of whom are boys.
    {BGB, BBG, GBB}
3*(.49)*(.51^2)
## [1] 0.382347
  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). Writing a sample space can be quite tedious.

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?
choose(9,2)*((.15)^3)*(.85)^7
## [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%, events are independant
  2. 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?
    because a negative binomial distribution looks for probabbility of nth success out of kth times, where as b is asking for the probability for success, which is 15%