(a) Z > −1.13

87.1 percent of a standard normal distribution is in the area above Z = −1.13.

normalPlot(mean = 0, sd = 1, bounds = c(-1.13, Inf), tails = FALSE)

(b) Z < 0.18

57.1 percent of a standard normal distribution is in the area below Z = 0.18.

normalPlot(mean = 0, sd = 1, bounds = c(-Inf, 0.18), tails = FALSE)

(c) Z > 8

Approximately 0 percent of a standard normal distribution is in the area above Z = 8. In a normal distribution any value more than 3 standard distributions away from the mean is extremely rare and over 4 standard distributions away from the mean is so rare as to be negligible.

normalPlot(mean = 0, sd = 1, bounds = c(8, Inf), tails = FALSE)

(d) |Z| < 0.5

38.3 percent of a standard normal distribution is in the area between Z = -0.5 and Z = 0.5.

normalPlot(mean = 0, sd = 1, bounds = c(-0.5, 0.5), tails = FALSE)

(a) Write down the short-hand for these two normal distributions.

• The finishing times of the Men, Ages 30 - 34 group has a mean of 4313 seconds with a standard deviation of 583 seconds. \[N(\mu=4313, \sigma=583)\]
• The finishing times of the Women, Ages 25 - 29 group has a mean of 5261 seconds with a standard deviation of 807 seconds. \[N(\mu=5261, \sigma=807)\]

(b) What are the Z-scores for Leo’s and Mary’s finishing times? What do these Z-scores tell you?

LeoZ <- (4948-4313)/583
round(LeoZ, 2)

## [1] 1.09

MaryZ <- (5513-5261)/807
round(MaryZ, 2)

## [1] 0.31

The Z-score for Leo’s finishing time is 1.09 which means he finished 1.09 standard deviations above the mean for his age group.

The Z-score for Mary’s time is 0.31 which means she finished 0.31 standard deviations above the mean for her age group.

(c) Did Leo or Mary rank better in their respective groups? Explain your reasoning.

Leo ranked better than Mary for his age group since his Z-score was much higher than Mary’s. The higher the Z-score the higher the percentile rank.

(d) What percent of the triathletes did Leo finish faster than in his group?

round(pnorm(LeoZ)*100, 2)

## [1] 86.2

Leo finished faster then 86.2 percent of the other competitors in his age group.

(e) What percent of the triathletes did Mary finish faster than in her group?

round(pnorm(MaryZ)*100, 2)

## [1] 62.26

Mary finished faster then 62.26 percent of the other competitors in her age group.

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

Yes, my answers to parts (b) - (e) would change if the distribution is not normal. You could still calculate a Z-score, but you cannot use Z-scores to estimate probabilities and percentile ranks in a non-normal distribution. In that case you would need to know more information about the distribution than just the mean and standard deviation. If you had the total number of competitors in their age groups and the number of competitors who did better or worse than Leo and Mary then you could calculate a percentile rank.

(a) 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.

Yes, the heights of female college students in this dataset do seem to follow the 68-95-99.7% Rule since they appear to be normally distributed. In fact they follow a normal pattern even more closely than the simulations.

set.seed(250)
qqnormsim(heights)

(b) Do these data appear to follow a normal distribution? Explain your reasoning using the graphs

provided below.

Yes the data represented by the graphs above appear to follow a normal distribution. It’s a little harder to tell based on the histogram than on the normal probability plot. In the histogram the shape created by the tops of the bars roughly follows the normal curve, but in the normal probability plot, it’s much clearer that the data follows a straight line.

(a) What is the probability that the 10th transistor produced is the first with a defect?

The probability that the 10th transistor produced is the first with a defect is 0.017.

# geometric distribution
p <- .02
n <- 10
P <- (1-p)^(n-1)*p
round(P, 3)

## [1] 0.017

(b) What is the probability that the machine produces no defective transistors in a batch of 100?

There is a 0.133 probability that the machine produces no defective transistors in a batch of 100.

round((1-p)^100, 3)

## [1] 0.133

(c) On average, how many transistors would you expect to be produced before the first with a

defect? What is the standard deviation?

You would expect on average that 50 transistors would be produced before the first with a defect with a standard deviation of 49.5.

EV <- 1/p
EV

## [1] 50

sd <- sqrt((1-p)/p^2)
round(sd, 2)

## [1] 49.5

(d) 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?

You would expect on average that 20 transistors would be produced before the first with a defect on this machine, with a standard deviation of 19.5.

p <- .05
EV <- 1/p
EV

## [1] 20

sd <- sqrt((1-p)/p^2)
round(sd, 1)

## [1] 19.5

(e) 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 an event (that is a Bernoulli random variable) makes both the expected value and the standard deviation lower.

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

There is a probability of 0.382 that exactly two of them will be boys.

# binomial distribution
p = .51
P = round(dbinom(2,3,p),3)
P

## [1] 0.382

# double check
n <- 3
k <- 2
p <- .51
factorial(n)/(factorial(k)*factorial(n-k))*p^k*(1-p)^(n-k)

## [1] 0.382347

(b) 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.

There are three possible orderings of the children. Each with a probability of .51*.51*.49

1. bbg
2. bgb
3. gbb

# the addition rule for disjoint outcomes
P <- .51*.51*.49 + .51*.49*.51 + .49*.51*.51
P

## [1] 0.382347

(c) 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).

In this approach there are 56 different ways you could have exactly 3 boys out of 8 kids, so you’d have to calculate the probability of having exactly 3 boys and then add it up 56 times!

# number of ways you can get k successes in n trials
fc=function(n,k){factorial(n)/(factorial(k)*factorial(n-k))}
fc(8, 3)

## [1] 56

(a) What is the probability that on the 10th try she will make her 3rd successful serve?

There is a 0.039 probability that she will make her 3rd successful serve on the 10th trial.

# first find the probability of exactly 2 successes in 9 trials
p = .15
P = dbinom(2,9,p)
# then multiply by the probability that she will be successful on her 10th try
round(P * .15,3)

## [1] 0.039

(b) Suppose she has made two successful serves in nine attempts. What is the probability that her 10th serve will be successful?

The probability that her 10th serve will be successful is still 0.15. The gambler’s fallacy makes you think that the outcome of prior trials might have an affect on the next trial, but if the trials are independent, each trial has exactly the same probability as every other trial.

(c) 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?

Part (b) only asks about the probability of the 10th trial whereas part (a) asks about the joint probability of all of the first 10 trials.