(a) The CDF of Y is F_Y(y) = [F(y)]^n, where F(y) is the CDF of the individual random variables X1, X2, ..., Xn.
(b) W = Y/θ is a pivotal quantity, as the distribution of W does not depend on θ.
(c) P(W < c) = 1 - α can be set by finding c using the inverse CDF of the underlying distribution, where θc = F^(-1)((1 - α)^(1/n)).
(a) To find the cumulative distribution function (CDF) of Y, we need to determine the probability P(Y ≤ y) for a given value y.
For Y = max(X1, X2, ..., Xn), where X1, X2, ..., Xn are independent and identically distributed random variables with CDF F(x), we have:
P(Y ≤ y) = P(max(X1, X2, ..., Xn) ≤ y)
Since the maximum value of a set of random variables occurs when all the variables are less than or equal to y, we can express this probability as the product of the individual probabilities:
P(Y ≤ y) = P(X1 ≤ y, X2 ≤ y, ..., Xn ≤ y)
Since X1, X2, ..., Xn are independent, we can rewrite this as:
P(Y ≤ y) = P(X1 ≤ y) * P(X2 ≤ y) * ... * P(Xn ≤ y)
Since each Xi has the same CDF F(x), we can substitute F(x) into the expression:
P(Y ≤ y) = [F(y)]^n
Therefore, the CDF of Y is given by:
F_Y(y) = [F(y)]^n
(b) To show that W = Y/θ is a pivotal quantity, we need to demonstrate that the distribution of W does not depend on θ.
We know that Y follows the distribution defined in part (a) with CDF F_Y(y) = [F(y)]^n.
Now, let's consider W = Y/θ. The CDF of W can be expressed as:
F_W(w) = P(W ≤ w) = P(Y/θ ≤ w) = P(Y ≤ θw)
Using the CDF of Y derived in part (a), we can substitute it into the expression:
F_W(w) = P(Y ≤ θw) = [F(θw)]^n
Since F(x) is the CDF of the underlying distribution and does not involve θ, we can see that the distribution of W does not depend on θ. Hence, W is a pivotal quantity.
(c) To set P(W < c) = 1 - α, where α is the desired level of confidence, we need to find the appropriate value of c.
Using the CDF of W derived in part (b), we can write the equation as:
[F(θc)]^n = 1 - α
Taking the inverse CDF (quantile function) on both sides, we get:
F(θc) = (1 - α)^(1/n)
Finally, solving for c, we have:
θc = F^(-1)((1 - α)^(1/n))
Therefore, P(W < c) = 1 - α can be set by finding the appropriate value of c using the inverse CDF of the underlying distribution.
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: For the following exercise, write a formula for the function that results when the graph of a given toolkit function is transformed as described. 1 The graph of f(x) is vertically compressed by a factor of units, then shifted to the left 3 units and down 6 units. g(x)= Preview TIP
The formula for the function g(x) is: g(x) = (1/a) * f(x + 3) - 6. To obtain the formula for the function g(x) resulting from the described transformations:
We start with the toolkit function f(x) and apply the given operations step by step.
Vertical compression by a factor of a units:
If the graph of f(x) is vertically compressed by a factor of a, we can achieve this by multiplying the function by 1/a. So, g(x) = (1/a) * f(x).
Shift to the left by 3 units:
To shift the graph of f(x) to the left by 3 units, we replace x with (x + 3) in the function. Therefore, g(x) = (1/a) * f(x + 3).
Shift down by 6 units:
To shift the graph of f(x) down by 6 units, we subtract 6 from the function. Thus, g(x) = (1/a) * f(x + 3) - 6.
Combining these transformations, the formula for the function g(x) is:
g(x) = (1/a) * f(x + 3) - 6.
Note that the specific form of the function f(x) would depend on the given toolkit function.
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Use Stokes' Theorem to evaluate F. dr where F(x, y, z) =xy i+xzj+x² k and C is the rhombus of with corners at (0, 0, 0), (5, 0, 5), (5, 5, 10), and (0,5,5) and has positive orientation when viewed from above.
We sum up the two integrals over triangles T1 and T2 to obtain the value of the surface integral ∫∫S -z dA. This will be equal to the line integral ∫C F · dr by Stokes' Theorem.
In this problem, we are asked to use Stokes' Theorem to evaluate the line integral ∫F · dr, where F(x, y, z) = xy i + xz j + x² k, and C is the rhombus with corners at (0, 0, 0), (5, 0, 5), (5, 5, 10), and (0, 5, 5), and it has positive orientation when viewed from above.
Stokes' Theorem relates a surface integral to a line integral around the boundary curve of the surface. It states that for a vector field F and a surface S with a positively oriented boundary curve C,
∫∫S (curl F) · dS = ∫C F · dr
To apply Stokes' Theorem, we need to calculate the curl of F and then evaluate the line integral ∫C F · dr.
Step 1: Calculate the curl of F:
The curl of F is given by the cross product of the gradient operator and F:
curl F = (∂F₃/∂y - ∂F₂/∂z) i + (∂F₁/∂z - ∂F₃/∂x) j + (∂F₂/∂x - ∂F₁/∂y) k
In this case, we have F(x, y, z) = xy i + xz j + x² k, so we can substitute the components of F into the curl formula:
curl F = (0 - z) i + (1 - 0) j + (y - y) k
= -z i + j
Step 2: Determine the orientation of the boundary curve C:
The rhombus C has corners at (0, 0, 0), (5, 0, 5), (5, 5, 10), and (0, 5, 5), and it has positive orientation when viewed from above. From the given corners, we can see that the boundary curve C is formed by the four line segments connecting these points.
Step 3: Evaluate the line integral ∫C F · dr:
Using Stokes' Theorem, we can rewrite the line integral ∫C F · dr as the surface integral ∫∫S (curl F) · dS, where S is a surface bounded by the boundary curve C.
Since the boundary curve C is planar and lies on the xy-plane, we can choose a surface S that lies in the xy-plane and has C as its boundary curve. One possible choice for S is the region enclosed by the rhombus C.
To evaluate the surface integral, we need to calculate the area vector dS, which is perpendicular to the surface S and has magnitude equal to the area of an infinitesimal element of S.
For a surface in the xy-plane, the area vector dS is in the positive z-direction, so dS = k dA, where dA is the area element in the xy-plane.
Since the boundary curve C is a rhombus, we can divide it into two triangles and calculate the area of each triangle using the formula A = (1/2) base × height.
Let's label the triangles as T1 and T2. Triangle T1 has vertices (0, 0, 0), (5, 0, 5), and (5, 5, 10), and triangle T2 has vertices (0, 0, 0), (5, 5, 10), and (0, 5, 5).
By calculating the area of each triangle, we can determine the total area of S.
Once we have the area vector dS, we can evaluate the surface integral ∫∫S (curl F) · dS. Since the curl of F is -z i + j, and dS = k dA, the surface integral simplifies to ∫∫S (-z i + j) · (k dA).
Since the area vector dS points in the positive z-direction, we have (-z i + j) · (k dA) = -z dA.
Therefore, the surface integral becomes ∫∫S -z dA.
Step 4: Calculate the surface integral:
To evaluate the surface integral ∫∫S -z dA, we integrate -z over the surface S.
Since S is the region enclosed by the rhombus C, we can express the integral as the sum of the integrals over triangles T1 and T2:
∫∫S -z dA = ∫∫T1 -z dA + ∫∫T2 -z dA
We can evaluate each integral separately by integrating -z over each triangle using the appropriate limits based on the vertices of the triangles.
Step 5: Evaluate the line integral:
Finally, we sum up the two integrals over triangles T1 and T2 to obtain the value of the surface integral ∫∫S -z dA. This will be equal to the line integral ∫C F · dr by Stokes' Theorem.
By evaluating the line integral ∫C F · dr, we obtain the desired result.
Note: To provide the specific numerical values of the integrals, the vertices of the triangles and the corresponding limits of integration need to be given.
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According to Glassdoor.com c7, the average salary of a registered nurse in the SF Bay Area (as of December 2021 ) is $114,327. Assume the annual salaries are normally distributed with a mean of $114,327 and a standard deviation of $14,000. If 100 registered nurses from the SF Bay Area are randomly selected, find the probability that their mean salary is greater than $120,000.
The task is to find the probability that the mean salary of 100 randomly selected registered nurses from the SF Bay Area is greater than $120,000. The salaries are assumed to be normally distributed with a mean of $114,327 and a standard deviation of $14,000.
To solve this problem, we can use the Central Limit Theorem, which states that the distribution of sample means approaches a normal distribution as the sample size increases. Given that the sample size is 100, we can assume that the distribution of sample means will be approximately normal.
First, we need to calculate the standard error of the mean (SE), which is the standard deviation of the population divided by the square root of the sample size. In this case, SE = $14,000 / sqrt(100) = $1,400.
Next, we can calculate the z-score corresponding to the value $120,000 using the formula z = (x - μ) / SE, where x is the value, μ is the mean, and SE is the standard error. Plugging in the values, we get z = (120,000 - 114,327) / 1,400 = 4.05.
Finally, we can find the probability that the mean salary is greater than $120,000 by calculating the area under the standard normal curve to the right of the z-score of 4.05. This can be done using a standard normal distribution table or a statistical software. The resulting probability is the desired answer.
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Consider the function g:R→R defined by
g(x)=(∫0sin(x)e^(sin(t))dt)^2.
Find g′(x) and determine the values of x for which g′(x)=0
Hint: e^x≥0for all x∈R
We have found the expression for g′(x) and determined that g′(x) = 0 when ∫0sin(x)e^(sin(t))dt = 0.
To find g′(x), the derivative of the function g(x), we can apply the chain rule and the fundamental theorem of calculus. By differentiating the integral with respect to x, we obtain g′(x) in terms of the integrand and its derivative. To determine the values of x for which g′(x) = 0, we set g′(x) equal to zero and solve for x.
The function g(x) is defined as g(x) = (∫0sin(x)e^(sin(t))dt)^2. To find g′(x), the derivative of g(x), we will apply the chain rule and the fundamental theorem of calculus.
By the chain rule, if F(x) is an antiderivative of the integrand sin(x)e^(sin(t)), then g′(x) can be expressed as:
g′(x) = [2(∫0sin(x)e^(sin(t))dt)] * [∫0sin(x)e^(sin(t))dt]',
where [∫0sin(x)e^(sin(t))dt]' denotes the derivative of the integral with respect to x.
Now, let's focus on finding the derivative of the integral [∫0sin(x)e^(sin(t))dt].
Applying the fundamental theorem of calculus, we have:
[∫0sin(x)e^(sin(t))dt]' = sin(x)e^(sin(x)) - 0e^(sin(0)) = sin(x)e^(sin(x)).
Substituting this result into our expression for g′(x), we get:
g′(x) = [2(∫0sin(x)e^(sin(t))dt)] * [sin(x)e^(sin(x))].
Simplifying further, we have:
g′(x) = 2sin(x)e^(sin(x)) * ∫0sin(x)e^(sin(t))dt.
To determine the values of x for which g′(x) = 0, we set g′(x) equal to zero:
2sin(x)e^(sin(x)) * ∫0sin(x)e^(sin(t))dt = 0.
Since e^x ≥ 0 for all x ∈ R (as given in the hint), the term 2sin(x)e^(sin(x)) cannot equal zero. Therefore, the only way for g′(x) to be zero is when ∫0sin(x)e^(sin(t))dt = 0.
To find the values of x for which ∫0sin(x)e^(sin(t))dt = 0, we need to solve the integral equation separately. Unfortunately, solving this equation analytically may not be possible as it involves an integral with a variable limit.
In summary, we have found the expression for g′(x) and determined that g′(x) = 0 when ∫0sin(x)e^(sin(t))dt = 0. However, we cannot explicitly determine the values of x for which g′(x) = 0 without further analysis or approximation techniques.
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The value of the integral fedz, C is the circle counterdockwise 141 172 A. 0 B. 2nl C. - D. -2m E. None of them. The function f(z) is analytic A. Everywhere B. Everywhere except at x 0 C. Everywhere except atz-Znal D. Everywhere except at z = nxi E. None of them.
The value of the integral ∫f(z)dz along the circle C depends on whether the function f(z) is analytic and where it has singularities. The options provided are:
A. 0
B. 2πi
C. -
D. -2πi
E. None of them
The options for the function f(z) being analytic are:
A. Everywhere
B. Everywhere except at z = 0
C. Everywhere except at z = znal
D. Everywhere except at z = nxi
E. None of them
To determine the value of the integral ∫f(z)dz along the circle C, we need to know if the function f(z) is analytic and whether it has singularities within the region enclosed by C.
If the function f(z) is analytic everywhere inside the circle C and has no singularities, then the integral will evaluate to 0 (option A).
If the function f(z) is analytic everywhere except at z = 0, the integral will evaluate to 2πi times the winding number of C around the singularity at z = 0. Since no winding number or specific information about the singularity is given, we cannot determine the value to be exactly 2πi (option B).
Options C, D, and E are not valid based on the given information.
Regarding the options for the function f(z) being analytic, we cannot determine the exact condition based on the information provided. Additional information about the function and its singularities is needed to make a specific determination.
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a similar question You can retry this question below 320 over the interval 0
The integral of the function below 320 over the interval 0 can be calculated using the step-by-step process of evaluating the definite integral.
Step 1: Set up the integral:
We want to find the integral of a function below 320 over the interval from 0. Let's denote the function as f(x). The integral can be set up as follows:
∫[0, a] f(x) dx
Step 2: Evaluate the integral:
To evaluate the integral, we need to determine the antiderivative of the function f(x). Once we have the antiderivative, we can use the fundamental theorem of calculus to find the value of the definite integral.
Step 3: Find the antiderivative:
Let's assume the function f(x) is known. Determine its antiderivative, denoted as F(x). The antiderivative is found by integrating f(x) with respect to x.
Step 4: Apply the fundamental theorem of calculus:
The fundamental theorem of calculus states that the definite integral of a function f(x) over an interval [a, b] can be found by subtracting the antiderivative evaluated at the lower bound from the antiderivative evaluated at the upper bound:
∫[0, a] f(x) dx = F(a) - F(0)
Step 5: Calculate the integral:
Substitute the upper bound (a) and the lower bound (0) into the antiderivative F(x) obtained in Step 3. Evaluate F(a) and F(0), and subtract the latter from the former to find the value of the definite integral.
By following these steps, you can calculate the integral of the function below 320 over the interval 0.
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Use differentials to approximate √24.6. b Express your answer as a t where a is the nearest integer. L C Answer √24.6
To approximate √24.6 using differentials, we can use the formula for differentials: df = f'(x) * dx.
In this case, f(x) = √x, so f'(x) is the derivative of √x, which is 1/(2√x). Let's consider x = 25 as the nearest perfect square to 24.6. Using differentials, we have: df = (1/(2√x)) * dx. Substituting x = 24.6 and dx = 24.6 - 25 = -0.4, we get: df = (1/(2√24.6)) * (-0.4). Simplifying, we have: df ≈ -0.2/√24.6. Now, we can approximate √24.6 as: √24.6 ≈ 5 + df ≈ 5 - 0.2/√24.6, hence a = 5 .
This gives us the approximation for √24.6 as 5 - 0.2/√24.6, where a = 5 is the nearest integer.
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5x1 + 2x2 ≥ 145. If we produce 15 units of x2 how many units of x1 need to be made to satisfy the equation?
To satisfy the equation 5x1 + 2x2 ≥ 145, producing 15 units of x2 would require making at least [rounded value] units of x1, based on the given equation and condition.
The given inequality is 5x1 + 2x2 ≥ 145, where x1 and x2 represent variables. We are given that 15 units of x2 will be produced. To determine the number of units of x1 needed to satisfy the equation, we substitute the value of x2 as 15 in the inequality.
Substituting x2 as 15, we have 5x1 + 2(15) ≥ 145. Simplifying further, we get 5x1 + 30 ≥ 145. Next, we subtract 30 from both sides of the inequality, resulting in 5x1 ≥ 115.
To find the number of units of x1 needed, we divide both sides of the inequality by 5. This gives us x1 ≥ 23. Therefore, to satisfy the equation and produce 15 units of x2, we need to make at least 23 units of x1.
In summary, if 15 units of x2 are produced, at least 23 units of x1 need to be made to fulfill the equation 5x1 + 2x2 ≥ 145, based on the calculation and substitution of values in the given inequality.
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We didn’t discuss the role of sample size in this chapter as thoroughly as we did for confidence intervals in the previous chapter, but more advanced books do include sample size formulas for hypothesis testing. Consider the situation where you are testing the null hypothesis that a population mean is less than or equal to 100 versus a one-tailed alternative. A sample size formula might indicate the sample size needed to make the power at least 0.90 when the true mean is 103. What are the trade-offs here? Essentially, what is the advantage of a larger sample size
The advantage of a larger sample size in hypothesis testing is that it increases the power of the test.
Power refers to the ability of the test to correctly reject the null hypothesis when it is false. With a larger sample size, the test has a higher probability of detecting a true difference or effect, making it more sensitive. This reduces the likelihood of a Type II error (failing to reject the null hypothesis when it is false) and increases the reliability of the test results. Additionally, a larger sample size allows for more precise estimation of population parameters and reduces the impact of sampling variability. However, a larger sample size may also increase the cost and time required to collect and analyze the data.
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Find the derivative. 12) f(x) = In (e4x + 6)
The derivative of the function f(x) = ln(e^(4x) + 6), is f'(x) = 4e^(4x) / (e^(4x) + 6) To find the derivative of the function f(x) = ln(e^(4x) + 6), we can use the chain rule.
The chain rule states that if we have a composition of functions, such as f(g(x)), then the derivative is given by the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function.
In this case, the outer function is ln(x) and the inner function is e^(4x) + 6.
To find the derivative, we can follow these steps:
1. Find the derivative of the outer function:
The derivative of ln(x) with respect to x is 1/x.
2. Find the derivative of the inner function:
The derivative of e^(4x) + 6 with respect to x is 4e^(4x).
3. Apply the chain rule:
Multiply the derivative of the outer function (1/x) by the derivative of the inner function (4e^(4x)).
So, the derivative of f(x) = ln(e^(4x) + 6) is:
f'(x) = (1 / (e^(4x) + 6)) * (4e^(4x))
Simplifying further, we can write:
f'(x) = 4e^(4x) / (e^(4x) + 6)
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(c) What is the probability that one random sample of 100 U.S. births in 2014 would have a mean birth order lower than 2.0? (Enter your answer rounded to four decimal places.) P(x < 2.0) = Incorrect
The probability that a random sample of 100 U.S. births in 2014 would have a mean birth order lower than 2.0 is determined by calculating the z-score and referring to the standard normal distribution.
By standardizing the sample mean using the population mean and standard deviation, we can find the corresponding z-score. Using statistical tables or software, we can then determine the probability associated with that z-score. The exact probability depends on the population mean and standard deviation, as well as the assumption of a normal distribution.
To find the probability, we first need to standardize the sample mean. Assuming a normal distribution of birth orders, we can use the population mean and standard deviation as the mean and standard deviation of the sampling distribution of the sample mean, respectively. Let's assume the population mean birth order is μ = 2.5, and the population standard deviation is σ = 0.5.
To calculate the z-score, we use the formula:
z = (sample mean - population mean) / (population standard deviation / sqrt(sample size))
In this case, the sample mean is 2.0, the population mean is 2.5, the population standard deviation is 0.5, and the sample size is 100. Plugging these values into the formula, we get:
z = (2.0 - 2.5) / (0.5 / sqrt(100)) = -10
Next, we consult the standard normal distribution table or use statistical software to find the probability associated with a z-score of -10. However, such an extreme z-score is beyond the range covered by standard normal distribution tables. It indicates that the sample mean of 2.0 is significantly lower than the population mean of 2.5. The probability in this case would be extremely close to 0, indicating an extremely rare occurrence.
In summary, due to the extremely low probability associated with a z-score of -10, it is likely that the given value for the probability, P(x < 2.0), is incorrect.
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The market price of a stock is $23.95 and it is expected to pay
a dividend of $1.05 next year. The required rate of return is
11.91%. What is the expected growth rate of the dividend?
To compute the expected growth rate of a dividend, we can use the formula below: Expected growth rate of dividend = (Dividend for next year / Current dividend) - 1We are given the current dividend and the price of the stock in the problem statement
but we need to use the required rate of return as a discount rate to calculate the current dividend. We can use the dividend discount model (DDM) to find the current dividend. The DDM model is shown below:P0 = D1 / (r - g)Where:P0 = current stock priceD1 = dividend for next yearr = required rate of returng = expected growth rate of dividend Substituting the given values:P0 = 23.95D1 = 1.05r = 11.91% (convert to decimal: 0.1191)Solve for D0:D0 = D1 / (1 + g)Plugging in the values:D0 = 1.05 / (1 + g)Substitute D0 and other values into the DDM formula:23.95 = 1.05 / (0.1191 - g)Rearrange the formula to isolate g:(0.1191 - g) = 1.05 / 23.95g = 0.0811 or 8.11%The expected growth rate of the dividend is 8.11
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Derive the following wave equation in a linear, isotropic and homogeneous (LIH) medium of conductivity a, permittivity & and permeability μ: JE a²E VE-ou- 5 μ- -0. dr at² [15 marks] Hence show that if the field H-IH, exp jat-kz), the associated electric field is given by k E- -|H|} . jo-we [15 marks] (iii) In the case of a non-conducting LIH dielectric medium, the wavenumber k-weu. More realistically, a lossy dielectric may attenuate electromagnetic fields through two distinct processes. The first is associated with dielectric polarization and the second with a finite conductivity. This may be modelled by transforming the permittivity into a complex, frequency dependent quantity, such that ε = E'-je-jº Making use of this definition of & and defining jk = a + jß, show that a is an attenuation constant and derive the expression Hort 2 where &" =+=
The wave equation in a linear, isotropic, and homogeneous medium of conductivity α, permittivity ε, and permeability μ. It also involves showing the relationship between the magnetic field H and the associated electric field E in the case of plane wave propagation.
1) Deriving the wave equation: Start with Maxwell's equations and apply the appropriate relationships for a linear, isotropic, and homogeneous medium. By manipulating the equations and using vector calculus, you can derive the wave equation in terms of the electric field E. The specific steps may involve substitutions, differentiation, and rearrangement.
2) Showing the relationship between H and E: Assuming a harmonic plane wave solution of the form H = H0 exp(jωt - jkz), where H0 is the amplitude and ω and k are the angular frequency and wavenumber respectively, you can derive the associated electric field E. Apply the relationships between H, E, and the wave equation to express E in terms of H. Simplify the expression and obtain the desired relationship.
3) Considering a lossy dielectric medium: Introduce a complex, frequency-dependent permittivity ε = ε' - jε'' to account for attenuation in the dielectric medium. Define the wavenumber k as a complex number k = α + jβ, where α represents the attenuation constant. Using this definition and the relationships for the wave equation, derive the expression for the attenuation constant α. This may involve manipulating the complex permittivity and solving for α.
The specific mathematical steps and equations involved in each part will depend on the context and equations provided in the question.
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find the Z-score that separate the middle 89% of the distribution
from the area in thr tails of the standard normal
distribution
The z-score that separates the middle 89% of the distribution from the area in the tails of the standard normal distribution can be found using Find the area in the tails of the standard normal distributionSince the middle 89% of the distribution is already given, the area in the tails can be found by subtracting.
it from 1. 1 - 0.89 = 0.11So, the area in the tails is 0.11.Step 2: Divide the area in the tails by 2 to find each tail's area0.11 / 2 = 0.055Step 3: Look up the z-score that corresponds to the area 0.055 in the standard normal distribution table.The z-score for 0.055 is -1.645.
The area to the left of -1.645 in the standard normal distribution is 0.055, and the area to the right of 1.645 is also 0.055.Therefore, the z-score that separates the middle 89% of the distribution from the area in the tails of the standard normal distribution is -1.645.
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The tallest living man at one time had a height of 258 cm. The shortest living man at that time had a height of 124.4 cm. Heights of men at that time had a mean of 176.07 cm and a standard deviation of 7.32 cm. Which of these two men had the height that was more extreme? What is the z score for the tallest man? What is the z score for the shortest man?
The z-score for the tallest man is approximately 11.18, and the z-score for the shortest man is approximately -7.03.
To determine which of the two men had a more extreme height, we can compare their respective z-scores.
The z-score measures the number of standard deviations a data point is away from the mean.
To calculate the z-score, we'll use the following formula:
z = (x - μ) / σ
where:
z = z-score
x = individual value
μ = mean
σ = standard deviation
For the tallest man:
x = 258 cm
μ = 176.07 cm
σ = 7.32 cm
Calculating the z-score for the tallest man:
z(tallest) = (258 - 176.07) / 7.32
For the shortest man:
x = 124.4 cm
μ = 176.07 cm
σ = 7.32 cm
Calculating the z-score for the shortest man:
z(shortest) = (124.4 - 176.07) / 7.32
To determine which height is more extreme, we compare the absolute values of the z-scores.
The larger the absolute value, the more extreme the height is relative to the mean.
Now let's calculate the z-scores:
z(tallest) = (258 - 176.07) / 7.32 ≈ 11.18
z(shortest) = (124.4 - 176.07) / 7.32 ≈ -7.03
The absolute value of z(tallest) is 11.18, and the absolute value of z(shortest) is 7.03. Since 11.18 is larger than 7.03, the height of the tallest man is more extreme.
The z-score for the tallest man is approximately 11.18, and the z-score for the shortest man is approximately -7.03.
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A dragonologist is studying wild dragons in North West China. He hires a statistician to help him figure out the proportion of green dragons, compared to all other dragons. After surveying the land using a SRS tactic, the statistician found 15 out of 100 to be green dragons. Calculate the lower bound to the 95% confidence interval (round to 4 decimals)
To find the lower bound to the 95% confidence interval, we can use the formula:Lower Bound = p-hat - z*(√(p-hat*q-hat)/n)where:
p-hat = sample proportion of green dragons = 15/100 = 0.15q-hat = 1 - p-hat = 1 - 0.15 = 0.85n = sample size = 100z = z-score at 95% confidence level = 1.96 (from standard normal distribution table).
By plugging in the given values in the formula, we get:Lower Bound = 0.15 - 1.96*(√(0.15*0.85)/100)Lower Bound = 0.15 - 0.0775Lower Bound = 0.0725.
In this given problem, a dragonologist is studying wild dragons in North West China and he has hired a statistician to help him figure out the proportion of green dragons, compared to all other dragons. After surveying the land using a SRS tactic, the statistician found that out of 100 dragons, 15 were green dragons.
To calculate the lower bound of 95% confidence interval, we used the formula
Lower Bound = p-hat - z*(√(p-hat*q-hat)/n)
where p-hat represents the sample proportion of green dragons, q-hat represents the proportion of other dragons, n represents the sample size and z represents the z-score at 95% confidence level. Plugging in the given values in the formula, we obtained a lower bound of 0.0725 rounded off to 4 decimal places. Therefore, the lower bound to the 95% confidence interval is 0.0725.
Therefore, the lower bound to the 95% confidence interval is 0.0725.
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(6+6+6+6=24 pts ) Let X 1
,X 2
,…,X n
be a random sample from the distribution with pdf f(x;θ)=e θ−x
I (θ,[infinity])
(x) (a) Show that S=X (1)
is sufficient for θ. (b) Find the pdf for X (1)
. (c) Show that S=X (1)
is a complete statistic for estimating θ. (d) Find the UMVUE for θ.
(a) To show that S = X(1) is sufficient for θ, we need to show that the conditional distribution of the sample, given S = s and θ, depends only on S and θ.
and hence S is sufficient for θ. Therefore, we have shown that S = X(1) is sufficient for θ. (b) The pdf for X(1) is$$[tex]f_{X_{(1)}}(x
)=nf_X(x)(1-F_X(x))^{n-1}$$where f_X (x)
= e^(θ-x) I(θ,∞) (x)[/tex]. Hence,$$f_{X_{(1)}}(x
)=ne^{\theta-x}I(\theta, x)(1-e^{\theta-x})^{n-1}$$So, this is the pdf for X(1). (c) To show that S
= X(1) is a complete statistic, we need to show that if E[g(X(1))]
= 0 for all θ, then g(x) = 0 almost everywhere. Now, if we take g(x) to be such that E[g(X(1))]
= 0 for all θ, then we have$$0
=\int_{-\infty}^\infty g(x)f_{X_{(1)}}(x)dx
=n\int_\theta^\infty g(x)e^{\theta-x}(1-e^{\theta-x})^{n-1}dx$$Letting t
= e^(θ - x), dt
= -e^(θ - x) dx, we get$$0
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An experimental surgical procedure is being studied as an alternative to the existing method. Thirteen surgeons each perform the operation on two different patients matched by sex, age and other relevant factors. The time to complete the procedure for each technique is listed in the table below.
Problem Definition: Determine whether or not the new procedure is faster than the existing procedure. Alpha is .05 Utilize the six-step process to test the hypothesis the new procedure is faster the than the old procedure. One tailed test to the left.
Time to Complete
1 2 3 4 5 6 7 8 9 10 11 12 13
New Procedure
14 12 14 13 10 13 12 10 13 13 18 27 16
Old Procedure
29 32 30 39 31 27 29 33 28 26 29 25
Utilize all three statistical testing techniques in your Conclusion (critical value/critical ratio, confidence interval and pvalue. Remember to include the numbers from the Minitab output for each technique when developing your Conclusion or points will be deducted.
For Interpretation state if the new technique is significantly faster on average than the older technique.
Add a section labeled Assumption in which you will place your boxplot. Below the boxplot discuss how you know whether or not normality may be assumed. Be sure to discuss whisker lengths, skew in addition to assessing normality.
Perform a one-tailed paired t-test to compare the mean time to complete the new procedure with the mean time to complete the existing procedure and determine if the new procedure is faster.
A one-tailed paired t-test can be conducted to compare the mean time to complete the new surgical procedure with the mean time to complete the existing procedure. The null hypothesis would be that there is no difference in the mean times, while the alternative hypothesis would state that the new procedure is faster. The significance level (alpha) is set at 0.05. By analyzing the paired data from the 13 surgeons and their respective patients, the t-test can determine if there is sufficient evidence to reject the null hypothesis and conclude that the new procedure is indeed faster than the existing one.
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Explain what bivariate data is and the purpose of the linear
regression model.
Bivariate data involves two variables analyzed together to determine their relationship, while the linear regression model is used to find a straight line that best fits the data points on a scatter plot and make predictions based on that relationship.
Bivariate data refers to a set of two variables that can be analyzed together to determine the relationship between them.
For example, if we are studying the relationship between a person's height and weight, we would collect data on both of these variables for each individual in our sample.
Linear regression modeling is a statistical technique used to analyze bivariate data and determine the relationship between the two variables. The purpose of the linear regression model is to find a straight line that best fits the data points on a scatter plot. This line can then be used to make predictions about the relationship between the variables, such as predicting a person's weight based on their height.
Overall, the linear regression model is an important tool in analyzing bivariate data because it allows us to understand the relationship between two variables and make predictions based on that relationship.
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There are 500 employees in a firm, 45% are female. A sample of 60 employees is selected randomly. a. What is the probability that less than 25 females are in this sample. b. What is the probability that the sample proportion (proportion of females) is between 0.40 and 0.55?
a. The probabilities for each value of k from 0 to 24 and sum them up to get the desired probability.
b. The probability using the standard normal distribution table or a calculator P(0.40 < p < 0.55) = P(z1 < Z < z2)
To solve this problem binomial distribution to calculate probabilities related to the number of females in the sample.
Probability that less than 25 females are in the sample:
Total number of employees in the firm (N) = 500
Proportion of females in the firm (p) = 0.45
Sample size (n) = 60
To find the probability of getting less than 25 females in the sample calculate the cumulative probability for the binomial distribution.
P(X < 25) = P(X = 0) + P(X = 1) + ... + P(X = 24)
The binomial probability formula to calculate each individual probability:
P(X = k) = (n choose k) ×p²k × (1 - p)²(n - k)
Where:
(n choose k) = n! / (k! × (n - k)!)
p = proportion of females in the population
k = number of females in the sample
n = sample size
To find the probability that the sample proportion is between 0.40 and 0.55, use the normal distribution approximation.
The sample proportion follows an approximately normal distribution due to the large sample size this to approximate the probability.
calculate the mean and standard deviation of the sample proportion:
Mean (μ) = p = 0.45
Standard Deviation (σ) = sqrt((p × (1 - p)) / n) = √((0.45 × (1 - 0.45)) / 60)
use the z-score formula to standardize the values:
z1 = (0.40 - μ) / σ
z2 = (0.55 - μ) / σ
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Use De Morgan's laws to write negations for the following. (a) An integer a is divisible by 12 if it is divisible by 3 and 4. (b) No integer is both even and odd.
In part (a), we are given a statement about divisibility of an integer a by 12. Using De Morgan's laws, we can write the negation of this statement.
In part (b), we are given a statement about the evenness and oddness of integers. We can also use De Morgan's laws to write the negation of this statement.
(a) The given statement states that an integer a is divisible by 12 if it is divisible by 3 and 4. To write the negation of this statement, we can apply De Morgan's laws, which state that the negation of a conjunction (AND) is the disjunction (OR) of the negations. Therefore, the negation of the statement is: "An integer a is not divisible by 12 if it is not divisible by 3 or it is not divisible by 4."
(b) The given statement states that no integer is both even and odd. To write the negation of this statement, we can once again apply De Morgan's laws. The negation of the statement is: "There exists an integer that is both even and odd." This means that the negation states the existence of at least one integer that is simultaneously even and odd, which is a contradiction since even and odd integers are mutually exclusive.
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The median for the given set of six ordered data values is 29.5. 5 12 27 41 49 - What is the missing value? The missing value is
The missing value in the given set of ordered data values is 34. To find the missing value, we examine the set of ordered data values and focus on determining the median.
The median is the middle value when the data is arranged in ascending order. Given that the median is specified as 29.5, we can conclude that the missing value falls between the third and fourth values in the set. The third value is 27, and the fourth value is 41. Since the missing value needs to be greater than 27 and less than 41 to maintain the given median, the only value that satisfies this condition is 34. Hence, the missing value in the set is 34.
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The mean monthly mortgage paid by all home owners in a town is $2365 with a standard deviation of $340. a. Using Chebyshev's theorem, find the minimum percentage of all home owners in this town who pay a monthly mortgage of $1685 to $3045. b. Using Chebyshev's theorem, find the interval that contains the monthly mortgage payments of at least 84% of all home owners in this town.
According to Chebyshev's theorem, we can determine the minimum percentage of home owners in the town who pay a monthly mortgage between $1685 and $3045, as well as the interval that contains the monthly mortgage payments of at least 84% of all home owners.
a. Chebyshev's theorem states that for any data set, regardless of its shape, at least (1 - 1/k^2) percent of the data falls within k standard deviations of the mean, where k is any positive constant greater than 1. In this case, we want to find the percentage of home owners whose monthly mortgage falls within $1685 and $3045, which is within 1 standard deviation of the mean in either direction. The standard deviation given is $340, so the range for 1 standard deviation would be $2365 ± $340. Therefore, the minimum percentage of home owners in this town who pay a monthly mortgage of $1685 to $3045 is at least (1 - 1/1^2) = 0% according to Chebyshev's theorem.
b. To find the interval that contains the monthly mortgage payments of at least 84% of all home owners, we need to determine the range within which 84% of the data falls. Chebyshev's theorem tells us that at least (1 - 1/k^2) percent of the data lies within k standard deviations of the mean. We want to find the range within which at least 84% of the data falls, so we need to solve the inequality (1 - 1/k^2) ≥ 0.84.
Rearranging the inequality, we get 1/k^2 ≤ 0.16, which implies k^2 ≥ 6.25. Taking the square root of both sides, we find k ≥ 2.5. Thus, 84% of the data lies within 2.5 standard deviations of the mean. Multiplying the standard deviation of $340 by 2.5 gives us a range of $850. Therefore, the interval that contains the monthly mortgage payments of at least 84% of all home owners in this town is $2365 ± $850, or $1515 to $3215.
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Find parametric equations for the line through (5, 4, 4) that is perpendicular to the plane x − y + 2z = 5. (Use the parameter t.) (x(t), y(t), z(t)) = (b) In what points does this line intersect the coordinate planes? xy-plane (x, y, z) = yz-plane (x, y, z) = xz-plane (x, y, z) =
The parametric equations for the line through (5, 4, 4) that is perpendicular to the plane x - y + 2z = 5 are: x(t) = 5 + t, y(t) = 4 - t, z(t) = 4 + 2t
To find the parametric equations, we can consider the direction vector of the line, which is perpendicular to the given plane. The coefficients of x, y, and z in the plane equation x - y + 2z = 5 give us the direction vector of the line, which is (1, -1, 2).
Since we have a point on the line (5, 4, 4) and its direction vector (1, -1, 2), we can use the point-slope form of the line equation to obtain the parametric equations. Using the parameter t, we add the respective components of the direction vector to the coordinates of the given point.
Now, let's find the points of intersection with the coordinate planes:
xy-plane: To find the intersection with the xy-plane, we set z(t) = 0 and solve for t. From the parametric equations, we have 4 + 2t = 0, which gives t = -2. Substituting this value into the parametric equations, we find that the line intersects the xy-plane at the point (-1, 6, 0).
yz-plane: To find the intersection with the yz-plane, we set x(t) = 0 and solve for t. From the parametric equations, we have 5 + t = 0, which gives t = -5. Substituting this value into the parametric equations, we find that the line intersects the yz-plane at the point (0, 9, -6).
xz-plane: To find the intersection with the xz-plane, we set y(t) = 0 and solve for t. From the parametric equations, we have 4 - t = 0, which gives t = 4. Substituting this value into the parametric equations, we find that the line intersects the xz-plane at the point (9, 0, 12).
Therefore, the line intersects the xy-plane at (-1, 6, 0), the yz-plane at (0, 9, -6), and the xz-plane at (9, 0, 12).
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The shape of the distribution of the sime required to get an of through (c) ta 15-minute of-change facility is skewed right However, records indicate that the mean time is 16.4 minutes, and the standard deviation is 36 minutes Complete parts (a) (a) To compute probabilites regarding the sample mean using the normal model, what sie sample would be required? A Any sample size could be used B. The sample size needs to be less than or equal to 30. C. The normal model cannot be used if the shape of the distribution is skewed right D. The sample size needs to be greater than or equal to 30
Given that the shape of the distribution of the time required to get an office through (c) ta 15-minute change facility is skewed right. However, records indicate that the mean time is 16.4 minutes, and the standard deviation is 36 minutes.
To compute probabilities regarding the sample mean using the normal model, The normal model can be used if the sample size is greater than or equal to 30. The answer is option D.The central limit theorem states that if the sample size is large enough, the distribution of the sample means will be approximately normal, regardless of the shape of the population distribution, and the mean of the sample means is equal to the mean of the population.
The standard deviation of the sample means is equal to the standard deviation of the population divided by the square root of the sample size. When the sample size increases, the standard deviation of the sample means decreases, and the distribution of the sample means becomes more and more normal.
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Pierce Manufacturing determines that the daily revenue, in dollars, from the sale of x lawn chairs is R(x) = 0.004x³ +0.05x² +0.3x. Currently, Pierce sells 70 lawn chairs daily. a) What is the current daily revenue? b) How much would revenue increase if 74 lawn chairs were sold each day? c) What is the marginal revenue when 70 lawn chairs are sold daily? d) Use the answer from part (c) to estimate R(71), R(72), and R(73). a) The current revenue is $
Using the revenue function;
a. The current daily revenue is $1638
b. The revenue increase is $278.896 when 74 lawn chairs are sold.
c. The marginal revenue when 70 chairs are sold daily is $66.1.
d. The current revenue are R(71) ≈ $1,638 + $66.1, R(72) ≈ R(71) + $66.1, R(73) ≈ R(72) + $66.1
What is the current daily revenue?To find the current daily revenue, we can substitute the value of x = 70 into the revenue function R(x) = 0.004x³ + 0.05x² + 0.3x.
a) Current daily revenue:
R(70) = 0.004(70)³ + 0.05(70)² + 0.3(70)
= 0.004(343,000) + 0.05(4,900) + 21
= 1,372 + 245 + 21
= $1,638
Therefore, the current daily revenue is $1,638.
b) To find the increase in revenue if 74 lawn chairs were sold each day, we can calculate the revenue for x = 74 and subtract the current revenue.
Revenue for x = 74:
R(74) = 0.004(74)³ + 0.05(74)² + 0.3(74)
= 0.004(405,224) + 0.05(5,476) + 22.2
= 1,620.896 + 273.8 + 22.2
= $1,916.896
Increase in revenue:
Increase = R(74) - R(70)
= $1,916.896 - $1,638
= $278.896
Therefore, the revenue would increase by $278.896 if 74 lawn chairs were sold each day.
c) Marginal revenue represents the rate of change of revenue with respect to the number of lawn chairs sold. The marginal revenue can be found by taking the derivative of the revenue function R(x).
R(x) = 0.004x³ + 0.05x² + 0.3x
Differentiating both sides with respect to x:
dR/dx = d/dx (0.004x³ + 0.05x² + 0.3x)
= 0.012x² + 0.1x + 0.3
Now, substitute x = 70 into the derivative equation to find the marginal revenue at 70 lawn chairs:
Marginal revenue for x = 70:
dR/dx = 0.012(70)² + 0.1(70) + 0.3
= 0.012(4,900) + 7 + 0.3
= 58.8 + 7 + 0.3
= $66.1
Therefore, the marginal revenue when 70 lawn chairs are sold daily is $66.1.
d) To estimate R(71), R(72), and R(73), we can use the marginal revenue calculated in part (c). We know that marginal revenue represents the change in revenue for a small increase in the number of lawn chairs sold.
Estimated revenue for x = 71:
R(71) ≈ R(70) + (marginal revenue at 70) = $1,638 + $66.1
Estimated revenue for x = 72:
R(72) ≈ R(71) + (marginal revenue at 70) = R(71) + $66.1
Estimated revenue for x = 73:
R(73) ≈ R(72) + (marginal revenue at 70) = R(72) + $66.1
Using the calculations from part (a) and (c), we can estimate the values:
R(71) ≈ $1,638 + $66.1
R(72) ≈ R(71) + $66.1
R(73) ≈ R(72) + $66.1
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a) The current daily revenue is $1,638.
b) The daily revenue would increase by $206.4 if 74 lawn chairs were sold each day.
c) The marginal revenue when 70 lawn chairs are sold daily is $66.1.
d) An estimate for R(71), R(72), and R(73) include:
R(71) = $1704.1.R(72) = $1770.2.R(73) = $1836.3.How to determine the current daily revenue?In order to determine the current daily revenue, we would substitute the number of lawn chairs that Pierce sold daily as follows;
R(x) = 0.004x³ +0.05x² +0.3x.
R(70) = 0.004(70)³ +0.05(70)² +0.3(70).
R(70) = $1,638.
Part b.
Assuming 74 lawn chairs were sold each day, the increase in the daily revenue would be calculated as follows;
Increase = R(74) - R(70)
Increase = 0.004(73)³ + 0.05(73)² +0.3(73) - $1,638.
Increase = $1844.418 - $1,638.
Increase = $206.4.
Part c.
We would differentiate the revenue function in order to determine the marginal revenue as follows;
R(x) = 0.004x³ + 0.05x² + 0.3x.
R'(x) = 0.012x² + 0.1x + 0.3
R'(70) = 0.012(70)² + 0.1(70) + 0.3
R'(70) = $66.1
Part d.
Lastly, we would determine the estimate as follows;
R(71) = R(70) + R'(70)
R(71) = 1,638 + 66.1
R(71) = $1704.1
R(72) = R(70) + 2R'(70)
R(72) = 1,638 + 2(66.1)
R(72) = $1770.2
R(73) = R(70) + 3R'(70)
R(73) = 1,638 + 3(66.1)
R(73) = $1836.3.
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For the specified margin of error and confidence level, obtain a sample size that will ensure a margin of error of at most the one specified margin of error = 0.02; confidence level = 99% Click here to view Page 1 of the table of areas under the standard normal curve. Click here to view Page 2 of the table of areas under the standard normal curve. (Round up to the nearest integer.)
The sample size needed to ensure a margin of error of at most 0.02 with a 99% confidence level is 1327.
The formula for the sample size that will give a maximum margin of error is:
n = (Zα/2 / E)²
where Zα/2 is the critical value from the standard normal distribution, E is the margin of error, and n is the sample size.
Here, the margin of error E = 0.02 and confidence level = 99%.
Thus, determine the critical value of Zα/2 that corresponds to a 99% confidence level.
Using the standard normal table, find that the critical value for the 99% confidence level is
Zα/2 = 2.576.
Substituting the values of Zα/2 = 2.576 and E = 0.02 into the formula,
n = (Zα/2 / E)²
= (2.576 / 0.02)²
= 1326.13
Rounding up to the nearest integer, the final answer:
n = 1327
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Question 6
Consider the following data
X y
0 2
5 4
4 3
What is the sample variance of x? What is the sample standard deviation of x?
O Sample variance is 7, and sample standard deviation is 2.65
O Sample variance is 7, and sample standard deviation is 2.16
O Sample variance is 4.67, and sample variance is 2.16
O Sample variance is 7, and sample standard deviation is 4.67
The correct statement regarding the sample variance and standard deviation is given as follows:
Sample variance is 7, and sample standard deviation is 2.65.
How to obtain the sample variance and standard deviation?The sample for this problem is given as follows:
(0, 5, 4).
The mean of the sample is given as follows:
(0 + 5 + 4)/3 = 3.
The sum of the differences squared of the sample is given as follows:
(0 - 3)² + (5 - 3)² + (4 - 3)² = 14.
The sample variance is given by the sum of the differences squared, divided by one less than the sample size, hence:
14/(3 - 1) = 7.
The sample standard deviation is the square root of the sample variance, hence it is given as follows:
2.65.
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A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.55 hours, with a standard deviation of 2.33 hours. A random sample of 40 adults with children under the age of 18 daily leisure time of 4.12 hours, with a standard deviation of 1.89 hours. Construct and interpret a 95% confidence interval for the mean difference in leisure time between adults with no children and adults within (2) La represent the mean leisure hours of adults with no children under the age of 18 and py represent the mean lesure hours of adults with children under the age of 18 The 95% confidence interval for (4-2) is the range from hours to hours (Round to two decimal places as needed) What is the interpretation of this confidence interval? A. There is a 95% probability that the difference of the means is in the interval Conclude that there is a significant difference in the number of leisure hours. B. There is a 95% probability that the difference of the means is in the interval Conclude that there is insufficient evidence of a significant difference in the number of leisure hours C. There is 95% confidence that the difference of the means is in the interval. Conclude that there is insufficient evidence of a significant difference in the number of leisure hours. D. There is 95% confidence that the difference of the means is in the interval.
Given data: A Random sample of 40 adults with no children under the age of 18 years results
In a mean daily leisure time of 5.55 hours, with a Standard Deviation of 2.33 hours.
A random sample of 40 adults with children under the age of 18 daily Leisure time of 4.12 hours,
With a standard deviation of 1.89 hours.
Construct and interpret a 95% confidence interval for the mean difference in leisure time between adults with no children and adults with.
The difference in the means of leisure time for adults with no children under the age of 18 and adults with children under the age of 18 is given by `(La - Lb)`
Where La is the mean leisure hours of adults with no children under the age of 18 and Lb represent the mean leisure hours of adults with children under the age of 18.
The point estimate of the difference in the mean leisure time for both the groups is: `(La - Lb) = (5.55 - 4.12) = 1.43`
Now, we can compute the standard error as follows: `standard error = sqrt[(s1^2/n1) + (s2^2/n2)]` `= sqrt[(2.33^2/40) + (1.89^2/40)]` `= sqrt[0.1359 + 0.0893]` `= sqrt(0.2252)` `= 0.4747`
Therefore, the 95% confidence interval for the difference in the mean leisure time for both groups is given by: `(La - Lb) ± (t-value) x (standard error)`
The t-value for a 95% confidence interval with 38 degrees of freedom is 2.0244.
We need to use the t-value since the population standard deviation is not given.
We use the sample standard deviation in this case.
The lower limit of the Confidence Interval is `1.43 - (2.0244)(0.4747) = 0.47`
The upper limit of the confidence interval is `1.43 + (2.0244)(0.4747) = 2.39`
Therefore, the 95% confidence interval for the difference in the mean leisure time for both groups is from 0.47 hours to 2.39 hours.
The interpretation of this confidence interval is that there is 95% confidence that the difference in the mean leisure time
For adults with no children under the age of 18 and adults with children under the age of 18 is between 0.47 hours and 2.39 hours.
Therefore, There is 95% confidence that the difference of the means is in the interval.
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In a test of the hypothesis H0: μ=50 versus Ha: μ>50, a sample of n=100 observations possessed mean x=49.2 and standard deviation s=5.3. Find and interpret the p-value for this test.
The given hypothesis is:[tex]H0: μ=50[/tex] (Null Hypothesis)Ha: μ>50 (Alternative Hypothesis)A sample of n=100 observations possessed mean x=49.2 and standard deviation s=5.3.we accept the alternative hypothesis.
To determine whether the null hypothesis is valid, we use the following formula:[tex]t= (x-μ)/(s/√n)[/tex]Let's plug in the values.[tex]t= (49.2-50)/(5.3/√100)= -1.88[/tex]
The P-value is calculated by looking up the area to the right of [tex]t = -1.88[/tex] in the t-distribution table with 99 degrees of freedom. The P-value is the probability of observing such or a more extreme value of the test statistic (t-value) given that the null hypothesis is true.
We have sufficient evidence to support the claim that the population mean is greater than 50. This means that the sample mean of 49.2 is statistically significant and is highly unlikely to have occurred by chance alone.
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