Chapter 3

3.2 Area under the curve, Part II

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

require(fastGraph)
## Loading required package: fastGraph
mean<- 0 
SD <- 1
x <- seq(-4, 4, length = 1000)
y <- dnorm(x, mean, SD)

(a) Z > -1.13 

plot(x,y, type="n")
lines(x, y)
shadeDist(-1.13, lower.tail = FALSE)

(b) Z < 0.18

shadeDist(0.18)

(c) Z > 8

shadeDist(8, lower.tail = FALSE)

(d) |Z| < 0.5

shadeDist( c( 0, 0.5 ), , 0, 1, lower.tail = FALSE )

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 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:

. The finishing times of the Men, Ages 30 - 34 group has a mean of 4313 seconds with a standard deviation of 583 seconds.

. The finishing times of the Women, Ages 25 - 29 group has a mean of 5261 seconds with a standard deviation of 807 seconds.

. The distributions of finishing times for both groups are approximately Normal. Remember: a better performance corresponds to a faster finish..

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

N(\(\mu\)=4313, \(s^2\)=583) \(\rightarrow\) Men, Ages 30 - 34

N(\(\mu\)=5261, \(s^2\)=807) \(\rightarrow\) Women, Ages 25 - 29

(b & c) What are the Z-scores for Leo’s and Mary’s finishing times? What do these Z-scores tell you? Did Leo or Mary rank better in their respective groups? Explain your reasoning.

zLeo <- (4948 - 4313) / 538
zMary <- (5513 - 5261) / 807

These z-scores highlight that while Leo may have had a faster time, his z-score places him in a lower percentile based on others inside his bracket as compared to Mary. Mary placed closer to the mean. Both Mary and Leo performed below average.

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

x <- seq(-4, 4, length = 1000)
yMen <- dnorm(x, 4313, 583)
plot(x,yMen, type="n")
lines(x, y)
shadeDist(zLeo, lower.tail = FALSE)

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

yWomen <- dnorm(x, 5261, 807)

shadeDist(zMary, lower.tail = FALSE)

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

Each of these answers would need to be reevaluated and recalculated based upon a non-normal or a normalized distribution. The rate of change along the curve would no longer fit the standard normal and therefore z-scores likely change.

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.

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))
sd<-4.58
mean<-61.52
x <- seq(min(heights), max(heights), length = length(heights))
quantile(heights,0.68)
## 68% 
##  63
quantile(x,0.68)
##   68% 
## 66.92
quantile(heights,0.95)
##  95% 
## 68.6
quantile(x,0.95)
##   95% 
## 72.05
quantile(heights,0.997)
##  99.7% 
## 72.712
quantile(x,0.997)
##  99.7% 
## 72.943

By creating a normal distribution and comparing the quantiles of the data with a normal approximation, we recognize that the quantile ranges of each standard deviation are different, and therefore this is not a standard normal distribution.

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

No, the data are irregular and only approximate a normal distribution. Its line differs from the bell curve. The QQ plot also shows this lack of a linear normal distribution. This data is confusing in that a positive and negative single standard deviation makes a normal fit look accurate, when the distribution in fact is weighted more heavily towards the left. This is difficult to ascertain by merely contrasting with a normal curve.

hist(heights, freq=F, col ="red", ylim=c(0,0.1))
curve(dnorm(x, 61.52, 4.58),min(heights), max(heights), add=T, col="blue")

qqnorm(heights); qqline(heights)

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.

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

defect<-0.02
success<-0.98
success^(9)*defect
## [1] 0.01667496

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

success^(100)
## [1] 0.1326196
dbinom(0, 100, .02)
## [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?

1/defect
## [1] 50
sqrt((1-defect)/defect^2)
## [1] 49.49747

(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

defect <- 0.05
1/defect
## [1] 20
sqrt((1-defect)/defect^2)
## [1] 19.49359

(e) Based on your answers to parts (c) and (d), how does increasing the probability of an event aff???ect the mean and standard deviation of the wait time until success?

Increasing probability reduces the mean and therefore, in this case, increases the likelihood for failure, earlier. The standard deviation is also reduced. These two factors combined mean that the predictability of outcomes becomes higher.

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.

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

totalCombination <- factorial(3)
twoBoysCombination <- factorial(2)
(totalCombination/twoBoysCombination)*(0.51^2)*(0.49)
## [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.

MMF,MFM, FMM

P(A1orA2)=P(A1)+P(A2)

pM<-0.51
pF<-0.49
(pM*pM*pF)*3
## [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 the above hypothetical, the possible combinations were limited. The calculation for part b was simple and short. This process would prove to be more cumbersome (by hand) given the more rare expectation that is 3 boys out of 8.

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.

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

pMake<-0.15
pMiss<-0.85
k<-2
n<-7

kFactorial<- factorial(k)
nFactorial<- factorial(n)
diff<-factorial(n-k)
pMakeTwo<-( nFactorial / (kFactorial * diff) ) * pMake^k * (1-pMake)^(n-k)
pMakeTwo*0.15
## [1] 0.03144761

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

15%

(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?

The probability for a single event is different than gauging the probability for various combinations.. A single event is the basis for sequential probability, whereas understanding combinations is a matter of added dimension. Question A requires both reasonings while B only requires a single value ratio.