Please note all answers are clearly marked at the end of each problem section.
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.)
sd <- sqrt(40000)
p <- pnorm(979, mean=1300, sd=sd, lower=FALSE)Answer. 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.)
sd <- sqrt(1960000)
p <- pnorm(8340, mean=11000, sd=sd, lower=FALSE)Answer. 0.9713
Suppose the mean income of firms in the industry for a year is 80 million dollars with a standard devication of 3 million dollars. If incomes for the industry are represented normally, what is the probability that a randomly selected firm with earn between 83 and 85 million dollars? (Round your answer to 4 decimal places.)
Find the probability \(p_{lower}\) that the firm earns less than $83 million, and the probability \(p_{upper}\) that the firm earns less than $85 million, and substract \(p_{upper} - p_{lower}\).
mu <- 80000000
sd <- 3000000
p_lower <- pnorm(83000000, mean=mu, sd=sd, lower=TRUE)
p_upper <- pnorm(85000000, mean=mu, sd=sd, lower=TRUE)
p <- p_upper - p_lowerAnswer. 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.
R has a function, related to pnorm() used above, for this type of problem: qnorm().
p <- qnorm(.14, mean=456, sd=123, lower=FALSE)Answer. 589
The length 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 seperate 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.
mu <- 6.13
sd <- 0.06
p_lower <- qnorm(.07, mean=mu, sd=sd, lower=TRUE)
p_upper <- qnorm(.07, mean=mu, sd=sd, lower=FALSE)Answer. Reject nails shorter than 6.04 cm and longer than 6.22 cm
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.
mu <- 78.8
sd <- 9.8
x_lower <- qnorm(.2, mean=mu, sd=sd, lower=TRUE)
x_upper <- qnorm(.45, mean=mu, sd=sd, lower=FALSE)Answer. C’s are between 71 and 80
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.
x <- qnorm(.45, mean=21.2, sd=5.4, lower=FALSE)Answer. Minimum score is 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 gradudate on time is 9%. Approximate the probability using the normal distribution. (Round your answer to 4 decimal places.)
We could model this as a binomial distribution (where success is graduating on time and failure is not):
pbinom(11, 151, .11, lower=TRUE)## [1] 0.08665496but since the question speifically requests approximating with normal function, convert the distribution parameters by setting the mean to \(n \pi\) and the variance to \(n \pi (1 - \pi)\) and use pnorm():
N <- 151
pi <- .09
mu <- N * pi
s <- N * pi * (1 - pi)
sd <- sqrt(s)
p <- pnorm(11, mean=mu, sd=sd, lower=TRUE)
p## [1] 0.230715This is much larger than the true propability—likely because \(N\) is too small or \(\pi\) is too close to 0 or both.
Answer. 0.2307
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.)
Here, since we’re considering probability of a sample, we need to use the standard error of the sample mean \(\sigma_m\) for standard deviation in pnorm(), not the population standard deviation:
\[\sigma_m = \frac{\sigma}{\sqrt{N}}\]
n <- 147
sd <- 7
sigma_m <- sd / sqrt(n)
p <- pnorm(48.82, mean=48, sd=sigma_m, lower=FALSE)Answer. 0.0778
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 93.45 months? (Round your answer to 4 decimal places.)
n <- 68
sd <- 10
sigma_m <- sd / sqrt(n)
p <- pnorm(93.54, mean=91, sd=sigma_m, lower=FALSE)Answer. 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.)
This question asks, what is the probability the sample mean is between 4 and 10 percent. Find the probability that it is less than 9, and the probability that it is less than 4, and subtract the latter from the former, using pnorm().
n <- 540
m <- .07
sigma_p <- sqrt( (m * (1 - m)) / n )
p_upper <- pnorm(.10, mean=m, sd=sigma_p, lower=TRUE)
p_lower <- pnorm(.04, mean=m, sd=sigma_p, lower=TRUE)
p_upper - p_lower## [1] 0.9937105This is a pretty large probability, which makes sense because “within three percent of the mean” is almost the same area as “within three \(\sigma_p\) (0.011) of the mean,” as calculated above.
Answer. 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 sample of 602 bottles would differ from the population proportion by greater than 4%?
Or, what the probability the sample \(p\) is below 19 percent or above 27 percent. As above:
n <- 602
m <- .23
sigma_p <- sqrt( (m * (1 - m)) / n )
p_upper <- pnorm(.19, mean=m, sd=sigma_p, lower=TRUE)
p_lower <- pnorm(.27, mean=m, sd=sigma_p, lower=FALSE)
p_upper + p_lower## [1] 0.01969493This is a small number, which makes sense because, like above, 4 percent away from the mean is almost three \(\sigma_p\) away from the mean.
Answer. 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 \(\bar{x} = 3.9\). Assume \(\sigma = 0.8\). 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.)
n <- 208
x_bar <- 3.9
sigma <- 0.8
z_80 <- 1.28
sigma_m <- sigma / sqrt(n)
lower <- x_bar - (z_80 * sigma_m)
upper <- x_bar + (z_80 * sigma_m)Answer. Lower: 3.8, upper: 4 lbs./week
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 \(\bar{x} = 16.6\). Assume \(\sigma = 11\). Construct the 98% confidence interval for the mean per capita income. (Round your answers to 1 decimal place.)
n <- 7472
x_bar <- 16.6
sigma <- 11
z_98 <- 2.06
sigma_m <- sigma / sqrt(n)
lower <- x_bar - (z_98 * sigma_m)
upper <- x_bar + (z_98 * sigma_m)Answer. Lower: 16.3 thousands/dollars, upper: 16.9 thousands/dollars
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.
Step 1. Choose the picture which best describes the problem.
Answer. The “northeast” picture best describes the problem
Step 2. Write your answer below.
Looking at the \(t\) table, find the cell associated with (t=.95 one-sided, df=26).
Answer. 1.706
The following measurements (in picocuries per liter) were recorded by a set of helium gas detectors installed in a labratory 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.
Step 1. Calculate the sample mean for the given sample data. (Round answer to 2 decimal places.)
x <- c(383.6, 347.1, 371.9, 347.6, 325.8, 337)
x_bar <- mean(x)
n <- length(x)Answer. Sample mean is 352.17 picocuries per liter
Step 2. Calculate the sample standard devitaion for the given sample data. (Round answer to 2 decimal places.)
R’s sd() calculates the sample standard deviation (from help file: “Like var this uses denominator \(n - 1\)”):
s <- sd(x)Answer. 21.68
Step 3. Find the critical value that should be used in constructing the confidence interval. (Round answer to 3 decimal places.)
Looking at a \(t\) table—since population variance is unknown—and assuming a two-tailed test, we need to find the cell associated with (t=.90 two-sided, df=6-1).
Answer. 2.015
Step 4. Construct the 90% confidence interval. (Round answer to 2 decimal places.)
t <- 2.015
se <- s / sqrt(n)
lower <- x_bar - (t * se)
upper <- x_bar + (t * se)Answer. Lower: 334.34, upper: 370 picocuries per liter
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.)
Find the cell in the \(t\) table corresponding to (t=.80 two-tails, df=16-1).
Answer. 1.341
Step 2. Construct the 80% confidence interval. (Round to 1 decimal place.)
n <- 16
x_bar <- 46.4
s <- 2.45
t <- 1.341
se <- s / sqrt(n)
lower <- x_bar - (t * se)
upper <- x_bar + (t * se)Answer. Lower: 45.6 bushels per acre, upper: 47.2 bushels per acre
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 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.)
Here, we need to use the confidence interval formulas in a reverse way as in previous problems. We know that \(SE = 0.13\), but we don’t know \(n\), which is what we want to calculate.
Margin of error is:
\[z \times SE = z \frac{\sigma}{\sqrt{n}}\]
Rearrange to find \(n\):
\[ n = \left( \frac{z \sigma}{SE} \right) ^2\]
z_99 <- 2.576
sigma <- 1.9
se <- 0.13
n <- ( (z_99 * sigma) / se)^2Answer. Sample size must be 1418 toys
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.)
Following the same strategy as previous problem:
v <- 3.61
sigma <- sqrt(v)
z_95 <- 1.96
se <- 0.19
n <- ( (z_95 * sigma) / se)^2Answer. Sample size must be 385 (units of bacteria?)
The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the eight grade level.
Step 1. Suppose a sample of 2089 tenth grades 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.)
p <- 1 - (1734 / 2089)Answer. 0.17
Step 2. 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.
n <- 2089
s_p <- sqrt( (p * (1 - p)) / n )
z_98 <- 2.326
lower <- p - (z_98 * s_p) - (0.5 / n)
upper <- p + (z_98 * s_p) + (0.5 / n)Answer. Lower: 0.151, upper: 0.189
An environmentalist wants to find out the fraction of oil tankers that have spills each month.
Step 1. 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.)
p <- 156 / 474Answer. 0.329
Step 2. 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.)
n <- 474
s_p <- sqrt( (p * (1 - p)) / n )
z_95 <- 1.960
lower <- p - (z_95 * s_p) - (0.5 / n)
upper <- p + (z_95 * s_p) + (0.5 / n)Answer. Lower: 0.286, upper: 0.372