Solution:
From the above we can deduct as follows:
Mean: \(\mu=1300 \:lbs\)
Variance: \(\sigma^2=4000\)
We need to calculate the Standard Deviation:
\(\sigma=\sqrt{4000}\)
What we need to find is as follows: \(P(x > 979 \:\: | \:\: \sigma)\)
Since we know that this is a normal distribution and we know the mean and standard deviation, we can use pnorm as follows:
x <- 979
m <- 1300
v <- 40000
sd <- sqrt(v)
# Since pnorm returns the probability less than that number, we can calculate the complement as follows: pnorm(x, mean, standard deviation)
round(1 - pnorm(x , m, sd),4)## [1] 0.9458# Or we can use
round(pnorm(x, m, sd, lower.tail = FALSE),4)## [1] 0.9458Answer: The probability that the weight of a randomly selected steer is greater than 979 lbs is 94.58%.
Solution:
From the above we can deduct as follows:
Mean: \(\mu=11000 \:hours\)
Variance: \(\sigma^2=1960000\)
We need to calculate the Standard Deviation:
\(\sigma=\sqrt{1960000}\)
What we need to find is as follows: \(P(x > 979 \:\: | \:\: \sigma)\)
Since we know that this is a normal distribution and we know the mean and standard deviation, we can use pnorm as follows:
x <- 8340
m <- 11000
v <- 1960000
sd <- sqrt(v)
# Since pnorm returns the probability less than that number, we can calculate the complement as follows: pnorm(x, mean, standard deviation)
round(1 - pnorm(x, m, sd),4)## [1] 0.9713# Or we can use
round(pnorm(x, m, sd, lower.tail = FALSE),4)## [1] 0.9713Answer: The probability that the life span of the monitor will be more than 8340 hour is 97.13%.
Solution:
From the above we can deduct as follows:
Mean: \(\mu=80000000\)
Standard Deviation: \(\sigma=3 000 000\)
What we need to find is as follows: \(P(83000000 \:\: < x \:\: < \:\: 85000000 \:\: | \:\: \sigma)\)
Since we know that this is a normal distribution and we know the mean and standard deviation, we can use pnorm as follows:
x1 <- 85000000
x2 <- 83000000
m <- 80000000
sd <- 3000000
# Since pnorm returns the probability less than that number, we can calculate the complement as follows: pnorm(x, mean, standard deviation)
# In this case I am going to calculate the Probability below $85 million.
Px1 <- round(pnorm(x1, m, sd),4)
# In this case I am going to calculate the Probability below $80 million.
Px2 <- round(pnorm(x2, m, sd),4)
#Then I will substract Px1 - Px2 to obtain our probability
Px1 - Px2## [1] 0.1109Answer: The probability that a randomly selected firm will earn between 83 and 85 million dollars is 11.09%.
Solution:
From the above we can deduct as follows:
Mean: \(\mu=456\)
Standard Deviation: \(\sigma=123\)
What we need to find is as follows: \(P(x < 86\% \:\: | \:\: \sigma)\)
Since we know that this is a normal distribution and we know the mean, standard deviation and the probability whose we want it to be the top 14%, we can use qnorm as follows:
#We know the top
x1 <- 0.14
#We nned to calculate the bottom maximum in order to get our top minimum.
x2 <- 1 - x1
m <- 456
sd <- 123
# The idea behind qnorm is that you give it a probability, and it returns the number whose cumulative distribution matches the probability being our top of the bottom 86%.
round(qnorm(x2, m, sd),0)## [1] 589Answer: The minimum score required for the job offer is 589.
Solution:
From the above we can deduct as follows:
Mean: \(\mu=6.13\) cm
Standard Deviation: \(\sigma=0.06\) cm
What we need to find is as follows: \(P(7\% \:\: x < 93\% \:\: | \:\: \sigma)\)
Since we know that this is a normal distribution and we know the mean, standard deviation and the probability whose we want it to be the top 93% and the bottom 7%, we can use qnorm as follows:
# Let's define our top and bottom as follows
x1 <- 0.93
x2 <- 0.07
m <- 6.13
sd <- 0.06
# Since qnorm returns the value given a probability less than that number, we can calculate as follows:
# Top number that will provide the top probablity.
round(qnorm(x1, m, sd),2)## [1] 6.22# Bottom number that will provide the bottom probablity.
round(qnorm(x2, m, sd),2)## [1] 6.04Answer: The two lengths that separate the top 7% and the bottom 7% are 6.04 cm and 6.22 cm.
Solution:
From the above we can deduct as follows:
Mean: \(\mu=78.8\)
Standard Deviation: \(\sigma=9.8\)
What we need to find is as follows: \(P(20\% \:\: x < 55\% \:\: | \:\: \sigma)\)
Since we know that this is a normal distribution and we know the mean, standard deviation and the probability whose we want it to be the top 55% and the bottom 20%, we can use qnorm as follows:
# Let's define our top and bottom as follows
x1 <- 0.55
x2 <- 0.20
m <- 78.8
sd <- 9.8
# Since qnorm returns the value given a probability less than that number, we can calculate as follows:
# Top number that will provide the top probablity.
round(qnorm(x1, m, sd),0)## [1] 80# Bottom number that will provide the bottom probablity.
round(qnorm(x2, m, sd),0)## [1] 71Answer: The numerical limits for a C grade are 71 and 80.
Solution:
From the above we can deduct as follows:
Mean: \(\mu=21.2\)
Standard Deviation: \(\sigma=5.4\)
What we need to find is as follows: \(P(x < 55\% \:\: | \:\: \sigma)\)
Since we know that this is a normal distribution and we know the mean, standard deviation and the probability whose we want it to be the top 55% in order to obtain the minimum for the 45% we are looking for; in this case we can use qnorm as follows:
# Let's define our top and bottom as follows
x <- 0.55
m <- 21.2
sd <- 5.4
# Since qnorm returns the value given a probability less than that number, we can calculate as follows:
round(qnorm(x1, m, sd),1)## [1] 21.9Answer: The minimum score required for admission is 21.9.
Solution:
From the above we can deduct as follows:
Since this is a normal distribution function, we can use the binomial distribution with parameters ‘size’ and ‘probability’ as follows:
This is conventionally interpreted as the number of ‘successes’ in size trials.
Size: \(N=151\)
Probability: \(P=0.09\)
What we need to find is as follows: \(P(x \le 10 \:\: | \:\: p)\)
Since we know that this is a normal distribution and we know the size and the probability whose we want it to be 9% in order to obtain the probability for less than 11; in this case we can use pbinom as follows:
# Let's define our top and bottom as follows
x <- 10
N <- 151
p <- 0.09
# Since qnorm returns the value given a probability less than that number, we can calculate as follows:
round(pbinom(x, N, p),4)## [1] 0.192Answer: The probability using the normal distribution will be 19.20%.
Solution:
We are not given information about the shape of the sample, however \(N \ge 30\) therefore we can apply the Central Limit Theorem. The distribution of sample means will have a mean of 48 and a standard deviation of $ $.
From the above we can deduct as follows:
Mean: \(\mu=48\)
Sample Size: \(N=147\)
Standard Deviation: $ $
What we need to find is as follows: \(P( x > 48.83 \:\: | \:\: N=147)\) or in other words: \(1 - P( x \le 48.83 \:\: | \:\: N=147)\)
Since we know that this is a normal distribution and we know the mean and standard deviation, we need to calculate the new standard deviation for that sample size We can use qnorm as follows:
x <- 48.83
m <- 48
N <- 147
sdM <- 7
sd <- sdM/sqrt(N)
# Since pnorm returns the probability less than that number, we can calculate the complement as follows: pnorm(N, mean, standard deviation)
round(1 - pnorm(x, m, sd),4)## [1] 0.0753Answer: the probability that the mean of the sample would be greater than 48.83 months is 7.53%.
Solution:
By employing the same as the above problem, we have as follows:
x <- 93.54
m <- 91
N <- 68
sdM <- 10
sd <- sdM/sqrt(N)
# Since pnorm returns the probability less than that number, we can calculate the complement as follows: pnorm(N, mean, standard deviation)
round(1 - pnorm(x, m, sd),4)## [1] 0.0181Answer: The probability that the mean of a sample of 68 computers would be greater than 93.54 months is 1.81%.
Preamble:
These kind of problems are better solved with the Pearson correlation coefficient and the Fisher transformation.
Definition Pearson’s correlation coefficient is the covariance of the two variables divided by the product of their standard deviations. The form of the definition involves a “product moment”, that is, the mean (the first moment about the origin) of the product of the mean-adjusted random variables; hence the modifier product-moment in the name.
For this we need to employ a series of transformations as follows:
Fisher Transformation or ‘r’.
\[F(r)= \frac{1}{2}ln\left(\frac{1+r}{1-r}\right)\] and the standard error
\[SE=\frac{1}{\sqrt{n-3}}\]
Solution:
Since we have a percentage proportion to by less than 3% from our original 7%; we can calculate as follows:
\(P(7\% - 3\% < x < 7\% + 3\%) \: \: | N\)
in other words
\(P(4\% < x < 10\% ) | N\)
Let’s program as follows:
FisherTransformation <- function(r) {1/2 * log((1+r)/(1-r))}
FisherStandarDeviation <- function(n) {1 / sqrt(n-3)}
# Lets find the values for 4% and 10%
x10 <- FisherTransformation(0.10)
x4 <- FisherTransformation(0.04)
m <- FisherTransformation(0.07)
N <- 540
sd <- FisherStandarDeviation(N)
# Since pnorm returns the probability less than that number, we can calculate the complement as follows: pnorm(N, mean, standard deviation)
p10 <- pnorm(x10, m, sd)
p4 <- pnorm(x4, m, sd)
#Now I will substract the probabilities P(10) - P(4)
round(p10 - p4,4)## [1] 0.5153Answer: 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% is 51.53%.
Solution:
By employing the above procedure we can do as follows:
Since we have a percentage proportion to by less than 4% from our original 23%; we can calculate as follows:
\(P(23\% - 4\% < x \:\: and \:\: x > 23\% + 4\%) \: \: | N\)
in other words
Complement of \(P(19 < x \:\: and \:\: x > 27) \: | N\) and then we will take the complement for all the bottles over 27%.
# Lets find the values for 19% and 27%
x27 <- FisherTransformation(0.27)
x19 <- FisherTransformation(0.19)
m <- FisherTransformation(0.23)
N <- 602
sd <- FisherStandarDeviation(N)
# Since pnorm returns the probability less than that number, we can calculate the complement as follows: pnorm(N, mean, standard deviation)
p27 <- 1-pnorm(x27, m, sd)
p19 <- pnorm(x19, m, sd)
#Now I will substract the probabilities P(19) + P(27)
round(p27 + p19,4)## [1] 0.301Answer: The probability that the proportion of defective bottles in a sample of 602 bottles would differ from the population proportion by greater than 4% is 30.1%.
Solution:
For this problem, we need to use as follows: Confidence Interval for a Population Mean with Unknown Standard Deviation.
The formula is as follows:
\[CI = \mu \pm t_{n-1}\frac{\sigma}{\sqrt{N}}\] From the above problem, we can deduct as follows:
\(\mu = 3.9\)
\(\sigma = 0.8\)
\(N=208\)
Confidence interval desired probability = 80%; this leave us to work with a 20% divided on two sides of the distribution.
From the above we can calculate as follows:
m <- 3.9
sd <- 0.8
N <- 208
ci <- 0.8
p <- ( 1 - ci ) / 2
# By using "qt(p, N-1)" we can calculate our desired t value.
t <- qt(p, N-1)
CI1 <- round(m + t * sd / sqrt(N),1)
CI2 <- round(m - t * sd / sqrt(N),1)
CI1## [1] 3.8CI2## [1] 4Answer: The 80% confidence interval for the mean number of lb. of beef per week among males over 48 is from 3.8 lbs per week to 4 lbs per week.
Solution:
\(\mu = 16.6\)
\(\sigma = 11\)
\(N=7472\)
Confidence interval desired probability = 98%; this leave us to work with a 2% divided on two sides of the distribution.
From the above we can calculate as follows:
m <- 16.6
sd <- 11
N <- 7472
ci <- 0.98
p <- ( 1 - ci ) / 2
# By using "qt(p, N-1)" we can calculate our desired t value.
t <- qt(p, N-1)
CI1 <- round(m + t * sd / sqrt(N),1)
CI2 <- round(m - t * sd / sqrt(N),1)
CI1## [1] 16.3CI2## [1] 16.9Answer: The 98% confidence interval for the mean per capita income is from 16.3 to 16.9 thousand of dollars.
Step 1. Choose the picture which best describes the problem.
Answer: The picture that best describes the problem is the top right.
Step 2.
Solution:
Degrees of freedom: \(N=26\)
Area under the curve to the left of t: \(p=0.05\)
N <- 26
p <- 0.05
# By using "qt(p, N-1)" we can calculate our desired t value.
t <- round(qt(p, N-1),4)
t## [1] -1.7081Answer: The value of t such that 0.05 of the area under the curve is to the left of t is \(t=-1.7081\) or \(-t=1.7081\)
Step 1. Calculate the sample mean for the given sample data. (Round answer to 2 decimal places).
Solution:
Let’s define as follows:
S = {383.6, 347.1, 371.9, 347.6, 325.8, 337}
s <- c(383.6, 347.1, 371.9, 347.6, 325.8, 337)
m <- round(mean(s),2)
m## [1] 352.17Answer: The sample mean for the given sample data is \(\mu=352.17\)
Step 2. Calculate the sample standard deviation for the given sample data. (Round answer to 2 decimal places).
Solution:
\[\sigma=\sqrt{\frac{\Sigma(x-\mu)^2}{n-1}}\]
sd <- round(sd(s, na.rm = FALSE),2)
sd## [1] 21.68Answer: The standard deviation for the given sample data is \(\sigma=21.68\)
Step 3. Find the critical value that should be used in constructing the confidence interval. (Round answer to 3 decimal places).
Solution:
N <- length(s)
ci <- 0.90
p <- ( 1 - ci ) / 2
# By using "qt(p, N-1)" we can calculate our desired critical t value.
t <- round(qt(p, N-1), 3)
t## [1] -2.015Answer: The critical value that should be used in constructing the confidence interval is \(t=-2.015\) and \(t=2.015\)
Step 4. Construct the 90% confidence interval. (Round answer to 2 decimal places).
Solution:
From above we have as follows:
CI1 <- round(m + t * sd / sqrt(N),2)
CI2 <- round(m - t * sd / sqrt(N),2)
CI1## [1] 334.34CI2## [1] 370Answer: The 90% confidence interval for the mean is from 334.34 to 370.
Step 1. Find the critical value that should be used in constructing the confidence interval. (Round answer to 3 decimal places).
Solution:
From above we have as follows:
\(N = 16\)
\(\mu = 46.4\)
\(\sigma = 2.45\)
m <- 46.4
sd <- 2.45
N <- 16
ci <- 0.80
p <- ( 1 - ci ) / 2
# By using "qt(p, N-1)" we can calculate our desired t value.
t <- qt(p, N-1)
round(t,3)## [1] -1.341Answer: The critical value that should be used in constructing the confidence interval is \(t=1.341\).
Step 2. Construct the 80% confidence interval. (Round answer to 1 decimal place).
Solution:
CI1 <- round(m + t * sd / sqrt(N),1)
CI2 <- round(m - t * sd / sqrt(N),1)
CI1## [1] 45.6CI2## [1] 47.2Answer: The 80% confidence interval is from 45.6 to 47.2 bushels per acre.
Solution:
From above, we can deduct as follows:
\(\mu=8\)
\(\sigma = 1.9\)
Confidence interval = 99%
Standard Error: \(SE=0.13\)
And we don’t have an N; we need to proceed to solve as follows:
The formula for the margin of error is as follows:
\[ES = \frac{z \cdot \sigma}{\sqrt{n}}\]
And solving in therms of n, we deduct:
\[n = \left( \frac{z \cdot \sigma}{SE} \right)^2\]
Note that based on a table for Z values, the value for a 99% of confidence the z value is: 2.576
m <- 8
sd <- 1.9
# Standard error
SE <- 0.13
#Z value for 99% of confidence
z <- 2.576
n <- round((z * sd / SE)^2,0)
n## [1] 1417Answer: The sample needs to be 1417 toys.
Solution:
Taking the above idea, we can deduct as follows:
Mean: \(\mu = 12.6\)
Variance: \(\sigma^2=3.61\)
SE = 0.19
CI = 95%
m <- 12.6
sd <- sqrt(3.61)
# Standard error
SE <- 0.19
#Z value for 95% of confidence
z <- 1.960
n <- round((z * sd / SE)^2,0)
n## [1] 384Answer: The sample size needs to be 384.
Step 1. 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)
Solution:
Let’s take as follows:
Sample Size: \(N = 2089\)
Read above 8ht grade: \(n = 1734\)
#for this we can find the probability for the students that read above 8th grade and take the comlement for the students who don't.
N <- 2089
n <- 1734
p <- round(1 - (n/N),3)
p## [1] 0.17Answer: The result for the students who don’t read above the 8ht grade level is 0.170
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. (Round your answers to 3 decimal places).
Solution:
\(N = 2089\)
\(n_{above} = 1734\)
\(n_{below} = 2089-1734\)
CI = 98% reading below 8ht grade level.
Since the formulas are:
\[SE = \sqrt{\frac{p(1-p)}{n}}\] \[CI = Sample \: Proportion \pm z \cdot \sigma_{sample \: proportion}\]
We can calculate as follows:
N <-2089
na <- 1734
nb <- N - na
# Find Standard Error for tenth graders reading at or below the eighth grade level
SEb <- sqrt((p*(1-p))/nb)
# z value for 98% confidense level
z <- 2.326
# By calculating the intervals, we have as follows:
CI1 <- round(p - z * SEb,3)
CI2 <- round(p + z * SEb,3)
CI1 ## [1] 0.124CI2## [1] 0.216Answer: The 98% of confidence interval for the tenth graders reading at or below the eighth grade level is 0.124 to 0.216
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).
Solution:
Let’s take as follows:
Sample Size: \(N = 474\)
had spills: \(n = 156\)
#for this we can find the probability for the students that read above 8th grade and take the comlement for the students who don't.
N <- 474
n <- 156
p <- round(n/N,3)
p## [1] 0.329Answer: The proportion of oil tankers that had spills is 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).
Solution:
\(N = 474\)
\(n = 156\)
CI = 95%
We can calculate as follows:
N <-474
n <- 156
# Find Standard Error for tenth graders reading at or below the eighth grade level
SE <- sqrt((p*(1-p))/nb)
# z value for 95% confidense level
z <- 1.960
# By calculating the intervals, we have as follows:
CI1 <- round(p - z * SEb,3)
CI2 <- round(p + z * SEb,3)
CI1 ## [1] 0.29CI2## [1] 0.368Answer: The 95% confidence interval for the population proportion of oil tankers that have spills each month is from 0.29 to 0.368.
Answers:
What is the probability density function?
In probability theory, a probability density function (PDF), or density of a continuous random variable, is a function, whose value at any given sample (or point) in the sample space (the set of possible values taken by the random variable) can be interpreted as providing a relative likelihood that the value of the random variable would equal that sample. In other words, while the absolute likelihood for a continuous random variable to take on any particular value is 0 (since there are an infinite set of possible values to begin with).
What is the expected value?
The value of the PDF at two different samples can be used to infer that, in any particular draw of the random variable, how much more likely it is that the random variable would equal one sample compared to the other sample.
The above function represents an exponential distribution.
From the above we can find as follows:
Probability Density Function: \(PDF = ae^{-ax}\) or the first derivative of the Cumulative Density Function:
Expected Value: \(E[x]=\frac{1}{a}\)
Variance: \(VA[x]=\frac{1}{a^2}\)
#Determine P(X<.5 | alpha =1)
x<- 0.5
a <- 1
PDF = a * exp(-a * x)
PDF## [1] 0.6065307Answer: \(P(X<0.5 \:\: | \:\: \alpha)=0.6065307\)
What is \(E(Y)\)? What is \(E(Y^2)\)? What is the variance?
Answers:
\(E(Y)\) represents the mean.
\(E(Y^2)\) represents the Variance.
The variance = b
Since \(b = E(Y) = E(Y^2)\)