The test statistic for ANOVA is called an F test statistic (or F-ratio), which is calculated by dividing the variance between the samples by the variance within the samples.
Explanation:The test statistic for ANOVA is called an F test statistic (also called an F-ratio). It is calculated by dividing the variance between the samples (variation due to treatment) by the variance within the samples (variation due to error or chance).
The F statistic follows an F distribution with (number of groups - 1) as the numerator degrees of freedom and (number of observations - number of groups) as the denominator degrees of freedom.
For ANOVA (Analysis of Variance), the test statistic is called an F test statistic (also called an F-ratio), which is the variance between samples (a.k.a., variation due to treatment) divided by the variance within samples (a.k.a., variation due to error or chance).
Fill in blank (2): between, Fill in blank (3): within
Analysis of Variance (ANOVA)
ANOVA is a statistical method used to compare the means of three or more samples to see if at least one of them is significantly different from the others. It does this by analyzing the variances within the data.
F-Test Statistic
The test statistic used in ANOVA is called the F-test statistic or the F-ratio. This F-ratio helps to determine whether the variances between the sample means are significantly larger than the variances within the samples. The F-ratio is calculated as follows:
[tex]\[ F = \frac{\text{variance between samples}}{\text{variance within samples}} \][/tex]
Variance Between Samples
The variance between samples (also known as between-group variance or treatment variance) measures the variability among the sample means. This variability reflects how much the group means differ from the overall mean. If the group means are very different from each other, the between-group variance will be large. This part of the variance is often attributed to the effect of the different treatments or conditions being compared.
- Blank (2): The term used here is between.
Variance Within Samples
The variance within samples (also known as within-group variance or error variance) measures the variability within each of the groups. This variability reflects how much the individual data points within each group differ from their respective group mean. This part of the variance is usually attributed to random error or chance.
- Blank (3): The term used here is within.
In the context of ANOVA:
- The F-test statistic (F-ratio) is used to compare the variance between samples to the variance within samples.
- The variance between samples represents the variation due to treatment.
- The variance within samples represents the variation due to error or chance.
The complete question is
For ANOVA, the test statistic is called an_ test statistic (also called an_-ratio), which is the variance (2) samples (a.k.a., variation due to treatment) divided by the variance _(3)_samples (a.k.a., variation due to error or chance) The first two blanks are completed with the letter this author uses for the ANOVA test statistic. What is this letter? Fill in blank (2): Fill in blank (3)
Consider two x distributions corresponding to the same x distribution. The first x distribution is based on samples of size n = 100 and the second is based on samples of size n = 225. Which x distribution has the smaller standard error? The distribution with n = 100 will have a smaller standard error. The distribution with n = 225 will have a smaller standard error. Explain your answer. Since σx = σ2/√n, dividing by the square root of 100 will result in a small standard error regardless of the value of σ2. Since σx = σ/n, dividing by 100 will result in a small standard error regardless of the value of σ. Since σx = σ/n, dividing by 225 will result in a small standard error regardless of the value of σ. Since σx = σ/√n, dividing by the square root of 100 will result in a small standard error regardless of the value of σ. Since σx = σ/√n, dividing by the square root of 225 will result in a small standard error regardless of the value of σ. Since σx = σ2/√n, dividing by the square root of 225 will result in a small standard error regardless of the value of σ2.
Answer:
The distribution with n = 225 will give a smaller standard error.
Since sigma x = sigma/√n, dividing by the square root of 225 will result in a small standard error regardless of the value of sigma.
Step-by-step explanation:
Standard error is given by standard deviation (sigma) divided by square root of sample size (√n).
The distribution with n = 225 would give a smaller standard error because the square root of 225 is 15. The inverse of 15 multiplied by sigma is approximately 0.07sigma which is smaller compared to the distribution n = 100. Square of 100 is 10, inverse of 10 multiplied by sigma is 0.1sigma.
0.07sigma is smaller than 0.1sigma
The value of a house is increasing by 1900 per year. If it is worth 180,000 today, what will it be worth in five years?
Answer: the population in 5 years is
187600
Step-by-step explanation:
The value of a house is increasing by 1900 per year. This means that it is increasing in an arithmetic progression. The formula for determining the nth term of an arithmetic sequence is expressed as
Tn = a + d(n - 1)
Where
a represents the first term of the sequence.
d represents the common difference.
n represents the number of terms in the sequence.
From the information given,
a = 180000
d = 1900
n = 5
We want to determine the value of the 5th term, T5. Therefore,
T5 = 180000 + 1900(5 - 1)
T5 = 180000 + 7600
T5 = 187600
In preparation for the upcoming school year, a teacher looks at raw test scores on the statewide standardized test for the students in her class. Instead of looking at the scores relative to the norms in the state, the teacher wants to understand the scores relative to the students who will be in the class. To do so, she decides to convert the test scores into z-scores relative to the mean and standard deviation of the students in the class. The mean test score in her upcoming class is 49, and the standard deviation is 20.7. The teacher wants to identify those students who may need extra challenges. As a first cut, she decides to look at students who have z-scores above z = 2.00 Identify the test score corresponding to a z-score of above z=2.00. Round to the nearest whole number.
Answer:
[tex] z = \frac{X -\mu}{\sigma}[/tex]
And we can solve for the value of X like this:
[tex] X = \mu + z*\sigma[/tex]
And since we know that z=2 we can replace and we got:
[tex] X = 49 +2*20.7= 90.4 \approx 90[/tex]
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the scores of a population, and for this case we can assume the distribution for X is given by:
[tex]X \sim N(49,20.7)[/tex]
Where [tex]\mu=49[/tex] and [tex]\sigma=20.7[/tex]
And for this case the z score is given by:
[tex] z = \frac{X -\mu}{\sigma}[/tex]
And we can solve for the value of X like this:
[tex] X = \mu + z*\sigma[/tex]
And since we know that z=2 we can replace and we got:
[tex] X = 49 +2*20.7= 90.4 \approx 90[/tex]
To find the test score corresponding to a z-score above z=2.00, use the formula x = (z * standard deviation) + mean. Plugging in the values, the test score is approximately 90 when rounded to the nearest whole number.
Explanation:The teacher wants to identify students who have z-scores above z=2.00. To find the corresponding test score,
we can use the formula for z-score:
z = (x - mean) / standard deviation
Rearranging the formula, we get:
x = (z * standard deviation) + mean
Substituting z=2.00, standard deviation=20.7, and mean=49, we have:
x = (2.00 * 20.7) + 49
Simplifying the equation, we get:
x = 41.4 + 49 = 90.4
Therefore, the test score corresponding to a z-score above z=2.00 is approximately 90 when rounded to the nearest whole number.
Express the function y(t)= 4 sin 2πt + 15 cos 2πt in terms of (a) a sine term only. (b) Determine the amplitude, the period, the frequency in hertz of the function. (c) Draw the function in time domain.
(a) Rewrite y(t) = 4sin(2πt) + 15cos(2πt) using a sine term only: y(t) = √241sin(2πt - 1.249)
(b) Amplitude: √241, Period: 0.5s, Frequency: 2Hz.
(c) Graph shows two sine waves, one with amplitude √241 and another with 15, out of phase, repeating every 0.5s.
(a) Expressing the function in terms of a sine term only:
To express the function y(t) = 4sin(2πt) + 15cos(2πt) in terms of a sine term only, we can use trigonometric identities to rewrite the cosine term in terms of sine:
cos(2πt) = sin(π/2 - 2πt)
Now, we can rewrite the function as follows:
y(t) = 4sin(2πt) + 15sin(π/2 - 2πt)
(b) Determining the amplitude, period, and frequency:
Amplitude (A): The amplitude of a sinusoidal function is the coefficient of the sine term. In this case, the amplitude is 4.
Period (T): The period of a sinusoidal function is the time it takes for one complete cycle. The period can be found using the formula T = 1/f, where f is the frequency. In this case, the frequency is 2, as we'll see in part (b). So, T = 1/2 = 0.5 seconds.
Frequency (f): The frequency of a sinusoidal function is the number of cycles per second (in hertz, Hz). In this case, the frequency is 2 Hz.
(c) Drawing the function in the time domain:
To draw the function y(t) = 4sin(2πt) + 15sin(π/2 - 2πt), you can follow these steps:
1. Create a set of axes with time (t) on the horizontal axis and y(t) on the vertical axis.
2. The amplitude of the first sine term is 4, so the first term oscillates between -4 and 4.
3. The second sine term has an amplitude of 15, but it is shifted in phase by π/2. This means it starts at its maximum value of 15 and goes down to -15.
4. The total function y(t) is the sum of these two sine terms. You'll see a combination of two sinusoidal waves, one with a smaller amplitude (4) and one with a larger amplitude (15), and they are out of phase with each other.
The period of the function is 0.5 seconds (as calculated in part (b)), so you'll see this pattern repeat every 0.5 seconds.
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54 and 67 use numbers and operations to write each phrase as an expression
The phrases you would like to be written as expressions are not listed. I would nevertheless, explain how to write phrases as expressions so that the same approach could be applied to you own question.
Phrases are dynamic, depending on the problem. They do not necessarily take a particular form.
The constant thing about phrases is the operators connecting the words in the phrases. Theses operators are:
Addition (+), Subtraction (-), Division (÷), and Multiplication (×).
In word problems, it is a matter of interpretation, these operators can be written in many ways.
ADDITION
plus
the sum of
increase
grow
add
profit
And so on.
SUBTRACTION
minus
loss
decrease
reduce
subtract
And so on
MULTIPLICATION
times
multiply
triple
And so on
DIVISION
split
share
divide
distribute
And so on.
Examples
(1) 56 is added to a number to give 100
Interpretation: x + 56 = 100
(2)The difference between Mr. A and Mr. B is 5
Interpretation: A - B = 5
(3) This load (L1) is three times heavier than that one (L2)
Interpretation: L1 = 3L2
(4) Share this orange (P) equally between the three children
Interpretation: P/3
To write expressions using the numbers 54 and 67 with operations, you can add (54 + 67), subtract (54 - 67), multiply (54 x 67), or divide (54 ÷ 67). The order of operations is important and parentheses may be used to dictate the sequence of calculations. Practice with expressions can be enhanced by working through problems symbolically and then substituting numbers.
Explanation:The question asks to write expressions using the numbers 54 and 67 with operations. When creating expressions, operations such as addition (+), subtraction (-), multiplication (x or ×), and division (÷) are typically used. For example, if we want to add 54 and 67, the expression would be 54 + 67.
If we wanted to multiply them, the expression would be 54 x 67 (or 54 × 67). Remember that in writing expressions, the order of operations is important, so if we want to perform different operations, we may need to use parentheses to ensure the correct sequence of calculations.
For more practice with expressions and operations, you can approach problems by covering up one number and solving for it using the remaining information (as mentioned in the given examples). This technique helps you solve various types of problems using the same principle. Moreover, when working with algebraic expressions, it's beneficial to solve the equation symbolically first and then substitute the numbers.
Substituting numbers and parentheses placement can significantly alter the result of the expression, as shown in the examples given in the question. The expressions created through experimentation, such as (74) to the third power, can be solved symbolically or using a calculator to understand how different operations and powers affect the result.
The files school1.dat, schoo12. dat and schoo13.dat contain data on the amount of time students from three high schools spent on studying or homework during an exam period. Analyze data from each of these schools separately, using the normal model with a conjugate prior -5, σ1-4,K0-1Po-2) and compute or distribution, in which f40 approximate the following: a) posterior means and 95% confidence intervals for the mean θ and standard deviation σ from each school b) the posterior probability that θǐ 〈 θj 〈 θk for all six permutations i,j,k of 1,2,3); c) the posterior probability that Yi 〈 Yk for all six permutations i, j, k} of {1,2,3], where , is a sample from the posterior predictive distribution of school i, d) Compute the posterior probability that is bigger than both θ2 and , and the posterior probability that Y1 is bigger than both Yo and 3
School1.dat:
2.11
9.75
13.88
11.3
8.93
15.66
16.38
4.54
8.86
11.94
12.47
11.11
11.65
14.53
9.61
7.38
3.34
9.06
9.45
5.98
7.44
8.5
1.55
11.45
9.73
School2.dat:
0.29
1.13
6.52
11.72
6.54
5.63
14.59
11.74
9.12
9.43
10.64
12.28
9.5
0.63
15.35
5.31
8.49
3.04
3.77
6.22
2.14
6.58
1.11
School3.dat:
4.33
7.77
4.15
5.64
7.69
5.04
10.01
13.43
13.63
9.9
5.72
5.16
4.33
12.9
11.27
6.05
0.95
6.02
12.22
12.85
Answer:
See answers below
Step-by-step explanation:
# prior
mu0 <- 5
k0 <- 1
s20 <- 4
nu0 <- 2
# read in data
dat <- list()
dat[1] <- read.table("school1.dat")
dat[2] <- read.table("school2.dat")
dat[3] <- read.table("school3.dat")
n <- sapply(dat, length)
ybar <- sapply(dat, mean)
s2 <- sapply(dat, var)
# posterior
kn <- k0 + n
nun <- nu0 + n
mun <- (k0 * mu0 + n * ybar)/kn
s2n <- (nu0 * s20 + (n - 1) * s2 + k0 * n * (ybar - mu0)^2/kn)/nun
# produce a posterior sample for the parameters
s.postsample <- s2.postsample <- theta.postsample <- matrix(0, 10000, 3, dimnames = list(NULL,
c("school1", "school2", "school3")))
for (i in c(1, 2, 3)) {
s2.postsample[, i] <- 1/rgamma(10000, nun[i]/2, s2n[i] * nun[i]/2)
s.postsample[, i] <- sqrt(s2.postsample[, i])
theta.postsample[, i] <- rnorm(10000, mun[i], s.postsample[, i]/sqrt(kn[i]))
}
# posterior mean and 95% CI for Thetas (the means)
colMeans(theta.postsample)
## school1 school2 school3
## 9.294 6.943 7.814
apply(theta.postsample, 2, function(x) {
quantile(x, c(0.025, 0.975))
})
## school1 school2 school3
## 2.5% 7.769 5.163 6.152
## 97.5% 10.834 8.719 9.443
# posterior mean and 95% CI for the Standard Deviations
colMeans(s.postsample)
## school1 school2 school3
## 3.904 4.397 3.757
apply(s.postsample, 2, function(x) {
quantile(x, c(0.025, 0.975))
})
## school1 school2 school3
## 2.5% 2.992 3.349 2.800
## 97.5% 5.136 5.871 5.139
The question calls for Bayesian inference with a normal model and a conjugate prior to derive posterior distributions and probabilities concerning means and standard deviations. Statistical software such as R or Python with libraries like PyMC3 is needed to perform the detailed computations, which are not provided here.
Explanation:The provided question involves advanced statistical analysis, specifically Bayesian inference with prior distributions and the computation of posterior probabilities. Unfortunately, without software or computational methods to process the provided datasets for schools school1.dat, school2.dat, and school3.dat, a comprehensive answer cannot be provided. However, the general approach to tackle this problem would involve:
Calculating the posterior means and 95% confidence intervals for the mean (θ) and standard deviation (σ) for each school's data.Determining the probability that one school's mean is less than another and establishing the likelihood hierarchy between them (θ1 < θ2 < θ3).Computing the posterior probability for predicted homework times across schools.Focusing on posteriors involving hypotheses about means and predictive distributions.To complete the analysis, one would typically utilize statistical software capable of Bayesian computation, such as R or Python with appropriate libraries (e.g., PyMC3).
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Consider the following data and corresponding weights. xi Weight (wi) 3.2 6 3.0 3 2.5 2 4.0 8 (a) Compute the weighted mean. (Round your answer to three decimal places.) (b) Compute the sample mean of the four data values without weighting. Note the difference in the results provided by the two computations.
Answer:
(a) 3.432
(b) 3.175
Step-by-step explanation:
The given data and corresponding weights is:
[tex]\begin{array}{cc}x_i&w_i\\3.2&6\\3.0&3\\2.5&2\\4.0&8\\\end{array}[/tex]
(a) The weighted mean is determined by the following expression:
[tex]M_W=\sum\frac{x_i*w_i}{w_i}\\M_W=\frac{3.2*6+3.0*3+2.5*2+4.0*8}{6+3+2+8}\\M_W=3.432[/tex]
(b) The simple mean of the given data is determined by adding up the four values dividing the result by 4:
[tex]M_S = \frac{3.2+3.0+2.5+4.0}{4}\\ M_S=3.175[/tex]
The value is lower than the weighted mean.
The weighted mean is 3.455, while the sample mean without weighting is 3.175.
Explanation:To compute the weighted mean, we multiply each data value by its corresponding weight, then add up the results. We then divide this sum by the sum of the weights. In this case, the weighted mean can be calculated as follows:
Weighted Mean = (3.2 * 6 + 3.0 * 3 + 2.5 * 2 + 4.0 * 8) / (6 + 3 + 2 + 8) = 3.455
To compute the sample mean without weighting, we simply add up all the data values and divide the sum by the number of data values. In this case, the sample mean can be calculated as follows:
Sample Mean = (3.2 + 3.0 + 2.5 + 4.0) / 4 = 3.175
The difference between the two computations is that the weighted mean takes into account the importance of each data value by assigning weights to them, whereas the sample mean without weighting treats all data values equally.
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The weights of steers in a herd are distributed normally. The standard deviation is 200lbs and the mean steer weight is 1300lbs. Find the probability that the weight of a randomly selected steer is between 939 and 1417lbs. Round your answer to four decimal places.
Answer:
Probability that the weight of a randomly selected steer is between 939 and 1417 lbs is 0.68389 .
Step-by-step explanation:
We are given that the weights of steers in a herd are distributed normally. The standard deviation is 200 lbs and the mean steer weight is 1300 lbs.
So, Let X = weights of steers in a herd ,i.e.; X ~ N([tex]\mu,\sigma^{2}[/tex])
Here, [tex]\mu[/tex] = population mean = 1300 lbs
[tex]\sigma[/tex] = population standard deviation = 200 lbs
The z score area distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
So, Probability that the weight of a randomly selected steer is between 939 and 1417 lbs = P(939 lbs < X < 1417 lbs)
P(939 lbs < X < 1417 lbs) = P(X < 1417) - P(X <= 939)
P(X < 1417) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{1417-1300}{200}[/tex] ) = P(Z < 0.58) = 0.71904
P(X <= 939) = P( [tex]\frac{X-\mu}{\sigma}[/tex] <= [tex]\frac{939-1300}{200}[/tex] ) = P(Z <= -1.81) = 1 - P(Z < 1.81)
= 1 - 0.96485 = 0.03515
Therefore, P(939 lbs < X < 1417 lbs) = 0.71904 - 0.03515 = 0.68389 .
The population of a community is known to increase at a rate proportional to the number of people present at time t. The initial population P0 has doubled in 5 years. Suppose it is known that the population is 8,000 after 3 years. What was the initial population P0? (Round your answer to one decimal place.) P0 = What will be the population in 10 years? (Round your answer to the nearest person.) persons How fast is the population growing at t = 10? (Round your answer to the nearest person.) persons/year
Answer:
5278.0
21112
2927
Step-by-step explanation:
P = Po[2^(t/5)]
8000 = Po(2^⅗)
Po = 5278.0
P = 5278(2^(10/5))
P = 21112
P = Po[2^(t/5)]
ln(P/Po) = (t/5)ln2
ln(P) - ln(Po) = (t/5)ln2
1/P . dP/dt = ln2/5
dP/dt = P(ln2)/5
At t = 10, P = 21112
dP/dt = 2927
Find the volume of a cylinder with a diameter of 6 inches and a height that is three times the radius use 3.14 for pie and round your answer to the nearest 10th and you may only enter numerals decimal points
Answer:
254.34 in^3
Step-by-step explanation:
Volume of a cylinder is V=pi*r^2*h
Since r=6/2=3, and h=3r=3*3=9, then we solve for V:
V=3.14*3^2*9
V=3.14*9*9
V=3.14*81
V=254.34
So the volume of the cylinder is 254.34 in^3
Answer: volume = 254.3 inches³
Step-by-step explanation:
The formula for determining the volume of a cylinder is expressed as
Volume = πr²h
Where
r represents the radius of the cylinder.
h represents the height of the cylinder.
π is a constant whose value is 3.14
From the information given,
Diameter = 6 inches
Radius = diameter/2 = 6/2 = 3 inches
Height = 3 × radius = 3 × 3 = 9 inches
Therefore,
Volume = 3.14 × 3² × 9
Volume = 254.3 inches³ to the nearest tenth
A box contains red marbles and green marbles. Sampling at random from the box five times with replacement, you have drawn a red marble all five times. What is the probability of drawing a red marble the sixth time
Answer: The probability of drawing a red marble the sixth time is 1/2
Step-by-step explanation:
Here is the complete question:
A box contains 10 red marbles and 10 green marbles. Sampling at random from the box five times with replacement, you have drawn a red marble all five times. What is the probability of drawing a red marble the sixth time?
Explanation:
Since the sampling at random from the box containing the marbles is with replacement, that is, after picking a marble, it is replaced before picking another one, the probability of picking a red marble is the same for each sampling. Probability, P(A) is given by the ratio of the number of favourable outcome to the total number of favourable outcome.
From the question,
Number of favourable outcome = number of red marbles =10
Total number of favourable outcome = total number of marbles = 10+10= 20
Hence, probability of drawing a red marble P(R) = 10 ÷ 20
P(R) = 1/2
Since the probability of picking a red marble is the same for each sampling, the probability of picking a red marble the sixth time is 1/2
The probability is [tex]\frac{1}{2}[/tex]
First, we have to calculate the probability of drawing a red marble on any given try using the formula of probability:
Probability of an event = [tex]\frac{Number\ of\ favorable\ outcomes}{Total\ number\ of\ outcomes}[/tex]
P(red) = [tex]\frac{Number\ of\ red\ marbles}{Total\ number\ of\ marbles} =\frac{10}{20} = \frac{1}{2}[/tex]
Since we are replacing after every time we pick up a marble, each event of picking a marble is an independent event
Thus, every time a marble is to be picked, the probabilities remain same.
Hence the probability of drawing a red marble the sixth time is = [tex]P(red) = \frac{1}{2}[/tex]
The complete question is:
A box contains 10 red marbles and 10 green marbles. Sampling at random from the box five times with replacement, you have drawn a red marble all five times. What is the probability of drawing a red marble the sixth time?
A publisher knows that from all the writers the company published, 20% wrote romantic novels and 40% wrote sci-fi books. If we look at the last 5 years, 40% published 2 books, and 30% published only 1 book. From the writers that did not published in the last 5 years, 20% wrote romantic novels and 40% sci-fi books. From the writers that wrote other types of books, 50% published 2 books. Finally, the number of sci-fi writers that published 1 and the number that published 2 books was the same. 1. What is the average number of books published in the last five years? 2. What proportion of writers are sci-fi writers and published 2 books during the last five years? 3. What is the probability that if we choose a romantic novels writer, he had not published in the last five years? 4. What is the probability that a writer did not publish romantic or sci-fi and did not publish exactly 1 book in the past five years? 5. If we select a writer that had published in the last five years, what is the probability that he writes romantic novels? 6. If we select 5 writers, what is the probability that they didn't publish any books in the past 5 years? 7. What is the probability that a writer is not a romantic novel writer and published more than 1 book in the past five years? 8. Are types of books and the number of books published in the past five years statistically independent? 9. Is writing sci-fi books and not publishing in the last years statistically independent? 10. What is the probability that if we choose 4 writers, 2 of them have published 1 romantic novel in the last 5 years? 11. What proportion of writers did not write sci-fi and published one or two books?
The answers to mentioned questions are conditional probabilities and proportions, requiring more detailed numerical data to provide exact values. But they can be calculated using simple formulas of probability and proportion.
Explanation:This is a statistics problem and we would need more details to fully answer these questions. But here are some general insights:
The average number of books published would require total number of books/total number of writers in the last 5 years.The proportion of sci-fi writers that published 2 books would be the total number of sci-fi writers with 2 books/total number of sci-fi writers.If we choose a romantic novels writer, the probability of them not having published in the last 5 years would be the number of romantic writers who didn't publish in last 5 years/total number of romantic writers.A writer not published romantic or sci-fi and did not publish exactly 1 book in the past five years would be calculated by first calculating the total of such authors and then dividing by total authors.If we know the writer published in the last five years, the probability they write romantic novels would be number of romantic writers that published in last 5 years / total number of writers who published in the last 5 years.Without explicit numbers provided for each category of writer, it's impossible to give numerical solutions to these problems. Instead, we can only provide the formulas to solve them. The same reasoning applies to all subsequent questions.
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Average number of books published in the last five years = 1.1 books
Proportion of sci-fi writers who published 2 books = 0.2
Probability that a romantic novel writer had not published in the last five years = 0.04
Probability a writer who did not publish romantic or sci-fi did not publish exactly 1 book = 0.3
Probability a writer who published in the last years writes romantic novels = 0.2
Probability that 5 writers didn't publish any books = .00243.
Probability that a non-romantic writer published more than 1 book= 0.32
Independence of book types and number published= independent events
Independence of writing sci-fi and not publishing = independent events.
Probability 2 of 4 have published romantic novels= 0.1536
Proportion of writers did not write sci-fi and published 1 or 2 books=0.42
Let's breakdown this complex problem step-by-step.
1. Average number of books published in the last five years
Given that 40% published 2 books and 30% published 1 book, we can find the average as follows:
Average = (0.4 * 2) + (0.3 * 1) + (0.3 * 0) = 0.8 + 0.3 + 0 = 1.1 books
2. Proportion of sci-fi writers who published 2 books
Since the number of sci-fi writers publishing 1 and 2 books is the same, and sci-fi comprises 40% of all writers:
Proportion = 0.4 * 0.5 = 0.2 or 20%
3. Probability that a romantic novel writer had not published in the last five years
20% of total writers wrote romantic novels, and 20% of those did not publish in the last five years:
Probability = 0.2 * 0.2 = 0.04 or 4%
4. Probability a writer who did not publish romantic or sci-fi did not publish exactly 1 book
If 60% did not publish romantic or sci-fi, half of these published 2 books:
Probability = 0.6 * 0.5 = 0.3 or 30%
5. Probability a writer who published in the last years writes romantic novels
40% published 2 books and 30% published 1 book, so 70% published. Romantic novelists comprise 20% of all writers:
Probability = (0.2 * 0.7) / 0.7 = 0.2 or 20%
6. Probability that 5 writers didn't publish any books
The probability that one writer didn't publish is 0.3:
Probability = 0.3^5 = 0.00243 or 0.243%
7. Probability that a non-romantic writer published more than 1 book
80% are non-romantic, and 40% of total published 2 books:
Probability = (0.4 * 0.8) = 0.32 or 32%
8. Independence of book types and number published
We need to see if P(A ∩ B) = P(A)*P(B). Since specific data does not align well, these events are not independent.
9. Independence of writing sci-fi and not publishing
40% wrote sci-fi and 30% did not publish:
P(A ∩ B) = 0.4 * 0.3 = 0.12 or 12%
These are independent events.
10. Probability 2 of 4 have published romantic novels
Using binomial distribution:
Probability P(X = 2) = 4C2 * (0.2)^2 * (0.8)^2 = 0.1536 or 15.36%
11. Proportion of writers did not write sci-fi and published 1 or 2 books
60% did not write sci-fi, and of those published, 70% published 1 or 2 books:
Proportion = 0.6 * 0.7 = 0.42 or 42%
A piano is hauled into space a distance 3 earth radii above the surface of the north pole and is then dropped. About how long will it take to splash down in the arctic ocean
Answer:
The answer to the question is
It would take about 167.021 s to splash down in the arctic ocean.
Step-by-step explanation:
The gravitational force is given by
[tex]F_G = \frac{G*m_1*m_2}{r^2}[/tex]
Where m₁ mass of the piano and
m₂ = mass of the Earth
r = 3·R where R = radius of the earth
as stated in the question we have varying acceleration due to the inverse square law
Therefore at 3 × Radius of the earth we have
[tex]F_G[/tex] = 109.083 N and the acceleration =1.09083 m/s²
If the body falls from 3·R to 2·R with that acceleration we have
S = u·t +0.5×a·t² = 0.5×a·t² as u = 0
That is 6371 km = 0.5·1.09083·t²
t₁ = 108.079 s and we have
v₁² = u₁² +2·a₁·s₁ = 2·a₁·s₁ = 117.895 m/s
For the next stage r₂ = 2R
Therefore F = 245.436 N and a₂ = F/m₁ = 2.45436 m/s²
Therefore the time from 2R to R is given by
S₂ =R=u·t+0.5·a₂·t² = v₁·t + 0.5·a₂·t²
or 6371 km = 117.895 m/s × t + 0.5 × 2.45436 × t²
Which gives 1.22718 × t² + 117.895 × t -6371 = 0
Factorizing we have (t+134.631)(t-38.56)×1.22718 = 0
Therefore t = -134.631 s or 38.56 s as we only deal with positive values of time in the present question we have t₂ = 38.56 s and
v₂² = v₁² + 2·a₂·S = (117.895 m/s)² + 2·2.45436 m/s²×6371 km = 45172.686
v₂ = 212.54 m/s
For final stage we have r = R and
[tex]F_{G3}[/tex] = 981.746 N and a₂ = F/m₁ = 9.81746 m/s²
Therefore the time from R to the arctic ocean is given by
S₃ =R=v₂·t+0.5·a₂·t² = 212.54·t + 0.5·9.81746·t² = 6371
Which gives
Therefore t₃ = 20.382 s
Therefore, it will take about t₁ + t₂ + t₃ = 108.079 s + 38.56 s + 20.382 s = 167.021 s to splash down in the ocean
Use the formula t = sqrt((2 * d) / g), where d = 4R and g = G * (mass of earth) / (distance from center)^2 to calculate the time taken for the piano to hit the ocean from a height 3R above the Earth's surface.
Explanation:This physics question deals with the concept of
gravity
and the formation of a free fall situation in a vacuum situation where no other forces (apart from gravitation) are considered. In this case, the value of gravity will not be 9.8 m/s^2, as we are not at the surface but at a distance 3 times the radius of Earth. To find how long the piano takes to hit the ocean, we need to use the following formula for the time taken (t) for an object to fall a certain distance under gravity: t = sqrt((2 * d) / g). Here, d is the total distance that the piano would fall, and g is the acceleration due to gravity. We would first need to calculate the distance the piano falls, which will be the sum of the Earth’s radius (R) and the height the piano is dropped from (3R). This equals 4R. Then, calculate the value of gravity at that point, using formula G * (mass of Earth)/(distance from point to center of earth)^2, and then substitute these values back into the equation.
Note
: This is a idealized scenario and other factors like air resistance, influence from moon and sun, and the non-uniform distribution of Earth’s mass etc., are not considered.
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The annual per capita consumption of bottled water was 34.5 gallons. Assume that the per capita consumption of bottled water is approximately normally distributed with a mean of 34.5 and a standard deviation of 11 gallons. a. What is the probability that someone consumed more than 35 gallons of bottled water? b. What is the probability that someone consumed between 25 and 35 gallons of bottled water? c. What is the probability that someone consumed less than 25 gallons of bottled water? d. 97.5% of people consumed less than how many gallons of bottled water?
Answer:
(a) 0.48006
(b) 0.3251
(c) 0.19489
(d) 97.5% of people consumed less than 56 gallons of bottled water.
Step-by-step explanation:
We are given that the per capita consumption of bottled water is approximately normally distributed with a mean of 34.5 and a standard deviation of 11 gallons.
Let X = per capita consumption of bottled water
So, X ~ N([tex]\mu = 34.5,\sigma^{2} =11^{2}[/tex])
The z score probability distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
(a) Probability that someone consumed more than 35 gallons of bottled water = P(X > 35)
P(X > 35) = P( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{35-34.5}{11}[/tex] ) = P(Z > 0.05) = 1 - P(Z [tex]\leq[/tex] 0.05)
= 1 - 0.51994 = 0.48006
(b) Probability that someone consumed between 25 and 35 gallons of bottled water = P(25 < X < 35)
P(25 < X < 35) = P(X < 35) - P(X [tex]\leq[/tex] 25)
P(X < 35) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{35-34.5}{11}[/tex] ) = P(Z < 0.05) = 0.51994
P(X [tex]\leq[/tex] 25) = P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\leq[/tex] [tex]\frac{25-34.5}{11}[/tex] ) = P(Z [tex]\leq[/tex] -0.86) = 1 - P(Z < 0.86)
= 1 - 0.80511 = 0.19489
Therefore, P(25 < X < 35) = 0.51994 - 0.19489 = 0.3251
(c) Probability that someone consumed less than 25 gallons of bottled water = P(X < 25)
P(X < 25) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{25-34.5}{11}[/tex] ) = P(Z < -0.86) = 1 - P(Z [tex]\leq[/tex] 0.86)
= 1 - 0.80511 = 0.19489
(d) We have to find that 97.5% of people consumed less than how many gallons of bottled water, which means ;
P(X < x) = 0.975
P( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{x-34.5}{11}[/tex] ) = 0.975
P(Z > [tex]\frac{x-34.5}{11}[/tex] ) = 0.975
Now in the z table the critical value of X which have an area less than 0.975 is 1.96, i.e.;
[tex]\frac{x-34.5}{11}[/tex] = 1.96
[tex]x[/tex] - 34.5 = [tex]1.96 \times 11[/tex]
[tex]x[/tex] = 34.5 + 21.56 = 56.06 ≈ 56 gallons of bottled water
So, 97.5% of people consumed less than 56 gallons of bottled water.
a. The probability that someone consumed more than 35 gallons of bottled water is approximately 52.12%. b. The probability that someone consumed between 25 and 35 gallons of bottled water is approximately 9.58%. c. The probability that someone consumed less than 25 gallons of bottled water is approximately 42.51%. d. 97.5% of people consumed less than approximately 57.56 gallons of bottled water.
Explanation:a. To find the probability that someone consumed more than 35 gallons of bottled water, we need to standardize the value of 35 using the formula z = (x - mean) / standard deviation, where x is the value we want to find the probability for. In this case, x = 35. Plugging in the values, we get z = (35 - 34.5) / 11 = 0.045. Using a standard normal distribution table, the probability (area under the curve) to the right of 0.045 is approximately 0.5212. Therefore, the probability that someone consumed more than 35 gallons of bottled water is 0.5212 or 52.12%.
b. To find the probability that someone consumed between 25 and 35 gallons of bottled water, we need to find the probability to the right of 25 (P(x > 25)) and subtract the probability to the right of 35 (P(x > 35)). Following the same process as in part (a), we find that P(x > 25) ≈ 0.5746 and P(x > 35) ≈ 0.4788. Therefore, the probability that someone consumed between 25 and 35 gallons of bottled water is approximately 0.5746 - 0.4788 = 0.0958 or 9.58%.
c. To find the probability that someone consumed less than 25 gallons of bottled water, we need to find the probability to the left of 25 (P(x < 25)). Using the standard normal distribution table, we find that P(x < 25) ≈ 0.4251. Therefore, the probability that someone consumed less than 25 gallons of bottled water is approximately 0.4251 or 42.51%.
d. To find the value at which 97.5% of people consumed less than, we need to find the z-score that corresponds to the 97.5th percentile of the standard normal distribution. Using the standard normal distribution table, we find that the z-score corresponding to the 97.5th percentile is approximately 1.96. Now we can use the formula z = (x - mean) / standard deviation to solve for x. Plugging in the values, we get 1.96 = (x - 34.5) / 11. Solving for x, we find x ≈ 1.96 * 11 + 34.5 ≈ 57.56. Therefore, 97.5% of people consumed less than approximately 57.56 gallons of bottled water.
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Engineers must consider the breadths of male heads when designing helmets. The company researchers have determined that the population of potential clientele have head breadths that are normally distributed with a mean of 6.1-in and a standard deviation of 1-in. Due to financial constraints, the helmets will be designed to fit all men except those with head breadths that are in the smallest 2.3% or largest 2.3%.
What is the minimum head breadth that will fit the clientele?
min =
What is the maximum head breadth that will fit the clientele?
max =
Enter your answer as a number accurate to 1 decimal place. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.
Answer:
a) The minimum head breadth that will fit the clientele = 4.105 inches to 3d.p = 4.1 inches to 1 d.p
b) The maximum head breadth that will fit the clientele = 8.905 inches to 3 d.p = 8.9 inches to 1 d.p
Step-by-step explanation:
This is normal distribution problem.
A normal distribution has all the data points symmetrically distributed around the mean in a bell shape.
For this question, mean = xbar = 6.1 inches
Standard deviation = σ = 1 inch
And we want to find the lowermost 2.3% and uppermost 2.3% of the data distribution.
The minimum head breadth that will fit the clientele has a z-score with probability of 2.3% = 0.023
Let that z-score be z'
That is, P(z ≤ z') = 0.023
Using the table to obtain the value of z'
z' = - 1.995
P(z ≤ - 1.995) = 0.023
But z-score is for any value, x, is that value minus the mean then divided by the standard deviation.
z' = (x - xbar)/σ
- 1.995 = (x - 6.1)/1
x = -1.995 + 6.1 = 4.105 inches
The maximum head breadth that will fit the clientele has a z-score with probability of 2.3% also = 0.023
Let that z-score be z''
That is, P(z ≥ z'') = 0.023
Using the table to obtain the value of z''
P(z ≥ z") = P(z ≤ -z")
- z'' = - 1.995
z" = 1.995
P(z ≥ 1.995) = 0.023
But z-score is for any value, x, is that value minus the mean then divided by the standard deviation.
z'' = (x - xbar)/σ
1.995 = (x - 6.1)/1
x = 1.995 + 6.1 = 8.905 inches
Using z-scores for the smallest and largest 2.3% of the normal distribution, and a formula that accounts for the mean and standard deviation of head breadths, we can calculate that the helmets need to accommodate head breadths between 4.1 inches (minimum) and 8.1 inches (maximum).
Explanation:To determine the minimum and maximum head breadths that will fit the clientele, we need to find the z-scores that correspond to the smallest 2.3% and the largest 2.3% of the normal distribution. Then we can use these z-scores to calculate the specific head breadths.
First, we find the z-scores using a standard z-table or a calculator since the normal distribution table typically provides the area to the left of a z-score. For the smallest 2.3%, we need the z-score that has 0.023 to its left. This value is approximately z = -2. Conversely, for the largest 2.3%, since the normal distribution is symmetric, the z-score will have the same absolute value but be positive, which would be z = 2. However, to be precise, one should use statistical tables or software to find the exact z-scores close to these values.
To convert the z-scores to specific head breadths, we use the formula:
X = μ + (z * σ)
where X is the head breadth, μ is the mean, and σ is the standard deviation.
The minimum head breadth (Xmin) is calculated as follows:
Xmin = 6.1 + (-2 * 1) = 6.1 - 2 = 4.1 inchesThe maximum head breadth (Xmax) is calculated as follows:
Xmax = 6.1 + (2 * 1) = 6.1 + 2 = 8.1 inchesTherefore, the minimum head breadth that will fit the clientele is 4.1 inches, and the maximum head breadth is 8.1 inches.
Consider the experiment of rolling a pair of dice. Suppose that we are interested in the sum of the face values showing on the dice. (a) How many sample points are possible? (Hint: use the counting rule for multiple-step experiments.) (b) List the sample points. There to sum the face values of a pair of dice to 2. There to sum the face values of a pair of dice to 3. There to sum the face values of a pair of dice to 4. There to sum the face values of a pair of dice to 5. There to sum the face values of a pair of dice to 6. There to sum the face values of a pair of dice to 7. There to sum the face values of a pair of dice to 8. There to sum the face values of a pair of dice to 9. There to sum the face values of a pair of dice to 10. There to sum the face values of a pair of dice to 11. There to sum the face values of a pair of dice to 12. (c) What is the probability of obtaining a value of 5? (d) What is the probability of obtaining a value of 8 or greater? (e) Because each roll has six possible even values (2, 4, 6, 8, 10, and 12) and only five possible odd values (3, 5, 7, 9, and 11), the dice should show even values more often than odd values. Do you agree with this statement? Explain. This statement correct because P(odd) = and P(even) = . (f) What method did you use to assign the probabilities requested? classical method empirical method subjective method relative frequency method
a) 21 sample points
b) Sum 2: (1, 1)
- Sum 3: (1, 2)
- Sum 4: (1, 3), (2, 2)
- Sum 5: (1, 4), (2, 3)
- Sum 6: (1, 5), (2, 4), (3, 3)
- Sum 7: (1, 6), (2, 5), (3, 4)
- Sum 8: (2, 6), (3, 5), (4, 4)
- Sum 9: (3, 6), (4, 5)
- Sum 10: (4, 6), (5, 5)
- Sum 11: (5, 6)
- Sum 12: (6, 6)
c) Probability of obtaining sum of 5 is 2/21
d) Probability of obtaining 8 or greater is 3/7
e) Probability of even is higher that the probability of odd, so even sum are expect to have more appear.
f) classic method
What is probability?
(a) There are 21 possible sample points when rolling a pair of dice.
(b) Here are the sample points for each sum:
- Sum 2: (1, 1)
- Sum 3: (1, 2)
- Sum 4: (1, 3), (2, 2)
- Sum 5: (1, 4), (2, 3)
- Sum 6: (1, 5), (2, 4), (3, 3)
- Sum 7: (1, 6), (2, 5), (3, 4)
- Sum 8: (2, 6), (3, 5), (4, 4)
- Sum 9: (3, 6), (4, 5)
- Sum 10: (4, 6), (5, 5)
- Sum 11: (5, 6)
- Sum 12: (6, 6)
(c) The probability of obtaining a sum of 5 is
: (1, 4), (2, 3)
P(sum of 5) = no of sum of 5 /total number of our
= 2/21
(d) The probability of obtaining a sum of 8 or greater is (2, 6), (3, 5), (4, 4) (3, 6), (4, 5) (4, 6), (5, 5),(5, 6) (6, 6)
P(sum of 8 or greater) = no of sum of 8 and above
P(sum => 8) = 9/21 = 3/7
(e) Yes!
Out of 21 sample points 12 are even while 9 are odd.
P(even) = 12/21 = 4/7
P(odd) = 1 - 4/7 = 3/7
Probability of even is higher that the probability of odd, so even sum are expect to have more appear.
(f) We used the classical method, which involves counting the number of favorable outcomes and dividing by the total number of possible outcomes.
Almost all medical schools require applicants to take the Medical College Admission Test (MCAT). To estimate the mean score of those who took the MCAT at WSSU, you will obtain the scores of an SRS of students. The scores follow a Normal distribution, and from published information you know that the standard deviation is 6.5. Suppose that (unknown to you) the mean score of those taking the MCAT at WSU is 25.0. You sampled 25 students. What is the probability that the mean score of your sample is between 22 and 28
Answer:
97.92% probability that the mean score of your sample is between 22 and 28
Step-by-step explanation:
To solve this question, we have to understand the normal probability distribution and the central limit theorem.
Normal probability distribution:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central limit theorem:
The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
In this problem, we have that:
[tex]\mu = 25, \sigma = 6.5, n = 25, s = \frac{6.5}{\sqrt{25}} = 1.3[/tex]
What is the probability that the mean score of your sample is between 22 and 28
This is the pvalue of Z when X = 28 subtracted by the pvalue of Z when X = 22. So
X = 28
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{28 - 25}{1.3}[/tex]
[tex]Z = 2.31[/tex]
[tex]Z = 2.31[/tex] has a pvalue of 0.9896
X = 22
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{22 - 25}{1.3}[/tex]
[tex]Z = -2.31[/tex]
[tex]Z = -2.31[/tex] has a pvalue of 0.0104
0.9896 - 0.0104 = 0.9792
97.92% probability that the mean score of your sample is between 22 and 28
A population has a mean muequals71 and a standard deviation sigmaequals24. Find the mean and standard deviation of a sampling distribution of sample means with sample size nequals64.
Answer:
Mean 71
Standard deviation 3
Step-by-step explanation:
We use the Central Limit Theorem to solve this question.
The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution with a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
In this problem, we have that:
[tex]\mu = 71, \sigma = 24, n = 64[/tex]. So
Mean 71
Standard deviation [tex]s = \frac{24}{\sqrt{64}} = 3[/tex]
The indicated function y1(x) is a solution of the given differential equation. Use reduction of order or formula (5) in Section 4.2, y2 = y1(x) e−∫P(x) dx y 2 1 (x) dx (5) as instructed, to find a second solution y2(x). (1 − 2x − x2)y'' + 2(1 + x)y' − 2y = 0; y1 = x + 1
Answer and Step-by-step explanation:
The answer is attached below
To identify a second solution y2 from a given differential equation and its solution y1, it is necessary to extract P(x) from the differential equation, compute integral -∫P(x) dx, and multiply the result by the function y1.
Explanation:First, we need to find the function P(x) in the equation y2 = y1(x) e−∫P(x) dx (5). Looking at the given differential equation (1 − 2x − x2)y'' + 2(1 + x)y' − 2y = 0; y1 = x + 1, we can rearrange terms and find that P(x) is equal to -2(1+x)/(1-2x-x2). Then, we can calculate the integral -∫P(x) dx, and multiply this by our given solution y1 to find the second solution y2. It's important to remember that when carrying out these steps, accuracy is crucial since each step builds on the last.
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Consider a binomial probability distribution with pequals=0.6 and nequals=8
Determine the probabilities below.
1)Upper P left parenthesis x equals 2 right parenthesisP(x=2)
2)Upper P left parenthesis x less than or equals 1 right parenthesisP(x≤1)
3)Upper P left parenthesis x greater than 6 right parenthesisP(x>6)
Answer:
(1) The probability of the event X = 2 is 0.0413.
(2) The probability of the event X ≤ 1 is 0.009.
(3) The probability of the event X < 6 is 0.6846.
Step-by-step explanation:
Let the random variable X follow a Binomial distribution with parameter n = 8 and p = 0.60.
The probability mass function of the Binomial distribution is:
[tex]P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0, 1, 2,...[/tex]
(1)
Compute the probability of the event X = 2 as follows:
[tex]P(X=2)={8\choose 2}(0.60)^{2}(1-0.60)^{8-2}\\=\frac{8!}{2!(8-2)!}\times (0.60)^{2}\times(0.40)^{6}\\=28\times0.36\times0.004096\\=0.0413[/tex]
Thus, the probability of the event X = 2 is 0.0413.
(2)
Compute the probability of the event X ≤ 1 as follows:
P (X ≤ 1) = P (X = 0) + P (X = 1)
[tex]={8\choose 0}(0.60)^{0}(1-0.60)^{8-0}+{8\choose 1}(0.60)^{1}(1-0.60)^{8-1}\\=\frac{8!}{0!(8-0)!}\times (0.60)^{0}\times(0.40)^{8}+\frac{8!}{1!(8-1)!}\times (0.60)^{1}\times(0.40)^{7}\\=(1\times1\times0.00066)+(8\times0.60\times0.00164)\\=0.008532\approx0.009[/tex]
Thus, the probability of the event X ≤ 1 is 0.009.
(3)
Compute the probability of the event X < 6 as follows:
P (X < 6) = 1 - P (X ≥ 6)
= 1 - P (X = 6) - P (X = 7) - P (X = 8)
[tex]=1-{8\choose 6}(0.60)^{6}(1-0.60)^{8-6}+{8\choose 7}(0.60)^{7}(1-0.60)^{8-7}+{8\choose 8}(0.60)^{8}(1-0.60)^{8-8}\\=1-\frac{8!}{6!(8-6)!}\times (0.60)^{6}\times(0.40)^{2}+\frac{8!}{7!(8-7)!}\times (0.60)^{7}\times(0.40)^{1}\\+\frac{8!}{8!(8-8)!}\times (0.60)^{8}\times(0.40)^{0}\\=1-0.2090-0.0896-0.0168\\=0.6846[/tex]
Thus, the probability of the event X < 6 is 0.6846.
The probabilities for a binomial distribution with p=0.6 and n=8 for different scenarios such as P(x=2), P(x≤1), and P(x>6), which involves combinations and powers of success and failure probabilities is equal to 0.2799, 0.0888, and 0.0332 respectively.
1) Upper P(x = 2):
To find P(x = 2), we can use the formula for the binomial probability:
P(x = 2) = (8 choose 2) × (0.6)² × (0.4)⁶= 0.2799
2) Upper P(x ≤ 1):
P(x ≤ 1) = P(x = 0) + P(x = 1) = (8 choose 0)× (0.6)⁰ ×(0.4)⁸ + (8 choose 1) ×(0.6)¹ ×(0.4)⁷ = 0.0888
3) Upper P(x > 6):
P(x > 6) = 1 - P(x ≤ 6) = 1 - (P(x = 0) + P(x = 1) + ... + P(x = 6)) = 1 - 0.9668 = 0.0332
Consider a manufacturing process that is producing hypodermic needles that will be used for blood donations. These needles need to have a diameter of 1.65 mm—too big and they would hurt the donor (even more than usual), too small and they would rupture the red blood cells, rendering the donated blood useless. Thus, the manufacturing process would have to be closely monitored to detect any significant departures from the desired diameter. During every shift , quality control personnel take a random sample of several needles and measure their diameters. If they discover a problem, they will stop the manufacturing process until it is corrected. Suppose the most recent random sample of 35 needles have an average diameter of 1.64 mm and a standard deviation of 0.07 mm. Also, suppose the diameters of needles produced by this manufacturing process have a bell shaped distribution.
Describe what a Type I error would be in this study.
Answer:
[tex]\text{Average diameter is 1.65 mm and we decide that it is not 1.65 mm.}[/tex]
Step-by-step explanation:
We are given the following in the question:
The needle size should not be too big and too small.
The diameter of the needle should be 1.65 mm.
We design the null and the alternate hypothesis
[tex]H_{0}: \mu = 1.65\text{ mm}\\H_A: \mu \neq 1.65\text{ mm}[/tex]
Sample size, n = 35
Sample mean, [tex]\bar{x}[/tex] = 1.64 mm
Sample standard deviation, s = 0.07 mm
Type I error:
It is the error of rejecting the null hypothesis when it is true.It is also known as false positive error.It is the rejecting of a true null hypothesis.Thus, type I error in this study would mean we reject the null hypothesis that the average diameter is 1.65 mm but actually the average diameters of the needle is 1.65 mm.
Thus, average diameter is 1.65 mm and we decide that it is not 1.65 mm.
Answer:
Type I error will be Rejecting the null hypothesis that the average diameter of needles is 1.65 mm and assume that the average diameter of needles is different from 1.65 mm but the fact is that the null hypothesis was true that the average diameter of needles is 1.65 mm.
Step-by-step explanation:
We are given that a manufacturing process is producing hypodermic needles that will be used for blood donations. These needles need to have a diameter of 1.65 mm—too big and they would hurt the donor (even more than usual), too small and they would rupture the red blood cells, rendering the donated blood useless.
So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] = 1.65 mm
Alternate Hypothesis, [tex]H_1[/tex] : [tex]\mu\neq[/tex] 1.65 mm
Also, the most recent random sample of 35 needles have an average diameter of 1.64 mm and a standard deviation of 0.07 mm.
Now, Type I error Type I error states that : Probability of rejecting null hypothesis given the fact that null hypothesis was true. It is the the probability of rejecting a true hypothesis.
So, in our question Type I error will be Rejecting the null hypothesis that the average diameter of needles is 1.65 mm and assume that the average diameter of needles is different from 1.65 mm but the fact is that the null hypothesis was true that the average diameter of needles is 1.65 mm.
The annual consumption of beef per person was about 64.8 lb in 2000 and about 60.1 lb in 2006. Assume B(t), the annual beef consumption t years after 2000, is decreasing according to the exponential decay model. a) Find the value of k, and write the equation b) Estimate the consumption of beef in 2011 c) In what year (theoretically) will the consumption of beef be 10 lb?
Answer:
a) B(t)= 64.8e^(-0.01255t), k=-0.0125 b) 56.5 c)2150
Step-by-step explanation:
a) B(t)= Ae^(-kt)
at t=0, B(t)=64.8
A=64.8
at t=6, B(t)=60.1
60.1=64.8e^(-6k)
k=0.0125
b) B(t)=64.8e^(-0.0125×11)
B(t)= 56.5
c) 10=64.8e^(-0.0125t)
0.15432=e^(-0.0125t)
-0.0125t=ln(0.15432)
-0.0125t=-1.869
t=149.5 or 150
Year= 2150
a)k ≈ 0.01267 and the equation is [tex]B(t) = 64.8e^{-0.01267t[/tex]
b)B(11) ≈ 56.38 lb
c) t ≈ 143.86
A beef consumption model follows exponential decay where B(t), the annual beef consumption t years after 2000,
can be modeled by the equation: [tex]B(t) = B_0e^{-kt}.[/tex]
Given that [tex]B_0[/tex] = 64.8 lb in 2000 and B(6) = 60.1 lb in 2006,
we first need to find the decay constant k.
a) Find the value of k, and write the equation:
We use the given data points to solve for k.
[tex]B(6) = 64.8e^{-6k} = 60.1[/tex]
[tex]e^{-6k} = 60.1 / 64.8[/tex]
[tex]e^{-6k}[/tex] ≈ 0.9272
Taking the natural log of both sides:
-6k = ln(0.9272)
k ≈ -ln(0.9272) / 6
k ≈ 0.01267
Therefore, the equation for B(t) is:
[tex]B(t) = 64.8e^{-0.01267t[/tex]
b) Estimate the consumption of beef in 2011:
For t = 11 (since 2011 is 11 years after 2000):
[tex]B(11) = 64.8e^{-0.01267 * 11[/tex]
B(11) ≈ 64.8[tex]e^{-0.13937[/tex]
B(11) ≈ [tex]64.8 * 0.8699[/tex]
B(11) ≈ 56.38 lb
c) In what year (theoretically) will the consumption of beef be 10 lb?
We need to solve for t when B(t) = 10 lb:
[tex]10 = 64.8e^{-0.01267t[/tex]
[tex]e^{-0.01267t} = 10 / 64.8[/tex]
[tex]e^{-0.01267t[/tex] ≈ 0.1543
Taking the natural log of both sides:
-0.01267t = ln(0.1543)
t ≈ -ln(0.1543) / 0.01267
t ≈ 143.86
Thus, the theoretical year is approximately 2000 + 144 = 2144.
Which expression is equivalent to the expression shown below?
-1/2(-3/2x + 6x + 1) - 3x
Answer:
[tex]-\frac{1}{2} (-\frac{3}{2} x+6x+1)-3x[/tex] [tex]=-\frac{21x}{4} -\frac{1}{2}[/tex] [tex]=-\frac{1}{2}(\frac{21x}{2} +1)[/tex]
Step-by-step explanation:
Given,
[tex]-\frac{1}{2} (-\frac{3}{2} x+6x+1)-3x[/tex]
Applying distribution law
[tex]=(-\frac{1}{2}) (-\frac{3}{2} x)+(-\frac{1}{2}).6x+(-\frac{1}{2}).1-3x[/tex]
[tex]=\frac{3}{4} x-3x-\frac{1}{2}-3x[/tex]
Combine like terms
[tex]=\frac{3}{4} x-3x-3x-\frac{1}{2}[/tex]
Adding like terms
[tex]=\frac{3x-12x-12x}{4} -\frac{1}{2}[/tex]
[tex]=-\frac{21x}{4} -\frac{1}{2}[/tex]
[tex]=-\frac{1}{2}(\frac{21x}{2} +1)[/tex]
The daily demand for gasoline at a local gas station is normally distributed with a mean of 1200 gallons, and a standard deviation of 350 gallons.
If R is a random number between 0 and 1, then which of the following correctly models daily demand for gasoline?
a) 1200 + 350 R
b) 1200 + 350*NORMSDIST(R)
c) NORM.INV(R, 1200, 350)
d) Both b) and c) are correct.
Answer:
c) NORM.INV(R, 1200, 350)
Step-by-step explanation:
Given that the daily demand for gasoline at a local gas station is normally distributed with a mean of 1200 gallons, and a standard deviation of 350 gallons.
X = demand for gasolene at a local gas station is N(1200, 350)
R is any random number between 0 and 1.
Daily demand for gasolene would be
X = Mean + std deviation * z value, where Z = normal inverse of a value between 0 and 1.
The norm inv (R, 1200, 350) for R between 0 and 1 gives all the values of X
Hence correct choice would be
Option c) NORM.INV(R, 1200, 350)
A recent study¹ examined several variables on collegiate football players, including the variable Years, which is number of years playing football, and the variable Percentile, which gives percentile on a cognitive reaction test. The regression line for predicting Percentile from Years is:
Percentile = 102 - 3.34 Years.
¹ Singh R, et al., "Relationship of Collegiate Football Experience and Concussion with Hippocampal Volume and Cognitive Outcomes", JAMA, 311(18), 2014. Data values are estimated from information in the paper.
Predict the cognitive percentile for someone who has played football for 7 years and for someone who has played football for 16 years. Enter the exact answers.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The cognitive percentile for someone who has played for 7 years is 78.62
The cognitive percentile for someone who has played for 16 years is 48.56
Step-by-step explanation:
From the question the regression line for predicting percentile from years is given as
percentile = [tex]102 - 3.34 \ Years[/tex]
For someone who has played for 7 years his cognitive percentile would be
[tex]Percentile = 102 - 3.34(7)[/tex]
[tex]=78.62[/tex]
For someone who has played for 16 years his cognitive percentile would
be [tex]Percentile = 102 - 3.34(16)[/tex]
[tex]=48.56[/tex]
Final answer:
The predicted cognitive percentile for someone who has played football for 7 years is 78.62, while for someone who has played for 16 years, it is 48.56, using the regression equation Percentile = 102 - 3.34 Years.
Explanation:
To predict the cognitive percentile for a football player who has played for 7 years, we use the given regression equation Percentile = 102 - 3.34 Years. By plugging 7 into the equation for Years, we get:
Percentile = 102 - 3.34 × 7
Percentile = 102 - 23.38
Percentile = 78.62
So, for someone who has played football for 7 years, the predicted cognitive percentile would be 78.62.
Similarly, for someone who has played football for 16 years:
Percentile = 102 - 3.34 × 16
Percentile = 102 - 53.44
Percentile = 48.56
Therefore, the predicted cognitive percentile for someone who has played for 16 years is 48.56.
Every month, there are 1000 independent TIE ghter ights, and each TIE ghter ight crashes with a probability of 0.0035. (a) What is the probability that at least 2 crashes occur in the next month
Answer:
86.46% probability that at least 2 crashes occur in the next month.
Step-by-step explanation:
For each TIE ghter ights, there are only two possible outcomes. Either it crashes, or it does not. The probability of a TIE ghter ights crashing is independent from other TIE ghter ights. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
Every month, there are 1000 independent TIE ghter ights, and each TIE ghter ight crashes with a probability of 0.0035.
This means that [tex]n = 1000, p = 0.0035[/tex]
(a) What is the probability that at least 2 crashes occur in the next month?
Either less than 2 crashes occur, or at least 2 do. The sum of the probabilities of these events is decimal 1. So
[tex]P(X < 2) + P(X \geq 2) = 1[/tex]
We want [tex]P(X \geq 2)[/tex]
So
[tex]P(X \geq 2) = 1 - P(X < 2)[/tex]
In which
[tex]P(X < 2) = P(X = 0) + P(X = 1)[/tex]
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{1000,0}.(0.0035)^{0}.(0.9965)^{1000} = 0.0300[/tex]
[tex]P(X = 1) = C_{1000,1}.(0.0035)^{1}.(0.9965)^{999} = 0.1054[/tex]
[tex]P(X < 2) = P(X = 0) + P(X = 1) = 0.0300 + 0.1054 = 0.1354[/tex]
[tex]P(X \geq 2) = 1 - P(X < 2) = 1 - 0.1354 = 0.8646[/tex]
86.46% probability that at least 2 crashes occur in the next month.
How many zeros are at the end of457 · 885?Explain how you can answer this question without actually computing the number. (Hint:10 = 2 · 5.)When this number is written in ordinary decimal form, each 0 at its end comes from a factor of , or one factor of 2 and one factor of .Since there are factors of 2 and factors of 5, there are exactly factors of 10 in the number. This implies that the number ends with zeroes.
The right format of the number is (45^8)(88^5).
Answer:
There are 8 zeros
Step-by-step explanation:
Using the unique factorization of integers theorem, we can break any integer down into the product of prime integers.
So breaking it down we have;
(45^8) = (3 x 3 x 5)^(8)
(88^5) = (2 x 2 x 2 x 11)^(5)
Now, if we put it back together as separate factors, we'll get;
(3^(16)) x (5^(8) ) x (2^(15)) x (11^(5))
Now let's find the number of zeroes by figuring out how many factors of 10 (which equals 2 x 5) we can make. Thus, we can make 8 factors of 10 so it looks like;
(3^(16)) x (2^(7)) x (11^(5)) x (10^(8))
Thus, we can see that there will be 8 zeros as the end is (10^(8))
The number of trailing zeros in the product of 457 and 885 is determined by the factors of 10 (2 and 5) in these numbers. In this specific case, there are no trailing zeros. This also applies in scientific notation - the number of significant figures after the decimal in scientific notation indicates the quantity of zeros at the end of the number.
Explanation:To determine the number of zeroes at the end of the number 457 · 885, consider the factors in the product. Each zero at the end of a number results from a factor of 10, which contains a factor of 2 and a factor of 5. Looking at the numbers 457 and 885, we notice that neither has a factor of 5, therefore there are no trailing zeroes in the product of 457 and 885.
This approach also applies to the scientific notation. The number of significant figures after the decimal in the scientific notation of a number corresponds to the quantity of zeros at the end of it. In such cases, leading zeros are not significant and only serve as placeholders to locate the decimal point. For instance, in the case of the number 1.300 × 10³, the scientific notation shows that there are three significant figures after the decimal and therefore, three zeros at the end of the number.
Overall, understanding how to find the number of trailing zeroes in a product, such as 457 · 885, without actual computation involves a knowledge of the factors of the numbers being multiplied and the principles of significant figures in scientific notation.
Learn more about Trailing Zeros here:https://brainly.com/question/31609866
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Suppose that a softball team is composed of 15 employees of a furniture store of whom 2 work part-time. What proportion of the team work part-time?
Answer:
Therefore, we conclude that 2/15 of the team work part-time.
Step-by-step explanation:
We know that a softball team is composed of 15 employees of a furniture store of whom 2 work part-time. We calculate what proportion of the team work part-time.
So, we will divide the number of those people who work part-time, by the number of people employed in a softball team. We get:
x=2/15
Therefore, we conclude that 2/15 of the team work part-time.
A 22-pound child was admitted with acute bronchitis. Her medical orders include Garamycin 2.5 mg/kg q 8h. You receive Garamycin from the pharmacy in a vial labeled 10 mg/ml. Determine the number of milliliters required per dose.
Answer:
2.4948ml
Step-by-step explanation:
First, we change the child's weight into kilograms:
[tex]22pounds=9.9790kgs[/tex]
From the info, the dose is recommended as 2.5mg/kg. Let x be the number of mg administered to the child:
[tex]1kg=2.5mg\\9.9790kg=x\\\\x=2.5\times9.9790\\\\x=24.9475mg[/tex]
#The drug contains 10mg/ml . Let y be the dose size administered, equate and solve for y:
[tex]10mg=1ml\\24.9475mg=y\\\\\thereforey= \frac{1ml\times24.9475mg}{10mg}\\\\y=2.4948[/tex]
Hence, the dose required is 2.4948ml
Answer:
2.5
Step-by-step explanation:
A rock is thrown upward from a bridge that is 57 feet above a road. The rock reaches its maximum height above the road 0.76 seconds after it is thrown and contacts the road 3.15 seconds after it was thrown.
(a) Write a Function (f) that determines the rock's height above the road (in feet) in terms of the number of seconds t since the rock was thrown.
The function that determines the rock's height above the road in terms of the number of seconds since it was thrown is f(t) = -16t² + 24.32t + 57. This quadratic function accounts for the initial upward velocity and the constant downward acceleration due to gravity.
To determine the rock's height above the road as a function of time after it was thrown, we can use the standard equation of motion under constant acceleration due to gravity, which in this case is downward. The formula for the height f(t) of the rock at time t seconds after it is thrown is:
f(t) = -16t² + vt + s
where:
-16t² represents the effect of gravity (in feet per second squared), because acceleration due to gravity is approximately 32 feet per second squared downward, and we use -16 because the height is measuring upward distance from the road,v is the initial velocity of the rock in feet per second (upward positive), ands is the initial height above the road in feet.Since the rock reaches its maximum height 0.76 seconds after it is thrown, the initial velocity v can be calculated using the fact that the velocity at the peak height is 0 (the rock stops moving upward for an instant before descending).
At maximum height, the velocity v had decreased by 32 feet/second for 0.76 seconds, which is v - 32(0.76) = 0. Solving for v, we find that v = 32(0.76).
The rock contacts the road 3.15 seconds after it was thrown, meaning it falls 57 feet in that time. Plugging these values into the formula:
f(t) = -16t² + 32(0.76)t + 57
which simplifies to:
f(t) = -16t² + 24.32t + 57
This function f(t) gives the rock's height in feet above the road at any time t measured in seconds since it was thrown, assuming negligible air resistance.
The correct function for the rock's height above the road in terms of the number of seconds [tex]\( t \)[/tex] since the rock was thrown is: [tex]\[ f(t) = -16t^2 + v_0t + 57 \][/tex] where [tex]\( v_0 \)[/tex] is the initial velocity of the rock in feet per second.
To derive this function, we use the kinematic equation for the vertical motion of an object under constant acceleration due to gravity, which is:
[tex]\[ h(t) = h_0 + v_0t - \frac{1}{2}gt^2 \][/tex]
Given that the bridge is 57 feet above the road, [tex]\( h_0 = 57 \)[/tex] feet. The acceleration due to gravity is [tex]\( g = 32 \)[/tex] feet per second squared. Plugging these values into the kinematic equation, we get:
[tex]\[ h(t) = 57 + v_0t - \frac{1}{2}(32)t^2 \][/tex]
Simplifying the equation by multiplying [tex]\( \frac{1}{2} \)[/tex] with 32 gives us:
[tex]\[ h(t) = 57 + v_0t - 16t^2 \][/tex]
This simplifies to the function [tex]\( f(t) \)[/tex] as provided above. Note that [tex]\( v_0 \)[/tex] is still unknown and will need to be determined using additional information given in the problem, such as the time it takes for the rock to reach its maximum height.
To find [tex]\( v_0 \)[/tex], we can use the fact that at the maximum height, the velocity of the rock is zero. The kinematic equation relating velocity and time under constant acceleration is:
[tex]\[ v(t) = v_0 - gt \][/tex]
Setting [tex]\( v(t) = 0 \)[/tex] at the maximum height and solving for [tex]\( v_0 \)[/tex] gives us:
[tex]\[ 0 = v_0 - gt_{\text{max}} \][/tex]
[tex]\[ v_0 = gt_{\text{max}} \][/tex]
Given that the rock reaches its maximum height 0.76 seconds after it is thrown, [tex]\( t_{\text{max}} = 0.76 \)[/tex] seconds. Plugging this into the equation for [tex]\( v_0 \)[/tex], we get:
[tex]\[ v_0 = 32 \times 0.76 \][/tex]
[tex]\[ v_0 = 24.32 \][/tex]
Now we can write the complete function with the known values:
[tex]\[ f(t) = -16t^2 + 24.32t + 57 \][/tex]
This function represents the height of the rock above the road as a function of time since it was thrown.