HW3_606

Q.3.2 Be sure to draw a graph. (please check the graph in another pdf file.)

1. Z > -1.13

1-pnorm(-1.13,mean=0,sd=1)

## [1] 0.8707619

2. Z < 0.18

pnorm(0.18,mean=0,sd=1)

## [1] 0.5714237

3. Z > 8

1-pnorm(8,mean=0,sd=1)

## [1] 6.661338e-16

4. |Z| < 0.5

pnorm(0.5,mean=0,sd=1)-pnorm(-0.5,mean=0,sd=1)

## [1] 0.3829249

Q.3.4

1. Write down the short-hand for these two normal distributions.

#Leo score 4948 seconds, with mean of 4313 seconds with sema of 583 seconds.
#marry score 4948 seconds, with mean of 5513 seconds with sema of 807 seconds.

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

#Z-scores for Leo 
z1<-(4948-4313)/583
z1

## [1] 1.089194

#Z-score for Mary 
z2<-(5513-5261)/807
z2

## [1] 0.3122677

# z=(sample - mean)/standar_dev
#Z is a number of standar_dev fall above or below the mean.

3. Did Leo or Mary rank better in their respective groups? Explain your reasoning.

# Mary rank better in her respective group than Leo, since her Z score lower than Leo's and which means there are lesser percentage of teamate faster than her.

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

1-pnorm(4948,mean=4313,sd=583)

## [1] 0.1380342

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

1-pnorm(5513,mean=5261,sd=807)

## [1] 0.3774186

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

#Leo: 1-Pr(1.089)=1- 0.86 = 0.14  VS 0.14
#Marry: 1-Pr(0.3123)=1-0.62 = 0.38 VS 0.38
#approximatly 0 changed.

Q.3.18 (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.

#68% : between (61.52-4.58, 61.52+4.58) = (56.94,66.1)
#95% : between (61.52-2*4.58, 61.52+2*4.58) = (52.36,70.68)
#99.7% : between (61.52-3*4.58, 61.52+3*4.58) = (47.78,75.26)

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

#It's a normal distribution.
#many of the data either on the line or very closed to the line. And there are a few outliners at two tails. 

Q.3.22 p=0.02 (defect) q=0.98 (a) What is the probability that the 10th transistor produced is the first with a defect?

#0.98^9*0.02=0.016675

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

#0.98^100=0.13262

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

#mean1=1/p1=50
#standard_dev1=sqrt((1-p1)/p1^2)=49.4975

4. 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?

#mean2=1/0.05= 20
#standard_dev2=sqrt((1-p2)/p2^2)=19.4936

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

# sample 1 with 2% defective; sample 2 with 5% defective.
# z1:(X1-50)/49.4975
# z2: (x2-20)/19.4936
#since z2 > z1, the probability of the defective in machine tow is higher.

Q.38 (a) Use the binomial model to calculate the probability that two of them will be boys.

#C(3,2)*0.51^2*(1-0.51)=0.38235

2. 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.

#b:boy; g:girl
#{bbg,bgb,gbb} => probaility = 1/3

3. 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).

#since in part b, even listed one by one and order is matter.

Q.42 (a) What is the probability that on the 10th try she will make her 3rd successful serve?

#negative binomial with k=3,n=10
#C(9,2)*0.15^3*(1-0.15)^7=0.03895

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

#1-0.15=0.85

3. 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?

#In part a, it eastimate a series combination events happen.
#In part b, it estimate next successful even, Also the previous events already exist, so the last one is indepandent with previous' 9 event.