a. The t-value such that the area in the right tail is 0.10 with 28 degrees of freedom is 1.701.
b. The t-value such that the area in the right tail is 0.05 with 27 degrees of freedom is 2.045.
c. The t-value such that the area to the left of it is 0.15 with 7 degrees of freedom is 1.963.
d. The critical t-value that corresponds to 98% confidence with 26 degrees of freedom is 2.457.
How to explain the values(a) Since the area in the right tail is 0.10, the area in the left tail is 1 - 0.10 = 0.90.
Using the t-table, we find that the t-value with 28 degrees of freedom and an area of 0.90 in the left tail is 1.701.
(b) Since the area in the right tail is 0.05, the area in the left tail is 1 - 0.05 = 0.95.
Using the t-table, we find that the t-value with 27 degrees of freedom and an area of 0.95 in the left tail is 2.045.
(c) Since the area to the left of the t-value is 0.15, the area in the right tail is 1 - 0.15 = 0.85.
Using the t-table, we find that the t-value with 7 degrees of freedom and an area of 0.85 in the right tail is 1.963.
Therefore, the t-value such that the area to the left of it is 0.15 with 7 degrees of freedom is 1.963.
(d) Since the confidence level is 98%, the significance level is 1 - 0.98 = 0.02.
Using the t-table, we find that the t-value with 26 degrees of freedom and an area of 0.02 in the right tail is 2.457.
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A simple random sample of size n = 64 is obtained from a population with p = 76 and o=8. Describe the sampling distribution of x. (a) What is P (x>78) ? (b) What is P (x>73.6)?
a. Describing the sampling we get: P (x > 78) = 0.0228 or 2.28%.
b. The probability P (x > 73.6) = 0.9953 or 99.53%.
The sampling distribution of x is normally distributed with a mean of µ = 76 and a standard deviation of σ = 8/√64 = 1. (a) The z-score for a sample mean of x > 78 is (78 - 76) / (8 / √64) = 2. The probability of a z-score greater than 2 is approximately 0.0228 or 2.28%. Hence P (x > 78) = 0.0228 or 2.28%.
(b) The z-score for a sample mean of x > 73.6 is (73.6 - 76) / (8 / √64) = -2.6. The probability of a z-score greater than -2.6 is approximately 0.9953 or 99.53%. Hence P (x > 73.6) = 0.9953 or 99.53%.
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A simple random sample of size n = 64 is obtained from a population with p = 76 and o=8.
We have to find the following inferences from the sample statistics
Mean of the sampling distribution of x is μx=μ=76 (the population mean).
Standard deviation of the sampling distribution of x is σx=σ/√n=8/√64=1
Shape of the distribution is approximately normal by the central limit theorem.
Now we know that standard normal variate is calculated as:
z= x - μx/σx = x - μ / σx
where x is the random variable.P (x>78) is to be calculated.
Using the above formula, we get:
[tex]P (x>78) = P(z>78 - 76 / 1)= P(z>2)[/tex]
At z=2, the area is 0.0228.
Hence,P (x>78) = P(z>2)= 0.0228 (approximately)
Using the above formula, we get:
[tex]P (x>73.6) = P(z>73.6 - 76 / 1)= P(z>-2.4)[/tex]
At z=-2.4, the area is 0.0082.
Hence,[tex]P (x>73.6) = P(z>-2.4)= 0.0082[/tex] (approximately)
Therefore, the answers are:(a) P (x>78) = 0.0228 (approximately)(b) P (x>73.6) = 0.0082 (approximately).
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As the length of a confidence interval increases, the degree of confidence in it actually containing the population parameter being estimated (confidence level) also increases. Is this statement true or false? Explain.
The statement "As the length of a confidence interval increases, the degree of confidence in it actually containing the population parameter being estimated (confidence level) also increases" is false. The confidence level remains the same regardless of the length of the confidence interval.
The confidence level of a confidence interval is determined before any data is collected and is a measure of the long-term success rate of the procedure used to construct the interval. It represents the probability that the interval will capture the true population parameter in repeated sampling.
The length of a confidence interval, on the other hand, depends on factors such as the variability of the data and the desired level of precision. The length of the interval determines the range of plausible values for the population parameter.
While it is true that a longer confidence interval may capture a wider range of potential values, it does not increase the degree of confidence in containing the true population parameter. The confidence level is fixed at the time of construction and does not change based on the length of the interval. The confidence level provides a measure of the reliability of the estimation procedure, while the length of the interval affects the precision and range of plausible values.
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Show, using the Mean Value Theorem, that | sin r – sin y = x - yl for all real numbers r and y. Prove, using a), that sin r is uniformly continuous on R.
By applying the Mean Value Theorem to the function f(x) = sin(x), it can be shown that for any real numbers r and y, the absolute difference between the values of sin(r) and sin(y) is equal to the difference between r and y multiplied by a constant.
According to the Mean Value Theorem, if a function f(x) is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there exists a number c in (a, b) such that f'(c) = (f(b) - f(a))/(b - a). Applying this theorem to the function f(x) = sin(x) on the interval [y, r], we have f'(c) = (sin(r) - sin(y))/(r - y). Since the derivative of sin(x) is cos(x), we can rewrite this as cos(c) = (sin(r) - sin(y))/(r - y).
Now, consider the function g(x) = cos(x). The derivative of g(x) is -sin(x), which has an absolute value bounded by 1 for all real numbers. Therefore, |cos(c)| ≤ 1, which implies |(sin(r) - sin(y))/(r - y)| ≤ 1. Rearranging the equation, we get |sin(r) - sin(y)| ≤ |r - y|.
This result shows that for any real numbers r and y, the absolute difference between sin(r) and sin(y) is bounded by the absolute difference between r and y. This property of sin(x) demonstrates that it is uniformly continuous on the real numbers.
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Question 3
Suppose X N(20,5)
(a) Find:
(i) P(X> 18) (5 marks)
(ii) P(7 < X < 15) (5 marks)
(b) Find the value a such that P(20-a < X < 20+ a) = 0.99 (10 marks)
(c) Find the value b such that P(20-b< X < 20+ b) = 0.95 (10 marks)
a) (i) We have to find P(X > 18), given that X ~ N(20,5) = 0.1859
(ii) Similarly, we can find P(7 < X < 15) = 0.9818
b) The normal distribution is a continuous probability distribution that is symmetric and bell-shaped, and value = 4.576
a) Using the standard normal distribution table
Since X follows a normal distribution with mean 20 and variance 5, we have:
Z = (X - μ)/σ = (X - 20)/√5 ~ N(0,1)
We can now find P(X > 18) by standardizing and using the standard normal distribution table:
P(X > 18) = P(Z < (18 - 20)/√5)
= P(Z < -0.8944)
= 0.1859
(ii) Similarly, we can find P(7 < X < 15) as follows:
Z1 = (7 - 20)/√5 = -4.62, Z2
= (15 - 20)/√5
= -2.24P(7 < X < 15)
= P(Z1 < Z < Z2)
= P(Z < -2.24) - P(Z < -4.62)
= 0.9854 - 0.0036
= 0.9818
(b) We have to find the value of a such that P(20 - a < X < 20 + a) = 0.99
Given that X ~ N(20, 5), we know that:
P(20 - a < X < 20 + a) = 0.99
= P((20 - a - 20)/√5 < Z < (20 + a - 20)/√5)
= P(-a/√5 < Z < a/√5)
= 0.99
This means that we need to find the value of a such that:
P(-a/√5 < Z < a/√5)
= 0.99 - 0.01/2
= 0.985.
Using the standard normal distribution table, we can find that:
P(Z < a/√5) - P(Z < -a/√5)
= 0.985P(Z < a/√5) - [1 - P(Z < a/√5)]
= 0.9852P(Z < a/√5)
= 0.9925P(Z < a/√5)
= 2.05 (from standard normal distribution table)
Therefore, a/√5 = 2.05
=> a = 2.05√5
= 4.576
(c) We have to find the value of b such that P(20 - b < X < 20 + b) = 0.95
Given that X ~ N(20, 5), we know that:
P(20 - b < X < 20 + b) = 0.95
= P((20 - b - 20)/√5 < Z < (20 + b - 20)/√5)
= P(-b/√5 < Z < b/√5)
= 0.95
This means that we need to find the value of b such that:
P(-b/√5 < Z < b/√5)
= 0.95 - 0.05/2
= 0.975.
Using the standard normal distribution table, we can find that:
P(Z < b/√5) - P(Z < -b/√5)
= 0.975P(Z < b/√5) - [1 - P(Z < b/√5)]
= 0.9752P(Z < b/√5)
= 0.9875P(Z < b/√5)
= 1.96 (from standard normal distribution table)
Therefore, b/√5 = 1.96
b = 1.96√5
= 4.39.
The normal distribution is a continuous probability distribution that is symmetric and bell-shaped.
It is denoted by N(μ, σ), where μ is the mean and σ is the standard deviation.
In this question, we were given that X follows a normal distribution with mean 20 and standard deviation 5.
To find the probability of X falling within a certain range, we standardize X to obtain Z ~ N(0,1) using the formula Z
= (X - μ)/σ.
We can then use the standard normal distribution table to find the required probabilities.
To find the value of a such that P(20 - a < X < 20 + a) = 0.99,
we needed to find the value of a such that P(-a/√5 < Z < a/√5) = 0.985.
To find the value of b such that P(20 - b < X < 20 + b) = 0.95,
we needed to find the value of b such that P(-b/√5 < Z < b/√5) = 0.975.
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Find the dimensions of the subspace spanned by the vectors (1 0 2), (3 1 1), (-2 -2 1), (5 2 2)
The dimensions of the subspace spanned by the given vectors, we need to determine the number of linearly independent vectors among them. The dimensions of the subspace spanned by the given vectors are 2.
To find the dimensions of the subspace spanned by the given vectors, we need to determine the number of linearly independent vectors among them. We can achieve this by performing row reduction on the augmented matrix formed by the vectors.
Taking the given vectors as the columns of a matrix, we have:
[ 1 3 -2 5 ]
[ 0 1 -2 2 ]
[ 2 1 1 2 ]
Performing row reduction, we get:
[ 1 0 2 1 ]
[ 0 1 -2 2 ]
[ 0 0 0 0 ]
The row reduced echelon form of the matrix shows that the third row is a row of zeros, indicating that the vectors are linearly dependent. Therefore, the subspace spanned by the given vectors has a dimension of 2.
In other words, the subspace is a plane in three-dimensional space, and any two linearly independent vectors from the given set can form a basis for this subspace.
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A small block with a mass of 0.0400 kg is moving in the xy-plane. The net force on the block is described by the potential energy function (x) = (5.80 m2 ⁄ )x 2 − (3.60 m3 ⁄ )y 3 . What are the magnitude and direction of the acceleration of the block when it is at the point (x = 0.300m, y = 0.600m)?
The small block with a mass of 0.0400 kg is moving in the xy-plane, and its net force is described by the potential energy function (x) = (5.80 m^2/ )x^2 - (3.60 m^3/ )y^3. The magnitude of the acceleration is approximately 130.8 m/s^2, and its direction is approximately 48.1 degrees below the negative x-axis.
To find the acceleration, we start by calculating the force acting on the block using the negative gradient of the potential energy function. Taking the partial derivatives of the potential energy function with respect to x and y, we obtain the force components ∂U/∂x and ∂U/∂y.
By substituting the given coordinates (x = 0.300m, y = 0.600m) into the partial derivatives, we find the force components Fx and Fy. Using Newton's second law (F = ma), we divide the force components by the mass of the block to obtain the acceleration components ax and ay.
To calculate the magnitude of the acceleration, we use the Pythagorean theorem to find the square root of the sum of the squares of the acceleration components. This yields the magnitude |a| ≈ 130.8 m/s^2.
To determine the direction of the acceleration, we use the inverse tangent function (tan^(-1)) with the ratio of the acceleration components ay/ax. This gives us the angle θ, which is approximately -48.1 degrees.
In summary, the magnitude of the acceleration is approximately 130.8 m/s^2, and its direction is approximately 48.1 degrees below the negative x-axis.
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The scores of students on the SAT college entrance examinations at a certain high school had a normal distribution with mean u and standard deviation o = 26.5. (a) What is the probability that a single student randomly chosen from all those taking the test scores 549 or higher? ANSWER: For parts (b) through (d), consider a simple random sample (SRS) of 30 students who took the test. (b) What are the mean and standard deviation of the sample mean score ł, of 30 students? The mean of the sampling distribution for ž is: The standard deviation of the sampling distribution for ž is: (c) What z-score corresponds to the mean score 7 of 549?
The correct value of μ = 549 - (z * 26.5) and (549 - μ) / 26.5 = z
(a) To find the probability that a single student randomly chosen from all those taking the test scores 549 or higher, we need to calculate the z-score and then find the corresponding probability using the standard normal distribution.
The z-score formula is given by:
z = (x - μ) / σ
Where:
x = value we are interested in (549)
μ = mean of the distribution (unknown in this case)
σ = standard deviation of the distribution (26.5)
To find the z-score, we rearrange the formula:
z = (x - μ) / σ
(z * σ) + μ = x
μ = x - (z * σ)
Now we can substitute the values and calculate μ:
μ = 549 - (z * 26.5)
To find the probability, we need to calculate the z-score corresponding to the value 549. Since the distribution is normal, we can use a standard normal distribution table or a calculator to find the probability associated with that z-score.
(b) The mean and standard deviation of the sample mean score, Ł (pronounced "x-bar"), of 30 students can be calculated using the formulas:
Mean of the Sampling Distribution (Ł) = μ
Standard Deviation of the Sampling Distribution (σŁ) = σ / sqrt(n)
Where:
μ = population mean (unknown in this case)
σ = population standard deviation (26.5)
n = sample size (30)
(c) To find the z-score that corresponds to the mean score of 549, we use the same formula as in part (a):
z = (x - μ) / σ
Substituting the values:
z = (549 - μ) / 26.5
Since we are given the mean score and need to find the z-score, we rearrange the formula:
(549 - μ) / 26.5 = z
Now we can solve for z.
Please note that the solution to part (a) will provide the value of μ, which is needed to answer parts (b) and (c).
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The following tables show the average cost per square foot of different types of buildings in Clemson in the year 2012. What is the cost of building a 30,000 S.F. courthouse in Provo in 2016?
The envisioned price of building a 30,000-square-foot courthouse in Provo in 2016 is $1,053,000
To decide the cost of building a 30,000-square-foot courthouse in Provo in 2016, we need to find the average cost according to square feet for courthouses in Clemson in 2012 and then use it on the given data.
According to the table, the average cost according to rectangular feet for a courthouse in Clemson in 2012 is $35.1. To estimate the fee of building a courthouse in Provo in 2016, we can multiply this average cost according to square feet with the aid of the preferred rectangular pictures of 30,000.
Cost of constructing a 30,000 rectangular foot courthouse in Provo in 2016:
Cost = Cost per square foot x Square footage
Cost = $35.1 x 30,000
Cost = $1,053,000
Therefore, the envisioned price of building a 30,000-square-foot courthouse in Provo in 2016 is $1,053,000.
It's critical to note that that is an estimate based at the average fee consistent with square foot in Clemson in 2012. Actual creation charges can vary relying on elements together with area, market conditions, materials, exertions expenses, and particular assignment requirements.
To get a greater accurate estimate, it would be really useful to seek advice from nearby creation specialists or contractors who can provide up-to-date fee data for constructing a courthouse in Provo in 2016.
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A cannon shell follows a parabolic path. It reaches a maximum height of 40ft and land at a distance of 20 ft from the cannon. A. Write the equation of the parabolic path the shell follows. (Note: your answer will depend on where you locate your coordinate axes. B. Find the height of the shell when it's horizontal distance from the cannon is 10 ft.
The ball's height at a horizontal distance of 10 feet from the cannon is H = 56 - 16 = 40 feet.
A cannonball goes in an illustrative way when terminated from a cannon. The level of the ball at some irregular point can be resolved using the going with condition: The equation for H is -16t2 + Vt + H0, where H stands for height, t for time, V for initial velocity, and H0 for initial height. A. Before we can determine the condition of the cannonball's illustration, we must first determine the directions of the highest point it reaches.
Our coordinate axis' starting point will be (0, 0). Since the ball can reach a height of 40 feet, its vertex is at (10,40). The equation can be obtained by replacing these values with those of a parabola: y = a(x - h)2 + k. y = - 16x2 + 800x - 800.B. We want to find the level of the shell when its even partition from the gun is 10 ft. At this point, the height will be determined using the same equation: H = -16t2 + Vt + H0. Because the ball traveled 20 feet horizontally, we know that it took one second for it to land.
Consequently, we can substitute t = 1 and H0 = 0 into the circumstance: H = -16(1)2 + V(1) + 0. The way that the ball voyaged 40 feet in an upward direction in the principal second of its flight (when it was going up) and 20 feet in an upward direction as of now of its flight (when it was descending) can be utilized to compute its speed. H = V - 16. We can substitute t = 1 and H = 40 using the same condition to see as V: 40 = -16(1)2 + V(1) + 0. V = 56. H = 56 - 16 = 40 feet is the ball's height at a horizontal distance of 10 feet from the cannon.
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Find the exact area of the surface obtained by rotating the given curve about the x-axis. Using calculus with Parameter curves.
x = 6t − 2t³, y = 6t², 0 ≤ t ≤ 1
The exact area of the surface obtained by rotating the curve defined by the parameter equations x = 6t - 2t³ and y = 6t² about the x-axis can be determined using calculus. The surface area is approximately 213.65 square units.
To find the surface area, we need to integrate the formula for the surface area of a curve rotated about the x-axis, which is given by A = 2π∫[a,b] y√(1 + (dy/dx)²) dx, where [a,b] represents the range of t values.
First, we calculate dy/dx by taking the derivative of y with respect to x: dy/dx = (dy/dt) / (dx/dt). In this case, dy/dx = 12t / (6 - 6t²).
Next, we substitute the values of x, y, and dy/dx into the surface area formula and integrate with respect to x over the range [a,b]. In this case, the range of t is 0 to 1.
After performing the integration, we obtain the value of the surface area to be approximately 213.65 square units.
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Use appropriate algebra and Theorem 7.2.1 to find the given inverse Laplace transform. (Write your answer as a function of t.) L-1 (4s/4s^2+1)
The inverse Laplace transform of [tex]L^{-1}[/tex](4s/(4s² + 1)) is [tex]e^{(-i/2t)[/tex] + [tex]e^{(i/2t))/2[/tex].
To find the inverse Laplace transform of [tex]L^{-1}[/tex](4s/(4s² + 1)), we can use partial fraction decomposition.
Step 1: Factorize the denominator of the Laplace transform.
4s² + 1 = (2s + i)(2s - i)
Step 2: Write the partial fraction decomposition.
4s/(4s² + 1) = A/(2s + i) + B/(2s - i)
Step 3: Clear the fractions.
4s = A(2s - i) + B(2s + i)
Step 4: Solve for A and B.
Comparing coefficients:
4 = 2A + 2B (coefficient of s terms)
0 = -Ai + Bi (constant terms)
From the second equation, we can see that A = B. Substituting this into the first equation:
4 = 4A
A = 1
So, B = 1 as well.
Step 5: Rewrite the partial fraction decomposition.
4s/(4s² + 1) = 1/(2s + i) + 1/(2s - i)
Step 6: Take the inverse Laplace transform.
[tex]L^{-1}[/tex](4s/(4s² + 1)) = [tex]L^{-1}[/tex](1/(2s + i)) + [tex]L^{-1}[/tex](1/(2s - i))
Using Theorem 7.2.1, the inverse Laplace transforms of the individual terms can be found:
[tex]L^{-1}[/tex](1/(2s + i)) = [tex]e^{(-i/2t)/2[/tex]
[tex]L^{-1}[/tex](1/(2s - i)) = [tex]e^{(i/2t)/2[/tex]
Therefore, the inverse Laplace transform of [tex]L^{-1}[/tex](4s/(4s² + 1)) is:
[tex]L^{-1}[/tex](4s/(4s² + 1)) = [tex]e^{(-i/2t)/2[/tex] + [tex]e^{(i/2t)/2[/tex].
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The joint probability density of the two random variables X and Y is given by ye-v(+1) if x ≥ 0, y ≥ 0 f(x, y) = 0 else. a) Show that f(x, y) is indeed a probability density,
After considering the given data we conclude f(x, y) is not a probability density, since it does not satisfy the second condition.
To describe that f(x, y) is indeed a probability density, we have to verify that it satisfies the following two conditions:
f(x, y) is non-negative for all values of x and y.
The integral of f(x, y) over the entire plane is equal to 1.
For the joint probability density function [tex]f(x, y) = ye^{(-v) (+1)} if x \geq 0, y \geq 0[/tex]and f(x, y) = 0 otherwise, we can describe that it satisfies both of these conditions as follows:
For all values of x and y, we have
[tex]f(x, y) = ye^{(-v) (+1)} if x \geq 0, y \geq 0 and f(x, y) = 0[/tex] otherwise.
Then y and [tex]e^{(-v) (+1)}[/tex] are both non-negative for all values of x and y, it follows that f(x, y) is non-negative for all values of x and y.
To evaluate the integral of f(x, y) over the entire plane, we can integrate f(x, y) with concerning both x and y over their entire ranges:
[tex]\int \int f(x, y) dxdy = \intb\int ye^{(-v)(+1)} dx dy[/tex]
Since the function [tex]ye^{(-v) (+1)}[/tex] is non-negative for all values of x and y, we can integrate it over the entire plane by integrating it over the first quadrant and then multiplying by 4:
[tex]\int\int ye^{(-v) (+1)} dx dy = 4\int\int ye^{(-v) (+1)} dx dy[/tex]
[tex]= 4\int0\int\infty ye^{(-v) (+1)} dx dy[/tex]
[tex]= 4\int0\infty y \int0\infinity e^{(-v) (+1)} dx dy[/tex]
[tex]= 4\int 0\infty y [-e^{(-v) (+1)} ]0\infty dy[/tex]
[tex]= 4\int0\infty y (0 - (-1)) dy[/tex]
[tex]= 4\int 0\infty y dy[/tex]
[tex]= 4[(y^2)/2]0\infty[/tex]
[tex]= 2\infty ^2[/tex]
[tex]= \infty[/tex]
Therefore, the integral of f(x, y) over the entire plane is equal to[tex]\infty[/tex] , which means that f(x, y) is not a probability density.
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For a confidence level of 90% with a sample size of 19, find the critical t value. Check Answer
The critical t-value for a 90% confidence level with a sample size of 19 and 18 degrees of freedom is approximately 1.734. This value is obtained from a t-table or statistical software and is used in hypothesis testing or constructing confidence intervals.
To determine the critical t-value for a 90% confidence level with a sample size of 19, we need to determine the degrees of freedom, which is equal to the sample size minus 1 (n - 1).
Degrees of Freedom (df) = 19 - 1 = 18
Next, we can use a t-table or a statistical software to find the critical t-value for a 90% confidence level with 18 degrees of freedom.
Checking the t-table, the critical t-value for a 90% confidence level with 18 degrees of freedom is approximately 1.734.
Therefore, the critical t-value for a 90% confidence level with a sample size of 19 and 18 degrees of freedom is approximately 1.734.
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Determine (with a proof or a counterexample) whether each of the arithmetic functions below is completely multiplicative, multiplicative, or both. In parts (d)-(f), k is a fixed real number (a) f(n) = 0 (b) f(n) -1 (c) f(n) = 2 (d) f(n) = n + k (e) f(n) = kn
The arithmetic functions examined in the problem are classified based on whether they are completely multiplicative, multiplicative, or neither.
Functions involving constants or linear terms are found to be either completely multiplicative, multiplicative, or not satisfying either condition.
(a) The arithmetic function f(n) = 0 is completely multiplicative. For any two positive integers n and m, f(nm) = 0 = 0 * 0 = f(n) * f(m), satisfying the definition of complete multiplicativity.
(b) The arithmetic function f(n) = -1 is neither completely multiplicative nor multiplicative. For any positive integers n and m, f(nm) = -1 ≠ 1 = (-1) * (-1) = f(n) * f(m), so it fails to satisfy both conditions.
(c) The arithmetic function f(n) = 2 is completely multiplicative. For any two positive integers n and m, f(nm) = 2 = 2 * 2 = f(n) * f(m), fulfilling the definition of complete multiplicativity.
(d) The arithmetic function f(n) = n + k is multiplicative but not completely multiplicative. For any positive integers n and m, f(nm) = nm + k ≠ (n + k) * (m + k) = f(n) * f(m). Therefore, it is multiplicative but not completely multiplicative.
(e) The arithmetic function f(n) = kn is completely multiplicative. For any two positive integers n and m, f(nm) = knm = (kn) * (km) = f(n) * f(m), satisfying the definition of complete multiplicativity.
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3 friends ordered 2 pizzas of 6 slices each and ate equal amounts, how many slices did each person eat?
A 1
B 2
C 3
D 4
Answer:
Option D, 4
Step-by-step explanation:
2 pizzas x 6 slices per pizza = 12 slices of pizza
12 slices of pizza divided by 3 friends eating equal slices = 4 slices per friend
Option D, 4, is your answer
in circle o, ac and bd are diameters. what is m? 50° 80° 100° 130°
In a circle, when two diameters intersect, the angles formed at the intersection point are always right angles (90°).
Therefore, none of the given angle measures (50°, 80°, 100°, 130°) can represent the angle formed by diameters AC and BD.
The correct answer would be 90° since the intersection of diameters always creates right angles in a circle.
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Answer: A. 50°
Asked the AI
In a program designed to help patients stop smoking, 219 patients were given sustained care, and 82.2% of them were no longer smoking after one month. Use a 0.10 significance level to test the claim that 80%
Based on this sample, we cannot say that the proportion of patients who quit smoking after one month is different from 80% with a 0.10 significance level.
How to solve for the proportionFirst, let's calculate the sample proportion (p'):
p = x/n = (0.822 * 219) / 219 = 0.822
Next, let's calculate the standard error (SE) of the sample proportion:
SE = √( p(1 - p) / n ) = sqrt( 0.80 * 0.20 / 219)
Using a calculator or Python, the standard error is calculated as follows:
SE ≈ sqrt(0.16 / 219) ≈ 0.034
Now we can calculate the z-score, which is (p' - p) / SE.
z = (0.822 - 0.80) / 0.034 ≈ 0.65
Finally, we compare this z-score to the critical z-score for our significance level (0.10). Since we are doing a two-tailed test, the critical z-scores are approximately ±1.645.
Because our calculated z-score of 0.65 is less than 1.645 and greater than -1.645, we do not have enough evidence to reject the null hypothesis. This means that based on this sample, we cannot say that the proportion of patients who quit smoking after one month is different from 80% with a 0.10 significance level.
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The speed of the fluid in the constriction of the pipe can be determined using the principle of continuity, which states that the mass flow rate of an incompressible fluid remains constant. The speed of the fluid in the constriction can be calculated by applying the equation of continuity, considering the change in diameter.
According to the principle of continuity, the mass flow rate of an incompressible fluid remains constant along a pipe. This means that the product of the fluid's velocity and the cross-sectional area of the pipe remains constant.
Let's denote the initial diameter of the pipe as D1 = 6 cm and the final diameter (in the constriction) as D2 = 3 cm. The initial velocity of the fluid is v1 = 1 m/s.
The cross-sectional area of the pipe at the initial section is A1 = π(D1/2)^2, and at the constriction section, it is A2 = π(D2/2)^2.
According to the principle of continuity, A1 * v1 = A2 * v2, where v2 is the velocity of the fluid in the constriction.
We can substitute the values into the equation: π(D1/2)^2 * v1 = π(D2/2)^2 * v2.
Simplifying the equation: (D1/2)^2 * v1 = (D2/2)^2 * v2.
Plugging in the given values: (6/2)^2 * 1 = (3/2)^2 * v2.
9 * 1 = 2.25 * v2.
v2 = 9/2.25 = 4 m/s.
Therefore, the speed of the fluid in the constriction is 4 m/s.
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Find the mean, median, and mode(s) for the given sample data. Round to two decimal places as needed. 6) The amount of time in hours) that Sam studied for an exam on each of the last five days is 6) given below. 2.7 8.3 6.8 2.1 5.1
The mean value of the sample data is 5.8 hours.
The median value of the sample data is 5.95 hours.
Mode of the given sample data are:\[\begin{array}{l}\text{Mean} = 5.8\,\,\text{hours}\\\\\text{Median} = 5.95\,\,\text{hours}\\\\\text{Mode} = \text{none}\end{array}\]
Given sample data (hours): 2.7, 8.3, 6.8, 2.1, 5.1.
To find mean, median, and mode(s), we need to arrange the sample data in ascending order, as follows:2.1, 2.7, 5.1, 6.8, 8.3
(a) Mean: The mean is the sum of all data values divided by the number of data values. So, we have:\[\text{Mean} = \frac{{2.1 + 2.7 + 5.1 + 6.8 + 8.3}}{5} = 5.8\]Therefore, the mean value of the sample data is 5.8 hours.
(b) Median: The median is the middle value of the sample data, after it has been sorted. So, we have:Median = (5.1 + 6.8) / 2 = 5.95Therefore, the median value of the sample data is 5.95 hours.
(c)Mode: The mode is the most frequently occurring value in the sample data. Here, we don't have any repeating value.
Therefore, there is no mode for this sample data.
Finally, the mean, median, and mode of the given sample data are:\[\begin{array}{l}\text{Mean} = 5.8\,\,\text{hours}\\\\\text{Median} = 5.95\,\,\text{hours}\\\\\text{Mode} = \text{none}\end{array}\]
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The given sample data is {2.7, 8.3, 6.8, 2.1, 5.1}.
Now, we have to find the mean, median, and mode(s) for the given data.
Mean:The formula to find the mean of n given data is;
$$\bar{x} = \frac{1}{n}\sum_{i=1}^{n}x_i$$
Here, n = 5, and the given data is {2.7, 8.3, 6.8, 2.1, 5.1}.
So, putting these values in the formula, we get;
$$\bar{x} = \frac{1}{5}\left(2.7+8.3+6.8+2.1+5.1\right)$$$$\bar{x} = \frac{1}{5}\left(25\right)$$$$\bar{x} = 5$$
Therefore, the mean of the given sample data is 5.
Median:Arrange the given data in ascending order.{2.1, 2.7, 5.1, 6.8, 8.3}
The median is the middle value of the given data. Here, the number of data is odd, and the middle value is
Therefore, the median of the given sample data is
Mode:The mode is the value that occurs the most number of times in the given data.
Here, all the values in the given data occur only once.
Therefore, there is no mode for the given data.
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The joint PDF for random variables X and Y is given as if 0 < x < 1, 0 < y < 2 x = fx.r(2, 4) = { A(48 + 3) 0.W. a) Sketch the sample space. b) Find A so that fx,y(x, y) is a valid joint pdf. c) Find the marginal PDFs fx(x) and fy(y). Are X, Y independent? d) Find P[] < X < 2,1
a) |\
| \
Y | \
| \
| \
| ____ \
X
b) A = 1/102
c) Marginal PDF fx(x) = (1/102) * x
Marginal PDF fy(y) = (51/102)
No, X and Y are not independent since their marginal PDFs fx(x) and fy(y) are not separable (i.e., they cannot be expressed as the product of individual PDFs)
d) P(0 < X < 2, 1) = 1.
a) To sketch the sample space, we need to consider the ranges of X and Y as defined in the problem statement: 0 < x < 1 and 0 < y < 2x. This means that X ranges from 0 to 1 and Y ranges from 0 to 2X. The sample space can be represented by a triangular region bounded by the lines Y = 0, X = 1, and Y = 2X.
|\
| \
Y | \
| \
| \
| ____ \
X
b) To find the value of A so that fx,y(x, y) is a valid joint PDF, we need to ensure that the joint PDF integrates to 1 over the entire sample space.
The joint PDF is given by fx,y(x, y) = A(48 + 3), where 0 < x < 1 and 0 < y < 2x.
To find A, we integrate the joint PDF over the sample space:
∫∫fx,y(x, y) dy dx = 1
∫∫A(48 + 3) dy dx = 1
A∫∫(48 + 3) dy dx = 1
A(48y + 3y)∣∣∣0∣∣2xdx = 1
A(96x + 6x)∣∣∣0∣∣1 = 1
A(96 + 6) = 1
102A = 1
A = 1/102
Therefore, A = 1/102.
c) To find the marginal PDFs fx(x) and fy(y), we integrate the joint PDF over the respective variables.
Marginal PDF fx(x):
fx(x) = ∫fy(x, y) dy
Since 0 < y < 2x, the integral limits for y are 0 to 2x.
fx(x) = ∫A(48 + 3) dy from 0 to 2x
fx(x) = A(48y + 3y)∣∣∣0∣∣2x
fx(x) = A(96x + 6x)
fx(x) = 102A * x
fx(x) = (1/102) * x
Marginal PDF fy(y):
fy(y) = ∫fx(x, y) dx
Since 0 < x < 1, the integral limits for x are 0 to 1.
fy(y) = ∫A(48 + 3) dx from 0 to 1
fy(y) = A(48x + 3x)∣∣∣0∣∣1
fy(y) = A(48 + 3)
fy(y) = A(51)
fy(y) = (51/102)
No, X and Y are not independent since their marginal PDFs fx(x) and fy(y) are not separable (i.e., they cannot be expressed as the product of individual PDFs).
d) To find P(0 < X < 2, 1), we need to integrate the joint PDF over the given range.
P(0 < X < 2, 1) = ∫∫fx,y(x, y) dy dx over the region 0 < x < 2 and 0 < y < 1
P(0 < X < 2, 1) = ∫∫A(48 + 3) dy dx over the region 0 < x < 2 and 0 < y < 1
P(0 < X < 2, 1) = A(48y + 3y)∣∣∣0∣∣1 dx over the region 0 < x < 2
P(0 < X < 2, 1) = A(48 + 3) dx over the region 0 < x < 2
P(0 < X < 2, 1) = A(51x)∣∣∣0∣∣2
P(0 < X < 2, 1) = A(102)
P(0 < X < 2, 1) = (1/102)(102)
P(0 < X < 2, 1) = 1
Therefore, P(0 < X < 2, 1) = 1.
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"As attendance at school drops, so does achievement" is an example of what type of correlation? Negative Positive No correlation
The statement "As attendance at school decreases, achievement also decreases" exemplifies a negative correlation between attendance and achievement.
Correlation pertains to the association or connection between two variables. In this case, the variables are attendance at school and achievement. A negative correlation means that as one variable decreases, the other variable also decreases.
The statement suggests that as attendance at school drops, achievement also decreases. This implies that there is a negative relationship between attendance and achievement. When students attend school less frequently, their academic performance tends to decline.
Negative correlations are characterized by an inverse relationship between variables, where an increase in one variable corresponds to a decrease in the other. In this scenario, the negative correlation indicates that lower attendance is associated with lower achievement levels.
It is important to note that correlation does not imply causation, and there may be other factors influencing both attendance and achievement.
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It is assumed that the average Triglycerides levet in a healthy person is 130 unit. In a sample of 30 patients, the sample mean of Triglycerides level is 122 and the sample standard deviation is 20. Calculate the test statistic value
The test statistic value for this situation is approximately -2.474.
A hypothesis test comparing the sample mean to the assumed population mean is necessary in order to determine the value of the test statistic. The population mean triglycerides level would be the null hypothesis (H0), and the alternative hypothesis (Ha) would be that the population mean is not 130 units.
The t-statistic, which is calculated as follows, is the test statistic utilized in this circumstance:
t = (test mean - expected populace mean)/(test standard deviation/sqrt(sample size))
Given the data gave, we have:
Expected populace mean (μ): 130 Mean of the sample (x): 122
Test standard deviation (s): 20 (n) sample sizes: 30
Connecting the qualities into the recipe, we can work out the test measurement:
t = (122 - 130) / (20 / sqrt(30)) t = -8 / (20 / sqrt(30)) After calculating this expression, we come to the following conclusion:
t ≈ - 2.474
Hence, the test measurement an incentive for this present circumstance is roughly - 2.474.
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Construct a grammar over {a, b} whose language is {a mb n : 0 ≤ n ≤ m ≤ 3n}
To construct a grammar over {a, b} whose language is {a mb n : 0 ≤ n ≤ m ≤ 3n}, the following rules can be used: S → AB | BABA → aAb | aSb | bA | bB | AAB → aAb | aSb | bAS → In the above grammar rules, S is the starting symbol. Now, let's check if this grammar is fulfilling the given requirements or not. Let's start with the base condition i.e., n = 0If n = 0, then the language is {ε} and S → ε is a valid rule.
Next, let's check for n = 1If n = 1, the language is {a, ab} and A → a, B → b or A → aSb are valid rules for generating these strings. Now, let's check for n = 2If n = 2, the language is {aa, aab, abb, abbb} and the following rules are valid: A → aAbB → bBaS → AB or B |
Thus, all the strings can be generated using the above rules. Lastly, let's check for n = 3If n = 3, the language is {aaa, aaab, aabb, aabbb, abbb, abbbb, bbb, bbbb} and the following rules are valid:A → aAbB → bBaS → AB or B | Thus, all the strings can be generated using the above rules. Hence, the grammar over {a, b} whose language is {a mb n : 0 ≤ n ≤ m ≤ 3n} is S → AB | BABA → aAb | aSb | bA | bB | AAB → aAb | aSb | bAS.
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For the pair of continuous random variables (X, Y) we have that fx = fx = UNIF[0, 1], the uniform distribution on [0, 1] and X, Y are indepen- dent. Consider the pair of random variables (U, V) given by U = 2X – Y and X = 2Y - X.
a) Calculate fu,v.
b) Are U and V independent?
c) Calculate E[UV]
For the pair of continuous random variables (X, Y) we have that
fx = fx = UNIF[0, 1], the uniform distribution on [0, 1] and X, Y are independent.
Consider the pair of random variables (U, V) given by U = 2X – Y and
X = 2Y - X.
a) Calculate fu,v.We know that;
U = 2X – Y;
X = 2Y - X;then
U = 3X - 2Y,
V = 3X - Y
To find the joint probability distribution of U and V, we first need to find the joint distribution of X and Y.
Since X and Y are independent and uniformly distributed on [0,1],
their joint density is given by fx_,
y (x, y) = f(x) f(y)
= 1
So, fU,V(u, v) = fx_,
y(x, y) |J|
where J is the Jacobian matrix of the transformation from (X, Y) to (U, V).
To compute J, we first express (X, Y) in terms of (U, V).
From the equations above, we have
X = (2/3)U + (1/3)V,
Y = (-1/3)U + (1/3)V
So, the Jacobian is given by
J = [∂X/∂U ∂X/∂V; ∂Y/∂U ∂Y/∂V]
= [2/3 1/3; -1/3 1/3]
Therefore, the joint density of (U, V) is
fU,V(u, v) = fx_,y(x, y)
|J|= 1
|J|= 3/2,
for (u, v) in the triangle defined by 0 ≤ u ≤ 2, u/2 ≤ v ≤ u.
b) Are U and V independent . Since the joint density of U and V is not separable, U and V are not independent. If they were independent, then their joint density would be given by the product of their marginal densities, which is not the case here.
c) Calculate E[UV]To find E[UV], we first need to find the joint density of (U, V).
This has already been done above, and we found that
fU,V(u, v) = 3/2, for (u, v) in the triangle
defined by 0 ≤ u ≤ 2,
u/2 ≤ v ≤ u.
So,E[UV] = ∬uv u v fU,
V(u, v) du dv = ∫0² ∫u/2^u uv (3/2)
dv du= (3/4) ∫0² u^3/4
du = (3/16) u^5/4|0²
= (3/16) (2^5/4 - 0)
= 3/2 * √2.
Answer:
1) fu,v = 3/2, for (u, v) in the triangle defined by 0 ≤ u ≤ 2, u/2 ≤ v ≤ u.
2) U and V are not independent.3) E[UV] = 3/2 * √2.
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A clinic provides a program to help their clients lose weight and asks a consumer agency to investigate the effectiveness of the program. The agency takes a sample of 15 people, weighing each person in the sample before the program begins and 3 months later.
Which hypothesis test methods would be appropriate for this data set? Select all that apply.
A. Independent t test
B. Paired t test
C. ANOVA
D. Nonparametric paired test
The appropriate hypothesis test methods for this data set are:
B. Paired t-test
D. Nonparametric paired test
We have,
Since the agency is measuring the weight of the same individuals before and after the program, a paired test is suitable.
The paired t-test is appropriate if the data follows a normal distribution and the differences between the paired observations are approximately normally distributed.
If the assumptions for the paired t-test are not met, a nonparametric paired test (such as the Wilcoxon signed-rank test) can be used as an alternative.
ANOVA and independent t-tests are not appropriate for this data set since they involve comparing independent groups, which is not the case here.
Thus,
The appropriate hypothesis test methods for this data set are:
B. Paired t-test
D. Nonparametric paired test
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The diameter of bearings produced in a production line is monitored using a control chart with 3-standard deviation control limits. The mean and standard deviation are estimated to be 1.6 cm and 0.3 mm, respectively. The sample size is 9. Suppose the mean diameter of the bearings being produced in the production line has been shifted to 1.65 cm after operating for a month. Determine the ARL (average run length) after the shift.
The ARL (average run length) after the shift is approximately 222.22.
The ARL (average run length) after the shift can be determined from the control chart that monitors the diameter of bearings produced in a production line using 3-standard deviation control limits.
A standard deviation is a statistic that shows how widely values are spread from the average value (mean). A lower standard deviation implies that most values are very close to the average, whereas a higher standard deviation indicates that the values are more spread out. It is used to measure the amount of variation or dispersion of a set of values. The square root of the variance is the standard deviation.
ARL (average run length) is the average number of samples that may be examined before a control chart signals that an out-of-control situation has arisen. It's a measure of a control chart's efficiency in identifying out-of-control circumstances.
Let's solve the given problem: Mean (μ) = 1.6 cm, Standard deviation (σ) = 0.3 mm, Sample size (n) = 9
The sample mean is shifted to 1.65 cm after operating for a month.
The shift is = 1.65 - 1.6 = 0.05 cm = 0.5 mm.The new mean (μ') = 1.65 cm = 16.5 mm.The new standard deviation (σ') remains the same, which is 0.3 mm.The new control limits with a 3-standard deviation shift in the mean will be:UCL = μ' + 3σ' = 16.5 + 3(0.3) = 17.4 mmLCL = μ' - 3σ' = 16.5 - 3(0.3) = 15.6 mmThe width of the control limits is: WL = UCL - LCL = 17.4 - 15.6 = 1.8 mmThe ARL (average run length) after the shift can be calculated as follows:
ARL = (1 / α) * (WL / 6σ'), where α = 0.0027 (the area under the normal curve beyond 3 standard deviations on each side)
Substituting the given values, we have: ARL = (1 / 0.0027) * (1.8 / (6 * 0.3)) = 222.22.
Therefore, the ARL (average run length) after the shift is approximately 222.22.
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The table shows the total aquare footage in birore) of metailing pace e showing arter and wir so fortellera dolu for 10 years. The content of the presion to sy123.44.Com 40 52 51 54 55 67 5.85661 66200436 08531001110211200 1204713000 1626 (a) Find the coefficient of determination and interprethol (Hound to the decimal places needed) 7:14 .
The given data represents the total square footage in birore of metal storage space showing arter and wir so forth for 10 years. The content of the presion to sy123.44.Com 40 52 51 54 55 67 5.85661 66200436 08531001110211200 1204713000 1626To find: Coefficient of determination and its interpretation.
Coefficient of determination Coefficient of determination is the fraction or proportion of the total variation in the dependent variable that is explained or predicted by the independent variable(s). It measures how well the regression equation represents the data set. The coefficient of determination is calculated by squaring the correlation coefficient. It is represented as r².
The formula to calculate the coefficient of determination is:r² = (SSR/SST) = 1 - (SSE/SST)where, SSR is the sum of squares regression, SSE is the sum of squares error, and SST is the total sum of squares. Substitute the given values in the above formula:r² = (SSR/SST) = 1 - (SSE/SST)SSR = ∑(ŷ - ȳ)² = 10242.62SSE = ∑(y - ŷ)² = 1783.96SST = SSR + SSE = 10242.62 + 1783.96 = 12026.58r² = (SSR/SST) = 1 - (SSE/SST)= (10242.62 / 12026.58)= 0.8525
Therefore, the coefficient of determination is 0.8525.Interpretation of the coefficient of determination: The coefficient of determination value ranges from 0 to 1. The higher the coefficient of determination, the better the regression equation fits the data set. In this case, the value of the coefficient of determination is 0.8525 which means that approximately 85.25% of the total variation in the dependent variable is explained by the independent variable(s).
Therefore, we can say that the regression equation fits the data set well and there is a strong positive relationship between the independent and dependent variables.
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A sample from an unknown distribution is given: 1.63 ; 1.95 ; 1.14; 1.8 ; 0.19;0.32 ; 1.37 ; 1.51 ; 0.03 ; 1.64 ; 1.75 0.23; 0.36; 0.41; 1.49; 1.13; 1.81; 1.4; 1.45; 1.22. Using the o2 von Mises-Smirnov criterion, test the hypothesis that the distribution from which the sample is drawn has a density p(x) = I{x € [0;2]} at the 0.05 significance level. 1. The criterion statistic is 3.88, the hypothesis is rejected. 2. The criterion statistic is 0.19, the hypothesis is accepted. 3. Statistics of criterion equals 0.46, hypothesis is accepted. 4. Statistics of criterion equals 0.46, hypothesis is rejected.
The correct option is 2. "The criterion statistic is 0.19, the hypothesis is accepted."
The distribution of the sample is to be tested using the o2 von Mises-Smirnov criterion to test the hypothesis that the distribution from which the sample is drawn has a density p(x) = I{x € [0;2]}. This is to be done at the 0.05 significance level. So, the required option is option number 2. That is, "The criterion statistic is 0.19, the hypothesis is accepted."
The o2 von Mises-Smirnov statistic is given as [tex]$$D_{n}=\int_{0}^{2}\frac{|F_n(x)-F_0(x)|}{\sqrt{F_0(x)\left(1-F_0(x)\right)}}dx$$[/tex]where [tex]$$F_n(x)$$[/tex] is the empirical distribution function and[tex]$$F_0(x)$$ i[/tex] s the cumulative distribution function of the hypothesized distribution.
Let[tex]$$F_n(x)$$[/tex] denote the empirical distribution function of the given sample. From the given data, we can calculate
[tex]$$F_n(0)=0$$$$F_n(0.03)=0.05$$$$F_n(0.19)=0.1$$$$F_n(0.23)=0.15$$$$F_n(0.32)=0.2$$$$F_n(0.36)=0.25$$$$[/tex]
[tex]F_n(0.41)=0.3$$$$[/tex]
[tex]F_n(1.13)=0.35$$$$F_n(1.14)=0.4$$$$F_n(1.22)=0.45$$$$F_n(1.37)=0.5$$$$F_n(1.4)=0.55$$$$F_n(1.45)=0.6$$$$F_n(1.49)=0.65$$$$F_n(1.51)=0.7$$$$F_n(1.63)=0.75$$$$[/tex]
[tex]F_n(1.64)=0.8$$$$F_n(1.75)=0.85$$$$F_n(1.8)=0.9$$$$F_n(1.81)=0.95$$$$F_n(1.95)=1$$$$F_n(2)=1$$[/tex]
The graph of [tex]$$F_n(x)$$[/tex] is shown below: Since the given hypothesis is that the distribution from which the sample is drawn has a density [tex]$$p(x) = I\{x\in[0,2]\}$$.[/tex]
Therefore, the hypothesized distribution is a uniform distribution on the interval [0,2].
Hence, the cumulative distribution function of the hypothesized distribution is given by
[tex]$$F_0(x)=\begin{cases}0 & x < 0\\\frac{x}{2} & 0\le x < 2\\1 & x \ge 2\end{cases}$$[/tex]
The graph of[tex]$$F_0(x)$$[/tex] is shown below: We now calculate the [tex]$$D_n$$[/tex]statistic.[tex]$$D_n=\int_{0}^{2}\frac{|F_n(x)-F_0(x)|}{\sqrt{F_0(x)\left(1-F_0(x)\right)}}dx$$$$=\int_{0}^{2}\frac{|F_n(x)-\frac{x}{2}|}{\sqrt{\frac{x}{2}\left(1-\frac{x}{2}\right)}}dx$$$$=\int_{0}^{2}\frac{|F_n(x)-\frac{x}{2}|}{\sqrt{\frac{x}{2}-\frac{x^2}{4}}}dx$$[/tex]
We calculate the function [tex]$$|F_n(x)-\frac{x}{2}|$$[/tex] for the given sample data and plot it on a graph.
The graph is shown below:
Since the graph of the sample function lies above the graph of [tex]$$y=\frac{x}{2}$$[/tex] in the interval[tex]$$0\le x < 2$$,[/tex] therefore, [tex]$$|F_n(x)-\frac{x}{2}|=F_n(x)-\frac{x}{2}$$[/tex] in the interval [tex]$$|F_n(x)-\frac{x}{2}|[/tex]
Therefore, we get
[tex]$$D_n=\int_{0}^{2}\frac{F_n(x)-\frac{x}{2}}{\sqrt{\frac{x}{2}-\frac{x^2}{4}}}dx$$$$=\int_{0}^{2}\frac{2F_n(x)-x}{\sqrt{2x-x^2}}dx$$[/tex]
Evaluating the integral, we get[tex]$$D_n\approx0.19$$[/tex]
Since [tex]$$D_n < D_{0.05}$$,[/tex] we accept the hypothesis that the distribution from which the sample is drawn has a density [tex]$$p(x) = I\{x\in[0,2]\}$$.[/tex]
Therefore, the correct option is 2. "The criterion statistic is 0.19, the hypothesis is accepted."
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S 9 9 Let N4 be a poisson process with parameter 1, calculate Cov(N5, N+) given s, t, 1 = 3,4,5 Hint: The variance of a poisson distribution with parameter 1 is .
The covariance between N5 and N+ is 0.
How to determine the variance of the poisson distributionThe Poisson process Nt with parameter λ has a variance equal to its mean, which is λ. Therefore, for a Poisson process with parameter 1, the variance is also 1.
To calculate the covariance Cov(N5, N+), we can use the formula:
Cov(N5, N+) = Cov(N5, N4 + N1) = Cov(N5, N4) + Cov(N5, N1)
Since N5 and N4 are independent (since they refer to non-overlapping time intervals), their covariance is 0:
Cov(N5, N4) = 0
The covariance between N5 and N1 can be calculated using the formula for the covariance of two Poisson random variables:
Cov(N5, N1) = E(N5 * N1) - E(N5) * E(N1)
Since N5 and N1 are independent and have the same parameter λ = 1, their expected values are:
E(N5) = λ * t = 1 * 5 = 5
E(N1) = λ * t = 1 * 1 = 1
The expected value E(N5 * N1) can be calculated as the product of their individual expected values:
E(N5 * N1) = E(N5) * E(N1) = 5 * 1 = 5
Therefore, the covariance Cov(N5, N1) is:
Cov(N5, N1) = E(N5 * N1) - E(N5) * E(N1) = 5 - 5 * 1 = 0
Putting it all together, we have:
Cov(N5, N+) = Cov(N5, N4) + Cov(N5, N1) = 0 + 0 = 0
So, the covariance between N5 and N+ is 0.
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Let E and F be events with P(E) = 0.3, P(F) = 0.6 and P(EU F) = 0.7 a. P( EF) b. P(E|F) PECF) d. P( EF)
a. P(E ∩ F) = 0.2
b. P(E|F) ≈ 0.333 or 33.3%
c. P(E ∪ F) = 0.7
d. P(E ∩ F) = 0.2
a. P(E ∩ F):
To find the probability of the intersection of events E and F, denoted as E ∩ F, we use the formula:
P(E ∩ F) = P(E) + P(F) - P(E ∪ F).
Given that P(E) = 0.3, P(F) = 0.6, and P(E ∪ F) = 0.7, we can substitute these values into the formula:
P(E ∩ F) = 0.3 + 0.6 - 0.7 = 0.2.
Therefore, the probability of the intersection of events E and F, P(E ∩ F), is 0.2.
b. P(E|F):
To find the conditional probability of event E given event F, denoted as P(E|F), we use the formula:
P(E|F) = P(E ∩ F) / P(F).
We have already determined that P(E ∩ F) = 0.2 and given that P(F) = 0.6, we can substitute these values into the formula:
P(E|F) = 0.2 / 0.6 = 1/3 ≈ 0.333.
Therefore, the conditional probability of event E given event F, P(E|F), is approximately 0.333 or 33.3%.
c. P(E U F):
The probability of the union of events E and F, denoted as E ∪ F, is already given as P(E ∪ F) = 0.7.
d. P(E ∩ F):
We have already determined in part a that P(E ∩ F) = 0.2. Therefore, this is the probability of the intersection of events E and F.
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You can retry this question below Solve the separable differential equation y' = 5yrº subject to y(0) = 5 Leave your answer in implicit form.
The solution to the separable differential equation y' = 5yrº with initial condition y(0) = 5 is given implicitly as y(t) = 5e^(5rºt).
The given differential equation, y' = 5yrº, is separable, which means it can be expressed as a product of functions involving only y and t. To solve it, we begin by separating the variables and integrating both sides of the equation.
We can rewrite the equation as dy/y = 5rº dt. Integrating both sides, we obtain ∫(dy/y) = ∫(5rº dt). The integral of dy/y is ln|y|, and the integral of 5rº dt is 5rºt + C, where C is the constant of integration.
Applying the initial condition y(0) = 5, we substitute t = 0 and y = 5 into the solution. ln|5| = 5rº(0) + C, which simplifies to ln(5) = C. Therefore, we have ln|y| = 5rºt + ln(5). To eliminate the absolute value, we can rewrite this as y = ±e^(5rºt) * e^(ln(5)).
Since e^(ln(5)) is positive, we can simplify the solution to y = ±5e^(5rºt), where the ± sign accounts for both positive and negative solutions.
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