3.6

a)The cutoff time for the fastest 5% of athletes in the men’s group, i.e. those who took the shortest 5% of time to finish.

95th percentile as there are only 5 percent higher than them

qnorm(.95,4313,583)
## [1] 5271.95

So for men the cutoff time for the fastest 5% is approximately 5272 seconds

b) The cutoff time for the slowest 10% of athletes in the women’s group.

qnorm(.10,5261,807)
## [1] 4226.788

So for women the cutoff time for the slowest 10% is approximately 4227 seconds

3.12

a) What percent of passenger vehicles travel slower than 80 miles/hour?

pnorm(80, 72.6, 4.78,lower.tail = TRUE)
## [1] 0.939203

Approximately 94% of passenger vehicles travel slower than 80 mph

b) What percent of passenger vehicles travel between 60 and 80 miles/hour?

pnorm(80,72.6, 4.78)-pnorm(60,72.6,4.78)
## [1] 0.9350083

Approximately 93.5% percent of passenger vehicles travel between 60 and 80 mph

c) How fast do the fastest 5% of passenger vehicles travel? 95th percentile

qnorm(0.95,72.6,4.78)
## [1] 80.4624

The fastest 5% travel at approximately 80.5 mph

d) The speed limit on this stretch of the I-5 is 70 miles/hour. Approximate what percentage of the passenger vehicles travel above the speed limit on this stretch of the I-5.

1-pnorm(70,72.6,4.78)
## [1] 0.7067562

Approximately 71% of passenger vehicles travel above speed limit.

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.

FH<- 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) ##Let FH be Female Height
mean(FH)
## [1] 61.52
sd(FH)
## [1] 4.583667
length(FH[FH<mean+sd&FH>mean-sd])/length(FH)*100
## [1] 68
length(FH[FH<mean+2*sd&FH>mean-2*sd])/length(FH)*100
## [1] 96
length(FH[FH<mean+3*sd&FH>mean-3*sd])/length(FH)*100
## [1] 100

Within the first standard deviation the observations exactly match the 68%. Howeve,when it comes to the second and third standard deviations it does not exactly match it but it is close enough that one could infer this data set does follow the 68-95-99.7% rule. It is important to note how within the second standard deviation there were observations than typically expected (1% compared to 0.3% of the third sd)

b) Do these data appear to follow a normal distribution? Explain your reasoning using the graphs provided below. Yes it does, while there are some observations that could be deemed as “outliers” the dataset is almost perfect symmetric, the histogram follows a bell curve.

3.24

a) A highway patrol officer is hidden on the side of the freeway. What is the probability that 5 cars pass and none are speeding? Assume that the speeds of the cars are independent of each other.

ns <- (72.6-70)/4.78 ##Let ns be no speeding
pns<- pnorm(ns) ##Let pns be Probability none speed
pns^5 ##It has to be 5 repeated events
## [1] 0.1763389

The probability is approximately 18%

b) On average, how many cars would the highway patrol officer expect to watch until the first car that is speeding? What is the standard deviation of the number of cars he would expect to watch?

nc <- 1/pns ##Let nc be number of cars
nc
## [1] 1.414915
Variance<-(1.4149*pns)*(1-pns)
sqrt(Variance)
## [1] 0.541517

He would expect approximately 1.4 cars. In regards to the standard deviation I was not sure as I believe formula is sqrt(n*p(1-p)),not sure if it is correct.

3.30

Pew Research reported in 2012 that the typical response rate to their surveys is only 9%. If for a particular survey 15,000 households are contacted, what is the probability that at least 1,500 will agree to respond?

1-pbinom(1499,15000,0.09)
## [1] 1.326331e-05

3.36

a) the first question she gets right is the 3rd question?

pqr<- 0.25 ##Let pqr be probability question is right
pqw<-(1-pqr) ##Let pqw be probability question is wrong
(pqw^2)*pqr
## [1] 0.140625

It is 14% likely that the first correct question will be the third b) she gets exactly 3 or exactly 4 questions right?

(dbinom(4,5,0.25))+(dbinom(3,5,0.25))
## [1] 0.1025391

c) she gets the majority of the questions right?

1-pbinom(2,5,0.25)
## [1] 0.1035156

3.42

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?

ps<- 0.15 ##Let p be probability of success
k<- 3
n<-10
factorial(n-1)/(factorial(k-1)*(factorial(n - k)))*ps^k*(1-ps)^(n-k)
## [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?

15% because the events 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?

The reason for this is that in a we are calculating for the 10th serve to be succesful, given that 2 of the previous 9 attempts were successful.However, for b we calculate the possibility that she has a succesful attempt, because they are independent events the probability at the start of each attempt will always be the same.