The weights of steers in a herd are distributed normally. The variance is 40,000 and the mean steer weight is 1300 lbs.
Find the probability that the weight of a randomly selected steer is greater than 979 lbs. (Round your answer to 4 decimal places)
Answer: variance = 40000
mean = 1300
standard deviation = sd = sqrt(Variance) = sqrt(40000)
z = (x-mean)/sd
if x = 979, then z = 979-1300/sqrt(40000) = -1.605
P(x>979) = P(z>-1.605) = 1 - round(pnorm(x, mean, sd), 4) = 0.9458 i.e, 94.58% probability
x <- 979
mean <- 1300
sd <- sqrt(40000)
z <- (x- mean)/sd
P <- 1 - round(pnorm(x, mean, sd), 4)
paste("The proability of weight greater than 979 lbs:", P)## [1] "The proability of weight greater than 979 lbs: 0.9458"SVGA monitors manufactured by TSI Electronics have life spans that have a normal distribution with a variance of 1,960,000 and a mean life span of 11,000 hours.
If a SVGA monitor is selected at random, find the probability that the life span of the monitor will be more than 8340 hours. (Round your answer to 4 decimal places)
Answer: variance = 1960000
mean = 11000
Standard Deviation = sd = sqrt(Variance) = sqrt(1960000)
z = (x-mean)/sd
if x = 8340, then z = (8340-11000)/sqrt(1960000) = -1.9
P(x>8340) = P(z>-1.9) = 1 - round(pnorm(x, Mean, sd), 4) = 0.9713 i.e, 97.13% probability
x <- 8340
mean <- 11000
sd <- sqrt(1960000)
z <- (x-mean)/sd
P <- 1 - round(pnorm(x, mean, sd), 4)
paste("The probability of life span exceed 8340 hours:", P)## [1] "The probability of life span exceed 8340 hours: 0.9713"Suppose the mean income of firms in the industry for a year is 80 million dollars with a standard deviation of 3 million dollars.
If incomes for the industry are distributed normally, what is the probability that a randomly selected firm will earn between 83 and 85 million dollars? (Round your answer to 4 decimal places)
Answer: mean = 80000000
standard deviation = sd = sqrt(Variance) = 3000000
variance = power(sd,2) = 9x10^12
z = (x-mean)/sd
x1 = 83000000
x2 = 85000000
P(83000000 > x > 85000000) = round((1-pnorm(x1, mean, sd))-(1-pnorm(x2, mean, sd)), 4) = 0.1109 i.e, 11.09% probability
x1 <- 83000000
x2 <- 85000000
mean <- 80000000
sd <- 3000000
z <- (x-mean)/sd
P1 <- 1-pnorm(x1, mean, sd)
P2 <- 1-pnorm(x2, mean, sd)
P <- round((P1-P2), 4)
paste("The probability of income within the limit range:", P)## [1] "The probability of income within the limit range: 0.1109"Suppose GRE Verbal scores are normally distributed with a mean of 456 and a standard deviation of 123. A university plans to offer tutoring jobs to students whose scores are in the top 14%.
What is the minimum score required for the job offer? Round your answer to the nearest whole number, if necessary.
Answer: mean = 456
standard deviation = sd = sqrt(variance) = 123
variance = power(sd,2) = 123^2
P(x) = 100%-14% = 1-0.14=0.86
x <- 0.86
mean <- 456
sd <- 123
P <- round(qnorm(x,mean,sd),0)
paste("The minimum score :", P)## [1] "The minimum score : 589"The lengths of nails produced in a factory are normally distributed with a mean of 6.13 centimeters and a standard deviation of 0.06 centimeters. Find the two lengths that separate the top 7% and the bottom 7%. These lengths could serve as limits used to identify which nails should be rejected. Round your answer to the nearest hundredth, if necessary.
Answer: mean = 6.13
standard deviation = sd = sqrt(variance) = 0.06
variance = power(sd,2) = 0.06^2 = 0.0036
upper_Percentile = 7% = 1-0.07 = 0.93
lower_Percentile = 7% = 0.07
upper_Limit = qnorm(upper_Percentile, mean, sd) = 6.22
lower_Limit = qnorm(lower_Percentile, mean, sd) = 6.04
mean <- 6.13
sd <- 0.06
upper_Percentile <- 0.93
lower_Percentile <- 0.07
paste("The upper limit length :", round(qnorm(upper_Percentile, mean, sd), 2), "centimeters")## [1] "The upper limit length : 6.22 centimeters"paste("The lower limit length :", round(qnorm(lower_Percentile, mean, sd), 2), "centimeters")## [1] "The lower limit length : 6.04 centimeters"An English professor assigns letter grades on a test according to the following scheme.
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.
Answer: mean = 78.8
standard deviation = sd = sqrt(variance) = 9.8
variance = power(sd,2) = 9.8^2
upper_percentile_C = 45% = 1-0.45 = 0.55
lower_percentile_C = 20% = 0.20
upper_Limit = qnorm(upper_percentile_C, mean, sd) = 80
lower_Limit = qnorm(lower_percentile_C, mean, sd) = 71
mean <- 78.8
sd <- 9.8
upper_percentile_C <- 1-0.45
lower_percentile_C <- 0.20
paste("The upper limit for C grade :", round(qnorm(upper_percentile_C, mean, sd), 0))## [1] "The upper limit for C grade : 80"paste("The lower limit for C grade :", round(qnorm(lower_percentile_C, mean, sd), 0))## [1] "The lower limit for C grade : 71"Suppose ACT Composite scores are normally distributed with a mean of 21.2 and a standard deviation of 5.4. A university plans to admit students whose scores are in the top 45%.
What is the minimum score required for admission? Round your answer to the nearest tenth, if necessary.
Answer: mean = 21.2
standard deviation = sd = sqrt(variance) = 5.4
variance = power(sd,2) = 5.4^2
upper_percentile = 45% = 1-0.45 = 0.55
upper_Limit = qnorm(upper_Percentile, mean, sd) = 21.9
mean <- 21.2
sd <- 5.4
upper_percentile <- 1-0.45
paste("The upper limit for C grade :", round(qnorm(upper_percentile, mean, sd), 1))## [1] "The upper limit for C grade : 21.9"Consider the probability that less than 11 out of 151 students will not graduate on time. Assume the probability that a given student will not graduate on time is 9%.
Approximate the probability using the normal distribution. (Round your answer to 4 decimal places.)
Answer: P(<11) = 10
Size = 151
P(fail) = 9% = 0.09
P = pbinom(P(<11), Size, P(fail)) = 0.192
PlessThan11 <- 10
Size <- 151
Pfail <- 0.09
paste("The probability :", round(pbinom(PlessThan11, Size, Pfail), 4))## [1] "The probability : 0.192"The mean lifetime of a tire is 48 months with a standard deviation of 7.
If 147 tires are sampled, what is the probability that the mean of the sample would be greater than 48.83 months? (Round your answer to 4 decimal places)
Answer: mean = 48
standard deviation = sd = sqrt(variance) = 7
variance = power(sd,2) = 7^2 = 49
size = 147
standard mean = sm = sd / sqrt(size)
limit = 48.83
P = 1-pnorm(limit, mean, sdm) = 0.0753
mean <- 48
sd <- 7
size <- 147
sm <- sd / sqrt(size)
limit <- 48.83
paste("The probability :", 1-round(pnorm(limit, mean, sm), 4))## [1] "The probability : 0.0753"The quality control manager at a computer manufacturing company believes that the mean life of a computer is 91 months, with a standard deviation of 10.
If he is correct, what is the probability that the mean of a sample of 68 computers would be greater than 93.54 months? (Round your answer to 4 decimal places)
Answer: mean = 91
standard deviation = sd = sqrt(variance) = 10
variance = power(sd,2) = 10^2 = 100
size = 68
standard mean = sm = sd / sqrt(size)
limit = 93.54
P = 1-pnorm(limit, mean, sm) = 0.0181
mean <- 91
sd <- 10
size <- 68
sm <- sd / sqrt(size)
limit <- 93.54
paste("The probability :", 1-round(pnorm(limit, mean, sm), 4))## [1] "The probability : 0.0181"A director of reservations believes that 7% of the ticketed passengers are no-shows.
If the director is right, what is the probability that the proportion of no-shows in a sample of 540 ticketed passengers would differ from the population proportion by less than 3%? (Round your answer to 4 decimal places)
Answer: P0 = 7% = 0.07
size = 540
limit = 3% = 0.03
standard error = se = sqrt(P0*(1-P0)/size)
Upper_Percentile = P0+limit=0.07+0.03=0.10
Lower_Percentile = P0-limit=0.07-0.03=0.04
Upper_Limit = pnorm(Upper_Percentile, P0, se) = 0.9968553
Lower_Limit = pnorm(Lower_Percentile, P0, se) = 0.003144737
Probability = Upper_Limit - Lower_Limit = 0.9937
P0 <- 0.07
size <- 540
se <- sqrt(P0*(1-P0)/size)
limit <- 0.03
Upper_Percentile = P0+limit
Lower_Percentile = P0-limit
Upper_Limit <- pnorm(Upper_Percentile, P0, se)
Lower_Limit <- pnorm(Lower_Percentile, P0, se)
paste("The probability :", round(Upper_Limit - Lower_Limit, 4))## [1] "The probability : 0.9937"A bottle maker believes that 23% of his bottles are defective.
If the bottle maker is accurate, what is the probability that the proportion of defective bottles in a sample of 602 bottles would differ from the population proportion by greater than 4%? (Round your answer to 4 decimal places)
Answer: P0 = 23% = 0.23
size = 602
limit = 4% = 0.04
standard error = se = sqrt(P0*(1-P0)/size)
Upper_Percentile = P0+limit=0.23+0.04=0.27
Lower_Percentile = P0-limit=0.23-0.04=0.19
Upper_Limit = 1-pnorm(Upper_Percentile, P0, se) = 0.009847463
Lower_Limit = pnorm(Lower_Percentile, P0, se) = 0.009847463
Probability = Upper_Limit + Lower_Limit = 0.0197
P0 <- 0.23
size <- 602
se <- sqrt(P0*(1-P0)/size)
limit <- 0.04
Upper_Percentile = P0+limit
Lower_Percentile = P0-limit
Upper_Limit <- 1-pnorm(Upper_Percentile, P0, se)
Lower_Limit <- pnorm(Lower_Percentile, P0, se)
paste("The probability :", round (Upper_Limit + Lower_Limit, 4))## [1] "The probability : 0.0197"A research company desires to know the mean consumption of beef per week among males over age 48. Suppose a sample of size 208 is drawn with x ?? = 3.9. Assume ® = 0.8 .
Construct the 80% confidence interval for the mean number of lb. of beef per week among males over 48. (Round your answers to 1 decimal place)
Answer: mean = 3.9
standard deviation = sd = sqrt(variance) = 0.8
size = 208
standard mean = sm = sd / sqrt(size)
confidence interval = ci = 80% = 0.80
p = (1-ci)/2
t = qt(p,size-1)
upper_Limit = mean - tsm = 3.971315
lower_Limit = mean + tsm = 3.828685
mean <- 3.9
sd <- 0.8
size <- 208
sm <- sd / sqrt(size)
ci <- 0.80
p <- (1-ci)/2
t <- qt(p,size-1)
upper_Limit <- mean - t*sm
lower_Limit <- mean + t*sm
paste("The upper limit :", round(upper_Limit, 1))## [1] "The upper limit : 4"paste("The lower limit :", round(lower_Limit, 1))## [1] "The lower limit : 3.8"An economist wants to estimate the mean per capita income (in thousands of dollars) in a major city in California. Suppose a sample of size 7472 is drawn with x ?? = 16.6. Assume ® = 11 . Construct the 98% confidence interval for the mean per capita income. (Round your answers to 1 decimal place)
Answer: mean = 16.6
standard deviation = sd = sqrt(variance) = 11
size = 7472
standard mean = sm = sd / sqrt(size)
confidence interval = ci = 98% = 0.98
p = (1-ci)/2
t = qt(p,size-1)
upper_Limit = mean - tsm = 16.8961
lower_Limit = mean + tsm = 16.3039
mean <- 16.6
sd <- 11
size <- 7472
sm <- sd / sqrt(size)
ci <- 0.98
p <- (1-ci)/2
t <- qt(p,size-1)
upper_Limit <- mean - t*sm
lower_Limit <- mean + t*sm
paste("The upper limit :", round(upper_Limit, 1))## [1] "The upper limit : 16.9"paste("The lower limit :", round(lower_Limit, 1))## [1] "The lower limit : 16.3"Find the value of t such that 0.05 of the area under the curve is to the left of t. Assume the degrees of freedom equals 26.
Choose the picture which best describes the problem.
Upper right
Write your answer below. Answer
x <- 26
y <- 0.05
round(abs(qt(y, x - 1)),4)## [1] 1.7081The following measurements ( in picocuries per liter ) were recorded by a set of helium gas detectors installed in a laboratory facility:
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.
Calculate the sample mean for the given sample data. (Round answer to 2 decimal places) Answer
helium <- c(383.6, 347.1, 371.9, 347.6, 325.8, 337)
mean <- sum(helium)/length(helium)
mean(helium)## [1] 352.1667paste("The mean :", round(mean, 2))## [1] "The mean : 352.17"Calculate the sample standard deviation for the given sample data. (Round answer to 2 decimal places) Answer
sd <- sd(helium)
StandardDeviation <- sqrt(var(helium))
paste("The standard deviation :", round(sd, 2))## [1] "The standard deviation : 21.68"Find the critical value that should be used in constructing the confidence interval. (Round answer to 3 decimal places)
Answer: size = 6
ci = 90% = 0.90
p = (1-ci)/2
t = qt(p,size-1)
size <- length(helium)
ci <- 0.90
p <- (1-ci)/2
t <- qt(p, size-1)
paste("The critical value :", round(abs(t), 3))## [1] "The critical value : 2.015"Construct the 90% confidence interval. (Round answer to 2 decimal places)
Answer: standard mean = sm = sd / sqrt(size)
upper_Limit = mean - tsm
lower_Limit = mean + tsm
sm <- sd / sqrt(size)
upper_Limit <- mean - t*sm
lower_Limit <- mean + t*sm
paste("The upper limit :", round(upper_Limit, 2))## [1] "The upper limit : 370"paste("The lower limit :", round(lower_Limit, 2))## [1] "The lower limit : 334.34"A random sample of 16 fields of spring wheat has a mean yield of 46.4 bushels per acre and standard deviation of 2.45 bushels per acre. Determine the 80% confidence interval for the true mean yield. Assume the population is normally distributed.
## Step 1. Find the critical value that should be used in constructing the confidence interval. (Round answer to 3 decimal places)
Answer: mean = 46.4
standard deviation = sd = sqrt(variance) = 2.45
size = 16
ci = 80% = 0.80
p = (1-ci)/2
t = qt(p,size-1)
mean <- 46.4
sd <- 2.45
size <- 16
ci <- 0.80
p <- (1-ci)/2
t <- qt(p, size-1)
paste("The critical value :", round(abs(t), 3))## [1] "The critical value : 1.341"Construct the 80% confidence interval. (Round answer to 1 decimal place)
Answer: standard mean = sm = sd / sqrt(size)
upper_Limit = mean - tsm
lower_Limit = mean + tsm
sm <- sd / sqrt(size)
upper_Limit <- mean - t*sm
lower_Limit <- mean + t*sm
paste("The upper limit :", round(upper_Limit, 1))## [1] "The upper limit : 47.2"paste("The lower limit :", round(lower_Limit, 1))## [1] "The lower limit : 45.6"A toy manufacturer wants to know how many new toys children buy each year. She thinks the mean is 8 toys per year. Assume a previous study found the standard deviation to be 1.9.
How large of a sample would be required in order to estimate the mean number of toys bought per child at the 99% confidence level with an error of at most 0.13 toys? (Round your answer up to the next integer)
Answer: mean = 8
standard deviation = sd = sqrt(variance) = 1.9
size = ?
standard mean = sm = sd / sqrt(size)
confidence interval = ci = 99% = 0.99
p = (1-ci)/2
t = qt(p, size-1)
upper_Limit = mean - tsm
lower_Limit = mean + tsm
z = qnorm(1-p)
standard error = se = 0.13
size = (z * sd/se)^2
mean <- 8
sd <- 1.9
ci <- 0.99
p <- (1-ci)/2
z <- qnorm(1-p)
se <- 0.13
size <- (z*sd/se)^2
paste("The sample size :", round(size, 0))## [1] "The sample size : 1417"A research scientist wants to know how many times per hour a certain strand of bacteria reproduces. He believes that the mean is 12.6. Assume the variance is known to be 3.61.
How large of a sample would be required in order to estimate the mean number of reproductions per hour at the 95% confidence level with an error of at most 0.19 reproductions? (Round your answer up to the next integer)
Answer: mean = 12.6
variance = 3.61
standard deviation = sd = sqrt(variance) = 1.9
size = ?
standard mean = sm = sd / sqrt(size)
confidence interval = ci = 95% = 0.95
p = (1-ci)/2
t = qt(p, size-1)
upper_Limit = mean - tsm
lower_Limit = mean + tsm
z = qnorm(1-p)
standard error = se = 0.19
size = (z * sd/se)^2
mean <- 12.6
variance <- 3.61
sd <- sqrt(variance)
ci <- 0.95
p <- (1-ci)/2
z <- qnorm(1-p)
se <- 0.19
size <- (z*sd/se)^2
paste("The sample size :", round(size, 0))## [1] "The sample size : 384"The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the eighth grade level.
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)
Answer: size_At10 = 2089
size_Above8 = 1734
r = (size_At10-size_Above8)/size_At10
size_At10 <- 2089
size_Above8 <- 1734
size_Below8 <- size_At10 - size_Above8
r <- size_Below8/size_At10
paste("The proportion :", round(r, 3))## [1] "The proportion : 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)
Answer: mean = ?
variance = ?
standard deviation = sd = sqrt(variance)
size = ?
size_At10 = 2089
size_Above8 = 1734
r = (size_At10-size_Above8)/size_At10
standard mean = sm = sd / sqrt(size)
confidence interval = ci = 98% = 0.98
p = (1-ci)/2
t = qt(p, size-1)
upper_Limit = mean - tsm
lower_Limit = mean + tsm
z = qnorm(1-p)
standard error = se = z * sqrt(r(1-r)/size_At10)
size = (z sd/se)^2
upper_Limit_r = r + se
lower_Limit_r = r - se
ci <- 0.98
p <- (1-ci)/2
z <- qnorm(1-p)
z## [1] 2.326348se <- z * sqrt(r*(1-r)/size_At10)
upper_Limit_r <- r + se
lower_Limit_r <- r - se
paste("The upper limit of at/below 8 level proportion :", round(upper_Limit_r, 3))## [1] "The upper limit of at/below 8 level proportion : 0.189"paste("The lower limit of at/below 8 level proportion :", round(lower_Limit_r, 3))## [1] "The lower limit of at/below 8 level proportion : 0.151"An environmentalist wants to find out the fraction of oil tankers that have spills each month.
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)
Answer: size_Tankers = 474
size_spilledTankers = 156
r = size_spilledTankers/size_Tankers
size_Tankers <- 474
size_spilledTankers <- 156
size_unspilledTankers <- size_Tankers - size_spilledTankers
r <- size_spilledTankers/size_Tankers
paste("The proportion :", round(r, 3))## [1] "The proportion : 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)
Answer: mean = ?
variance = ?
standard deviation = sd = sqrt(variance)
size = ?
size_Tankers = 474
size_spilledTankers = 156
r = size_spilledTankers/size_Tankers
standard mean = sm = sd / sqrt(size)
confidence interval = ci = 95% = 0.95
p = (1-ci)/2
t = qt(p, size-1)
upper_Limit = mean - tsm
lower_Limit = mean + tsm
z = qnorm(1-p)
standard error = se = z * sqrt(r(1-r)/size_Tankers)
size = (z sd/se)^2
upper_Limit_r = r + se
lower_Limit_r = r - se
ci <- 0.95
p <- (1-ci)/2
z <- qnorm(1-p)
z## [1] 1.959964se <- z * sqrt(r*(1-r)/size_Tankers)
upper_Limit_r <- r + se
lower_Limit_r <- r - se
paste("The upper limit of spilled proportion :", round(upper_Limit_r, 3))## [1] "The upper limit of spilled proportion : 0.371"paste("The lower limit of spilled proportion :", round(lower_Limit_r, 3))## [1] "The lower limit of spilled proportion : 0.287"