Graded: 3.2,3.4, 3.18, 3.22, 3.38, 3.42

3.2

Area under the curve, Part II. What percent of a standard normal distribution N(?? = 0, ! = 1) is found in each region? Be sure to draw a graph. (a) Z > ???1.13

x   <- seq(-3,3,length=1000)
y   <- dnorm(x,mean=0, sd=1)
plot(x,y, type="l", lwd=1)
x=seq(-1.13,3,length=100)
y=dnorm(x,mean=0,sd=1)
polygon(c(-1.13,x,3),c(0,y,0),col="red")

#since it is greater than Percentag is
100*(1 - pnorm(-1.13))  
## [1] 87.07619
  1. Z < 0.18
x   <- seq(-3,3,length=1000)
y   <- dnorm(x,mean=0, sd=1)
plot(x,y, type="l", lwd=1)
x=seq(-3,0.18,length=100)
y=dnorm(x,mean=0,sd=1)
polygon(c(-3,x,0.18),c(0,y,0),col="red")

#since it is greater than Percentag is
100*( pnorm(0.18))  
## [1] 57.14237
  1. Z > 8
x   <- seq(-3,3,length=1000)
y   <- dnorm(x,mean=0, sd=1)
plot(x,y, type="l", lwd=1)
x=seq(3,3,length=100)
y=dnorm(x,mean=0,sd=1)
polygon(c(3,x,3),c(0,y,0),col="red")

#since it is greater than Percentag is
100*(1- pnorm(8))  
## [1] 6.661338e-14
  1. |Z| < 0.5
x   <- seq(-3,3,length=1000)
y   <- dnorm(x,mean=0, sd=1)
plot(x,y, type="l", lwd=1)
x=seq(-0.5,0.5,length=100)
y=dnorm(x,mean=0,sd=1)
polygon(c(-.5,x,.5),c(0,y,0),col="red")

#since it is greater than Percentag is
100*( pnorm(.5)- pnorm(-.5))  
## [1] 38.29249

Q3.4

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.

Normal distribution for Men is \[ N ( \mu = 4313, \sigma = 583 ) \]

Normal distribution for Women is \[ N ( \mu = 5261, \sigma = 807 ) \]

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

\[ Zscore = \frac{(Obs - \mu)}{\sigma} \]

Leo’s Zscore :

(4948-4313)/583
## [1] 1.089194

Mary’s Zscore :

(5513-5261)/807
## [1] 0.3122677

THe Zscores tell me that Leo is the better athlete in his peer group compared to Mary

  1. Did Leo or Mary rank better in their respective groups? Explain your reasoning. Leo ranked better because of a Zscore of 1.08 is in 86th percentile while Mary’s Zscore of 0.312 is at the 62th percentile

  2. What percent of the triathletes did Leo finish faster than in his group? Leo finished faster than 100-86 = 14%
  3. What percent of the triathletes did Mary finish faster than in her group? Mary finished faster than 100-62 is 38%
  4. If the distributions of finishing times are not nearly normal, would your answers to parts
      1. change? Explain your reasoning. Part b will stay the same because Zscores are not restricted to normal distributions however part c to e, rely on the Zscores of the a normal distribution table for interpretation therefore answers to part c to e will have to change

Q3.18

Below are heights of 25 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.

url <- "https://raw.githubusercontent.com/jbryer/DATA606Fall2017/master/Data/Data%20from%20openintro.org/Ch%203%20Exercise%20Data/fheights.txt"
height <- read.table(url, header = TRUE, stringsAsFactors = FALSE)
head(height)
##   heighs
## 1     54
## 2     55
## 3     56
## 4     56
## 5     57
## 6     58
hmean <- 61.52
hsd <- 4.58
height$z <- (height$heighs - hmean)/hsd

one standard deviation should be 68 percent -1<=Z<=1

(sum(height$z<=1 & height$z>=-1)/nrow(height)) * 100
## [1] 66.66667

We get 67% which is close to 68

Let’s check 2 standard deviation should be 95% -2<=Z<=2

(sum(height$z<=2 & height$z>=-2)/nrow(height)) * 100
## [1] 95.83333

We get 95.83% which is close to 95

Let’s check 3 standard deviation should be 99.7% -3<=Z<=3

(sum(height$z<=3 & height$z>=-3)/nrow(height)) * 100
## [1] 100

We get 100% which is close to 99.7

Therefore we can say the distribution follows the 68-95-99.7 rule

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

The distribution seems to be unimodial and symmetric. So it is quite clear to be a normal distribution. When analyzing the qqplot it seems to follow a straight line with only one outlier

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?

P(10th Defect) = .98^9 x .02 =0.0167

  1. What is the probability that the machine produces no defective transistors in a batch of 100?

P(100 no defect) = .98^100 = 0.132

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

Mean or E(x) =1/0.02 =50 SD= sqrt(1-p/p^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 first with a defect? What is the standard deviation?

Mean or E(x) =1/0.05 =20 SD= sqrt(1-p/p^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 of an event decreases the expectation or the mean of variable and the standard deviation before success. We need less number of trials

3.38

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. P(k=2) = Number of scenarios x Pr(single scenario) Number of scenarios = 3!/2!x1! =3 Pr(k=2)single = 0.51^2 *(1-0.51) = 0.127

Pr(two boys) =3 x 0.127 =0.382

  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. Confirm that your answers from parts (a) and (b) match.

Scenario 1 = Pr(first child =boy) x Pr(Second child boy) x Pr(third child =girl) =0.51 x 0.51 x 0.49 Scenario 2 = Pr(first child =boy) x Pr(Second child girl) x Pr(third child =boy)= 0.51 x 0.49 x 0.51 Scenario 3 = Pr(first child =girl) x Pr(Second child boy) x Pr(third child =boy) = 0.49 x 0.51 x 0.51

Pr(two boys) = Pr(scenario 1) + Pr(scenario 2) + Pr(scenario 3) = 0.382

  1. 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). It would be more tedious because we have to calculate the probability for each scenario of having 3 boys in 8kid, which is 56 possibilities

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?

Pr(serve)= 0.15 No of scenarios =9!/(2!x7!) =36 Pr(3rd success at 10th) = #scenarios x 0.15^3 x 0.85^7 = 0.039 (b) Suppose she has made two successful serves in nine attempts. What is the probability that her 10th serve will be successful?

P(2 successive serves) = 8 x 0.15^3 x 0.85^7 = 0.008655 (c) 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? It is different because the number of different scenarios changed