3.2 Area under the curver, Part II. What percent if a standard normal distribution N(u = 0, s=1)is found in each region? Be sure to draw a graph

a. Z > -1.13: The area under the curve is 87.07%

1-pnorm(-1.13, mean = 0, sd = 1)
## [1] 0.8707619
normalPlot(mean = 0, sd = 1, bounds=c(-1.13, 4), tails = FALSE)

b. Z < 0.18: The area under the curve is 57.17%

pnorm(0.18, mean = 0, sd = 1)
## [1] 0.5714237
normalPlot(mean = 0, sd = 1, bounds=c(-4,0.18), tails = FALSE)

c. Z > 8: The area under the curve is approximately 0 or 6.66e-16

1-pnorm(8, mean = 0, sd = 1)
## [1] 6.661338e-16
normalPlot(mean = 0, sd = 1, bounds=c(8, 999999), tails = FALSE)

d. |Z| > 0.5: The area under the curve is 38.3%

normalPlot(mean = 0, sd = 1, bounds=c(-.5, .5), tails = FALSE)

3.4 Triathlon times, Part 1.

a. Writedown the short-hand for these two normal distributions.

Leo’s Group: \(N(\mu = 4313, \sigma =583)\) | Mary’s Group: \(N(\mu = 5261, \sigma =807)\)

b. What are the Z scores for Leo’s and Mary’s finishing times? What do Z scores tell you?

Leo was 1.09 sds from the mean and Mary was 0.31 sds from the mean. Neither Leo or Mary are real speedsters.

Z_leo <- (4948 - 4313) / 583
Z_leo
## [1] 1.089194
Z_mary <- (5513 -5261) / 807
Z_mary
## [1] 0.3122677

c. Did Leo or Mary rank better in thier respecive group.

Mary did ranked better than Leo. Only 13.8% of runners were slower than Leo, but 37.8% were slower than Mary.

1-pnorm(4948, mean = 4313, sd=583)
## [1] 0.1380342
1-pnorm(5513, mean = 5261, sd=807)
## [1] 0.3774186

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

Leo finished faster than 13.8% of triathletes.

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

Mary finished faster than 37.7% of triathletes.

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

The answers to c - e would change, because normality is a requirement of the calculations employed.

3.18 Heights of female college students

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 appear to follow the 68-95-99.7 rule (see below).

# Should be close to 0.68
pnorm(66.1, mean=61.52, sd=4.58) - pnorm(56.94, mean=61.52, sd=4.58)
## [1] 0.6826895
# Should be close to 0.95
pnorm(70.68, mean=61.52, sd=4.58) - pnorm(52.36, mean=61.52, sd=4.58)
## [1] 0.9544997
# Should be close to 0.997
pnorm(75.26, mean=61.52, sd=4.58) - pnorm(47.78, mean=61.52, sd=4.58)
## [1] 0.9973002

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

Yes, the data do seem to follow the normal distribution. The q-q plot tracks the straight line well with minimal skew in the tales (far left and right). Additionally the curves of the overlay fit the histrogram well. This also supports the data follow a normal distribution.’

3.22 Defective Rate

a. What’t the probability the 10th transistor produced is the first with a defect.

#geometric distribution
# (1-p)^(n-1) * p

P <- (1-.02)^(10-1)*.02
P
## [1] 0.01667496

b. What is the probability that the machine produces no defective transistors n a batch of 100

#(1-p)^n
P <- (1-.02)^100
P
## [1] 0.1326196

c. On average, how many transistors would you expect to be produced before the first with a defect? What is the standard deviation?

# expected value <- 1/p
ev <- 1/.02
ev
## [1] 50
# sd <- sqrt((1-p))/p*p)
sd <- sqrt((1-.02)/(.02*.02))
sd
## [1] 49.49747

d. Another machine that also produces transistors has a 5% defective rate each transistor is produced independent if the others. On average how many transitors would you expect to be produced with this machine before the first with a defect? What is the standard deviation?

# expected value <- 1/p
ev <- 1/.05
ev
## [1] 20
# sd <- sqrt((1-p))/p*p)
sd <- sqrt((1-.05)/(.05*.05))
sd
## [1] 19.49359

3.38 Male children

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

There is a 0.383 probability that exactly two children will be boys

p = .51
P = dbinom(2,3,p)
P
## [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 possibilities: bbg, bgb, gbb

#using addition rule 
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)

Using the choose function we know there are 56 combinations, thus making (b) a tedious and error prone approach compared to (a).

3.42 Serving in volleyball

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

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

p <- .15
n <- 10
k <- 3
P <- choose(n-1, k-1) *(1-p)^(n-k)*p^k
P
## [1] 0.03895012

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

The probability is still 0.15 because the servers are independent.

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 a is dealing with a joint probability and Part b is dealing the the probability of a single independent trial - the probability that her 10th serve will be successful.