`round(1-pnorm(979,1300,sd=200),4)`

`## [1] 0.9458`

`round(1-pnorm(8340,11000,sd = sqrt(1960000)),4)`

`## [1] 0.9713`

```
prob3 <- pnorm(85,80,3)- pnorm(83,80,3)
round(prob3,4)
```

`## [1] 0.1109`

```
GREmean <- 456
GREsd <- 123
minScore <- qnorm(.86)*GREsd + GREmean
round(minScore,0)
```

`## [1] 589`

```
NailMean <- 6.13
NailSD <- 0.06
BotLim <- qnorm(.07) * NailSD + NailMean
TopLim <- qnorm(.93) * NailSD + NailMean
sprintf("The acceptable range of nail lengths is %#.2f - %#.2f",BotLim, TopLim)
```

`## [1] "The acceptable range of nail lengths is 6.04 - 6.22"`

```
A: Top 13% of scores
B: Scores below the top 13% and above the bottom 55%
C: Scores below the top 45% and above the bottom 20%
D: Scores below the top 80% and above the bottom 9%
F: Bottom 9% of scores
Scores on the test are normally distributed with a mean of 78.8 and a standard deviation of 9.8. Find the numerical limits for a C grade. Round your answers to the nearest whole number, if necessary.
```

```
ScoreMean <- 78.8
ScoreSD <- 9.8
BotScLim <- qnorm(.2) * ScoreSD + ScoreMean
TopScLim <- qnorm(.55) * ScoreSD + ScoreMean
sprintf("The range of scores that would result in a C grade are %1.0f - %1.0f",BotScLim, TopScLim)
```

`## [1] "The range of scores that would result in a C grade are 71 - 80"`

```
ACTmean <- 21.2
ACTsd <- 5.4
minACTscore <- qnorm(.45,mean = ACTmean, sd = ACTsd, lower = FALSE)
round(minACTscore,1)
```

`## [1] 21.9`

```
N <- 151
Prob <- 0.09
StuMean <- N * Prob
var <- N * Prob * (1 - Prob)
StuSD <- sqrt(var)
ApxProb <- pnorm(11,mean = StuMean, sd = StuSD)
round(ApxProb,4)
```

`## [1] 0.2307`

```
TireMean <- 48
TireSD <- 7
Tires <- 147
SampleSD <- TireSD/sqrt(Tires)
1 - pnorm(48.83, mean = TireMean, sd = SampleSD)
```

`## [1] 0.07527348`

```
CompMean <- 91
CompSD <- 10
SamComp <- 68
SamCompSD <- CompSD / sqrt(SamComp)
CompChance <- 1-pnorm(93.54,mean = CompMean, sd = SamCompSD)
round(CompChance,4)
```

`## [1] 0.0181`

```
ProbNo = 0.07
N = 540
CheckVal11 <- .03
SampleError <- sqrt((ProbNo*(1-ProbNo))/N)
Prob11 <- pnorm((ProbNo + CheckVal11), mean = ProbNo, sd = SampleError) - pnorm (ProbNo - CheckVal11, mean = ProbNo, sd = SampleError)
round(Prob11,4)
```

`## [1] 0.9937`

```
BotDef <- .23
NBot <- 602
BotError <- sqrt((BotDef*(1-BotDef))/NBot)
CheckVal12 <- .04
Prob12Below <- pnorm (BotDef - CheckVal12, mean = BotDef, sd = BotError)
Prob12Above <- (1 - pnorm (BotDef + CheckVal12, mean = BotDef, sd = BotError))
Prob12 = Prob12Below + Prob12Above
round(Prob12,4)
```

`## [1] 0.0197`

```
BeefN <- 208
BeefMean <- 3.9
BeefVar <- 0.8
BeefSD <- sqrt(BeefVar)
ConfInt13 <- 0.8
#using t-distribution
Beef80t <- qt((1-ConfInt13)/2,BeefN-1)
lowerBeef <- BeefMean + Beef80t * BeefSD
upperBeef <- BeefMean - Beef80t * BeefSD
sprintf("The expected range of beef intake for males over the age of 48 (with an 80 confidence interval) is %#.1f - %#.1f",lowerBeef,upperBeef)
```

`## [1] "The expected range of beef intake for males over the age of 48 (with an 80 confidence interval) is 2.8 - 5.0"`

```
CalSam <- 7472
CalMean <- 16.6
CalVar <- 11
CalSD <- sqrt(CalVar)
ConfInt14 = .98
Cal98t <- qt((1 - ConfInt14) / 2, CalSam - 1)
Cal98t
```

`## [1] -2.326847`

```
lowerCal <- CalMean + Cal98t * CalSD
upperCal <- CalMean - Cal98t * CalSD
sprintf("Assuming a normal distribution, the expected range of mean per capita income in a major city in California (with a 98 confidence interval) is %#.1f - %#.1f",lowerCal,upperCal)
```

`## [1] "Assuming a normal distribution, the expected range of mean per capita income in a major city in California (with a 98 confidence interval) is 8.9 - 24.3"`

`print("The upper-right graph best represents the question")`

`## [1] "The upper-right graph best represents the question"`

`qt(0.05,26)`

`## [1] -1.705618`

383.6, 347.1, 371.9, 347.6, 325.8, 337 ### Using these measurements, construct a 90% confidence interval for the mean level of helium gas present in the facility. Assume the population is normally distributed. ### 16-1)

```
Samples <- c(383.6, 347.1, 371.9, 347.6, 325.8, 337)
SamMean16 <- round(mean(Samples),2)
print("Mean of samples:")
```

`## [1] "Mean of samples:"`

`SamMean16`

`## [1] 352.17`

```
SamSD16 <- round(sd(Samples),2)
print("Standard Deviation of samples :")
```

`## [1] "Standard Deviation of samples :"`

`SamSD16`

`## [1] 21.68`

```
CritVal16 <- abs(round(qt(0.05,length(Samples)-1),3))
print("Critical value for samples :")
```

`## [1] "Critical value for samples :"`

`CritVal16`

`## [1] 2.015`

```
PopSD16 <- SamSD16/sqrt(length(Samples))
Lower16 <- round(SamMean16 - CritVal16 * PopSD16, 2)
Upper16 <- round(SamMean16 + CritVal16 * PopSD16, 2)
print("Lower Bound")
```

`## [1] "Lower Bound"`

`Lower16`

`## [1] 334.34`

`print("Upper Bound")`

`## [1] "Upper Bound"`

`Upper16`

`## [1] 370`

```
N17 <- 16
Mean17 <- 46.4
SD17 <- 2.45
PopSD17 <- SD17/sqrt(N17)
CritVal17 <- abs(round(qt(.1,N17-1),3))
CritVal17
```

`## [1] 1.341`

```
Lower17 <- round(Mean17 - CritVal17 * PopSD17, 1)
Upper17 <- round(Mean17 + CritVal17 * PopSD17, 1)
print("Lower Bound")
```

`## [1] "Lower Bound"`

`Lower17`

`## [1] 45.6`

`print("Upper Bound")`

`## [1] "Upper Bound"`

`Upper17`

`## [1] 47.2`

```
ToyMean <- 8
ToySD <- 1.9
Conf99 <- abs(qt(.005,Inf))
SE18 <- 0.13
n18 <- ceiling((Conf99*ToySD/SE18)^2)
n18
```

`## [1] 1418`

```
Mean19 <- 12.6
Var19 <- 3.61
Conf95 <- abs(qt(.025,Inf))
SE19 <- 0.19
n19 <- ceiling((Conf95 * sqrt(Var19)/SE19)^2)
n19
```

`## [1] 385`

Suppose a sample of 2089 tenth graders is drawn. Of the students sampled, 1734 read above the eighth grade level. Using the data, estimate the proportion of tenth graders reading at or below the eighth grade level. (Write your answer as a fraction or a decimal number rounded to 3 decimal places)

```
Sample20 <- 2089
Prob20_1 <- 1 - (1734/Sample20)
round(Prob20_1,3)
```

`## [1] 0.17`

Suppose a sample of 2089 tenth graders is drawn. Of the students sampled, 1734 read above the eighth grade level.Using the data, construct the 98% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level. (Round your answers to 3 decimal places)

```
Conf98 <- abs(qt(0.01,Inf))
SE20 <- sqrt((Prob20_1 * (1 - Prob20_1))/Sample20)
Lower20 <- round(Prob20_1 - (Conf98 * SE20),3)
Upper20 <- round(Prob20_1 + (Conf98 * SE20),3)
print("Lower Bound")
```

`## [1] "Lower Bound"`

`Lower20`

`## [1] 0.151`

`print("Upper Bound")`

`## [1] "Upper Bound"`

`Upper20`

`## [1] 0.189`

Suppose a sample of 474 tankers is drawn. Of these ships, 156 had spills. Using the data, estimate the proportion of oil tankers that had spills. (Write your answer as a fraction or a decimal number rounded to 3 decimal places)

```
N21 <- 474
P21 <- 156/474
round(P21,3)
```

`## [1] 0.329`

Suppose a sample of 474 tankers is drawn. Of these ships, 156 had spills. Using the data, construct the 95% confidence interval for the population proportion of oil tankers that have spills each month. (Round your answers to 3 decimal places)

```
SE21 <- sqrt((P21*(1-P21)/N21))
Lower21 <- round(P21 - (Conf95 * SE21),3)
Upper21 <- round(P21 + (Conf95 * SE21),3)
print("Lower Bound")
```

`## [1] "Lower Bound"`

`Lower21`

`## [1] 0.287`

`print("Upper Bound")`

`## [1] "Upper Bound"`

`Upper21`

`## [1] 0.371`