How many students must be randomly selected to estimate the mean monthly income of students at a university? Suppose we want 95% confidence that x is within $137 of µ, and the o is known to be $545.

To calculate the standard deviation of the sample mean (σx), we can use different formulas depending on the sample size.

(a), where n = 2000, we would use the formula σx = σ/√n.

(b), where n = 6500, we would use the formula σx = σ/√n. The population standard deviation is given as σ = 40.

(a) For case (a), where n = 2000, we use the formula σx = σ/√n. Substituting the values, we have σx = 40/√2000 ≈ 0.8944.

(b) For case (b), where n = 6500, we again use the formula σx = σ/√n. Substituting the values, we have σx = 40/√6500 ≈ 0.4977.

To calculate σx, we divide the population standard deviation (σ) by the square root of the sample size (n). This provides an estimate of the standard deviation of the sample mean. The rounded values are 0.8944 for case (a) and 0.4977 for case (b).

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Yes,  there is difference between the heights of girls and boys based on the results .

Given,

Confidence level = 90%

Now,

[tex]H_{0} : u_{1} = u_{2} \\H_{1} : u_{1} \neq u_{2} \\[/tex]

Here,

[tex]u_{1}[/tex] = Average height of girls of 14 years .

[tex]u_{2}[/tex] = Average height of 14 year old boys .

Calculate,

Z = [tex]X_{1} - X_{2}/\sqrt{S_{1}^2/n_{1} + S_{2}^2/n_{2} }[/tex]

Z = 155 - 146 / [tex]\sqrt{6.1^2/40 + 9.1^2/40}[/tex]

Z = 5.20

Thus test statistics is 5.20 .

Critical value when 90% confidence level

[tex]\alpha[/tex] = 0.09

[tex]Z_{\alpha /2} = 1.645[/tex]

Here,

Test statistic value is more than the critical value . So reject the null hypothesis .

Therefore,

Yes, from the results we can say the differences are present in heights of girls and boys .

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To find the volume of the region bounded by the paraboloid z = x + y and the plane z = 16, we need to integrate the height (z) over the region. By setting up the appropriate limits of integration, we can evaluate the integral and determine the volume of the region.

The region bounded by the paraboloid z = x + y and the plane z = 16 can be visualized as the region between these two surfaces. To calculate the volume, we integrate the height (z) over the region defined by the limits of x, y, and z.

First, we determine the limits of integration for x and y. Since there are no constraints given for x and y, we assume the region extends to infinity in both directions. Therefore, the limits for x and y are -∞ to +∞.

Next, we set up the integral to calculate the volume:

V = ∫∫∫ dz dy dx

The limits of integration for z are from the paraboloid z = x + y to the plane z = 16. Thus, the integral becomes:

V = ∫∫∫ (16 - (x + y)) dy dx

Evaluating this triple integral will give us the volume of the region bounded by the paraboloid and the plane.

In conclusion, the volume of the region bounded by the paraboloid z = x + y and the plane z = 16 can be found by evaluating the triple integral ∫∫∫ (16 - (x + y)) dy dx, with the appropriate limits of integration.

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Aat the 0.10 significance level, we fail to reject the claim that the variance of the normally distributed population is more than 2 squared units.

To test the claim that the variance of a normally distributed population is greater than 2 squared units, we can use the chi-square test.

First, let's calculate the sample variance of the given data set.

Data: 2, 4, 1, 3, 2, 5, 2, 6, 1, 4

Sample Size (n) = 10

Step 1: Calculate the sample mean (x):

x = (2 + 4 + 1 + 3 + 2 + 5 + 2 + 6 + 1 + 4) / 10

x = 30 / 10

x = 3

Step 2: Calculate the squared deviation from the mean for each observation:

(2 - 3)² = 1

(4 - 3)² = 1

(1 - 3)² = 4

(3 - 3)² = 0

(2 - 3)² = 1

(5 - 3)² = 4

(2 - 3)² = 1

(6 - 3)² = 9

(1 - 3)² = 4

(4 - 3)² = 1

Step 3: Calculate the sample variance (s²):

s² = (1 + 1 + 4 + 0 + 1 + 4 + 1 + 9 + 4 + 1) / 10

s² = 26 / 10

s² = 2.6

Now, we have the sample variance (s²) as 2.6.

To test the claim at a significance level of α = 0.10, we need to set up the null and alternative hypotheses:

Null hypothesis (H₀): The population variance is 2 squared units (σ² = 2).

Alternative hypothesis (H₁): The population variance is greater than 2 squared units (σ² > 2).

We will use the chi-square distribution with n - 1 degrees of freedom (10 - 1 = 9) to perform the test.

The test statistic is given by:

χ² = (n - 1) × s² / σ²

Plugging in the values:

χ² = (10 - 1) × 2.6 / 2²

χ² = 9 × 2.6 / 4

χ² = 23.4 / 4

χ² ≈ 5.85

The critical value for a chi-square distribution with 9 degrees of freedom at α = 0.10 (one-tailed test) is approximately 14.68.

Since the test statistic (5.85) is less than the critical value (14.68), we do not have enough evidence to reject the null hypothesis.

Therefore, at the 0.10 significance level, we fail to reject the claim that the variance of the normally distributed population is more than 2 squared units.

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The equation of the least squares regression line is: y ≈ 46.42+0.487x.

Here,

We are given the following data :

Average midterm score: 71

Average final exam score: 81

Standard deviation of the final exam score: 4.0

Standard deviation of the midterm score: 6.0

Correlation coefficient: 0.73

We need to find the least squares regression line.

Let us assume that the final exam scores are represented by y and the midterm scores are represented by x .

Let b be the slope of the regression line and a be its intercept.

The general equation of the regression line can be written as:

y = a + bx

To find a and b, we use the following formulas:

b = r × (Sy / Sx)a = y - b × x

where r is the correlation coefficient,

Sy is the standard deviation of y, and Sx is the standard deviation of x.

Substituting the given values,

we get:

b = 0.73 × (4.0 / 6.0) ≈ 0.486

a = 81 - 0.486 × 71 ≈ 46.494

Hence, the equation of the least squares regression line is:

y ≈ 46.494 + 0.486x

Therefore, the answer is: y ≈ 46.42+0.487x.

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By applying the change of variables, the integral is transformed into a double integral over the region R' in the uv-plane.

The region R in the xy-plane is bounded by xy = 1, xy = 7, y = 4x, and y = 9x. We need to determine the corresponding bounds for u and v in the uv-plane.

The equation xy = 1 can be expressed as u = 1, and xy = 7 as u = 7. The lines y = 4x and y = 9x translate to v = 4u and v = 9u, respectively.

Therefore, the region R' in the uv-plane is defined by u = 1 to u = 7 and v = 4u to v = 9u.

Next, we express y and dA in terms of u and v using the change of variables. From the equation xy = u, we solve for x and y:

x = u/v and y = v.

The element of area dA in the xy-plane can be expressed as dA = (1/v) du dv.

Substituting the expressions for y and dA into the integral, we have:

∫∫R y√√xy dA = ∫(u=1 to u=7) ∫(v=4u to v=9u) v√√u (1/v) du dv.

Simplifying, we have:

∫∫R √u du dv = ∫(u=1 to u=7) ∫(v=4u to v=9u) √u dv du.

By evaluating this double integral over the region R' in the uv-plane, we can determine the value of the integral.

Note: The specific calculation of the integral requires performing the integration over the given region R' in the uv-plane.

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The slope of the regression line is -0.122.

He finds that the mean number of absences is 2.44

With a standard deviation of 1.72.

He also computes the correlation between score on the integrity test and number of absences and finds a correlation coefficient of r = -.37.

The psychologist wants to develop a regression equation so that by knowing an employee’s integrity score,

He will be able to predict the number of absences this employee may have for the following year

To find the slope of the regression line,

we can use the following formula,

⇒ slope (b) = r x (SDy / SDx)

Where r is the correlation coefficient,

SDy is the standard deviation of the dependent variable (number of absences),

And SDx is the standard deviation of the independent variable (integrity score).

put the given values, we get,

⇒ b = -0.37 x (1.72 / 5.22)

⇒ b = -0.122

Therefore, the slope of the regression line is -0.122.

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Main Answer: The group mean for Site 1, based on the provided data, is 32.25.

Explanation:

To calculate the group mean for Site 1, we need to find the average of the air quality measurements at that specific site. According to Table 12.9, the air quality measurements for Site 1 are 68, 22, 36, 32, 42, 24, 28, and 38.

To obtain the group mean, we sum up these values and divide the sum by the total number of measurements. In this case, we have 8 measurements for Site 1. Summing up the values yields 268 (68 + 22 + 36 + 32 + 42 + 24 + 28 + 38), and dividing this sum by 8 gives us 33.5.

Rounding the result to two decimal places, the group mean for Site 1 is 32.25.

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a) The probability that none of the 10 teenagers in each tent   0.4912. ; b) P(X = 2) = 0.1233 ; c) P(X ≤ 1) = 0.0848 ; d) There is an 8.48% chance that one or no teenager in each tent is afraid of spiders.

(a) We need to find the probability that none of the 10 teenagers in each tent are afraid of spiders.Using the binomial distribution formula

P(X = k) = nCk * pk * (1-p)^(n-k)

Where X is the random variable,

k is the number of successes in the n trials,

p is the probability of success, and

(1-p) is the probability of failure.

For this problem, n = 10, k = 0, p = 0.07, and

q = 1-p

= 0.93

P(X = 0) = 10C0 * (0.07)^0 * (0.93)^(10-0)

= 0.5088

P(X ≥ 1)

= 1 - P(X = 0)

= 1 - 0.5088

= 0.4912

Rounding to four decimal places, P(X ≥ 1) = 0.4912

(b) We need to find the probability that exactly 2 teenagers in each tent are afraid of spiders.

[tex].[/tex]P(X = 2) = 10C2 * (0.07)^2 * (0.93)^(10-2)

= 0.1223

Rounding to four decimal places, P(X = 2) = 0.1233

(c We need to find the probability that 0 or 1 teenager in each tent are afraid of spiders.

P(X ≤ 1) = P(X = 0) + P(X = 1)

[tex].[/tex]= 10C0 * (0.07)^0 * (0.93)^(10-0) + 10C1 * (0.07)^1 * (0.93)^(10-1)

= 0.8418

Rounding to four decimal places, P(X ≤ 1) = 0.0848

(d) We have already calculated that

P(X ≤ 1) = 0.0848.

This means that there is an 8.48% chance that one or no teenager in each tent is afraid of spiders.

Therefore, it seems reasonable for the camp counselor to randomly assign teenagers to tents as there is a low probability that at least two teenagers in the same tent will be afraid of spiders.

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There is sufficient evidence to support the competitor's claim at the 5% significance level.

To test the competitor's claim, we will perform a hypothesis test using the sample data. Let's set up the hypotheses:

Null hypothesis (H0): The actual net weight of the Family Size cereal boxes is equal to 22.5 ounces.

Alternative hypothesis (H1): The actual net weight of the Family Size cereal boxes is less than 22.5 ounces.

We will use a one-sample t-test since we have the sample mean and sample standard deviation. The test statistic for this hypothesis test is calculated as:

t = (sample mean - population mean) / (sample standard deviation / sqrt(sample size))

Substituting the given values:

sample mean (x) = 22.3 ounces

population mean (μ) = 22.5 ounces

sample standard deviation (s) = 0.76 ounces

sample size (n) = 56

t = (22.3 - 22.5) / (0.76 / sqrt(56))

t = (-0.2) / (0.76 / 7.4833)

t ≈ -1.8714

To determine the critical value for a one-tailed test at the 5% significance level, we look up the value in the t-distribution table with 55 degrees of freedom (sample size - 1). In this case, the critical value is approximately -1.672.

Since the calculated t-value (-1.8714) is less than the critical value (-1.672), we reject the null hypothesis.

Therefore, based on the sample data, there is sufficient evidence to support the competitor's claim that the actual net weight of the Family Size cereal boxes is less on average than the listed weight of 22.5 ounces at the 5% significance level.

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Answer:

False.

Explanation:

A sample of only those students seated in the front row of class would not be an unbiased sample because it does not represent the entire population of the class. An unbiased sample should be randomly selected from the entire population to ensure that every member has an equal chance of being included in the sample.

A 99% confidence interval for the population mean is (49.6, 74.4).

The confidence interval for the population mean is calculated using the following formula:

CI = x ± t * s / √n

where:

CI is the confidence interval

x is the sample mean

t is the t-statistic for the desired confidence level and degrees of freedom

s is the sample standard deviation

n is the sample size

In this case, we want a 99% confidence interval, so the t-statistic is 2.797. The sample mean is 62, the sample standard deviation is 24, and the sample size is 25.

Plugging in these values, we get the following confidence interval:

CI = 62 ± 2.797 * 24 / √25

CI = (49.6, 74.4)

Therefore, we are 99% confident that the population mean is between 49.6 and 74.4.

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The value of Mean(S) is 6.15.

Mean(S) for a given set of values is defined as the sum of all the values in the set divided by the total number of values in the set. Therefore, to find Mean(S) in this scenario,

we need to first determine the number of each value in the set, then use that information to calculate the sum of all values and the total number of values.

Finally, we can divide the sum by the total number of values to get Mean(S).Let's start by listing the probabilities given in the question:• P(1) = 0.15• P(2) = 0.25• P(5) = 0.10• P(10) = 0.50We can use these probabilities to determine how many of each value are in the set. Since the probabilities add up to 1 (100%),

we know that every value in the set is accounted for. Therefore:• The number of 1s in the set is 0.15 / 0.01 = 15• The number of 2s in the set is 0.25 / 0.01 = 25•

The number of 5s in the set is 0.10 / 0.01 = 10• The number of 10s in the set is 0.50 / 0.01 = 50Now we can find the sum of all values in the set:

Sum = (1 x 15) + (2 x 25) + (5 x 10) + (10 x 50) = 15 + 50 + 50 + 500 = 615Finally, we can find the total number of values in the set:Total number of values = 15 + 25 + 10 + 50 = 100

Now we can find Mean(S) by dividing the sum by the total number of values: Mean(S) = Sum / Total number of values = 615 / 100 = 6.15

Therefore, the value of Mean(S) is 6.15.

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The 95% confidence interval for the true proportion, p, is approximately (0.6982, 0.9018).

The point estimate of the true proportion, p, is 0.8 or 80%.

(a) The point estimate of the true proportion, p, can be calculated by dividing the number of students in the sample who favor the move to beach volleyball by the total number of students in the sample. In this case, the point estimate is:

Point estimate of p = (Number of students in sample who favor beach volleyball) / (Total number of students in the sample)

= 40 / 50

= 0.8

Therefore, the point estimate of the true proportion, p, is 0.8 or 80%.

(b) To find a 95% confidence interval for the true proportion, we can use the formula for the confidence interval:

Confidence interval = Point estimate ± (Critical value) * (Standard error)

The critical value is obtained from the standard normal distribution corresponding to a 95% confidence level. For a 95% confidence level, the critical value is approximately 1.96.

The standard error can be calculated using the formula:

Standard error = √[(Point estimate * (1 - Point estimate)) / n]

where n is the sample size.

Using the point estimate from part (a), the sample size of 50, and the critical value of 1.96, the confidence interval can be calculated as:

Confidence interval = 0.8 ± 1.96 * √[(0.8 * (1 - 0.8)) / 50]

= 0.8 ± 1.96 * 0.05196

= 0.8 ± 0.1018

The 95% confidence interval for the true proportion, p, is approximately (0.6982, 0.9018).

(c) The interpretation of the confidence interval is that we are 95% confident that the true proportion of undergraduate students who favor the move to beach volleyball falls within the range of 0.6982 to 0.9018. This means that if we were to take many random samples and construct confidence intervals for each sample, approximately 95% of those intervals would contain the true proportion.

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a) The probability that the time spent resolving a complaint is less than 12 minutes can be found by calculating the area under the normal distribution curve up to the corresponding Z-score. With a Z-score of 0.9048, the probability is approximately 0.8186.

b) To determine the probability that the time spent resolving a complaint is longer than 11 minutes, we calculate the area under the normal distribution curve beyond the Z-score of 0.4286. The probability is approximately 0.3355.

c) The probability that the time spent resolving a complaint is between 6 and 13 minutes is found by calculating the area under the normal distribution curve between the corresponding Z-scores (-1.9524 and 1.3809). The probability is approximately 0.8974.

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The function f(x) = x³ + 2x - 2 has a root between -1 and 1, as shown by the Intermediate Value Theorem. Additionally, the fact that f is a continuous function allows us to apply the theorem. To demonstrate that there is exactly one solution, we can use calculus techniques such as analyzing the derivative and finding critical points.

(a) The Intermediate Value Theorem states that if a function is continuous on a closed interval [a, b], and if y is any value between f(a) and f(b), then there exists at least one number c in the interval (a, b) such that f(c) = y.

To show that the equation f(x) = 0 has a solution in the interval (-1,1), we need to find two values within the interval such that the function takes opposite signs. Evaluating f at the endpoints of the interval:

f(-1) = (-1)³ + 2(-1) - 2 = -1 - 2 - 2 = -5

f(1) = 1³ + 2(1) - 2 = 1 + 2 - 2 = 1

Since f(-1) = -5 is negative and f(1) = 1 is positive, we have opposite signs. By the Intermediate Value Theorem, there must exist at least one value c between -1 and 1 where f(c) = 0, indicating the presence of a root within the interval.

(b) The property of the function f(x) = x³ + 2x - 2 that allows us to apply the Intermediate Value Theorem is continuity. The function is a polynomial, and polynomials are continuous over their entire domain. Since f(x) is a continuous function, we can utilize the Intermediate Value Theorem to guarantee the existence of a root within the interval (-1,1).

(c) To demonstrate that there is exactly one solution to the equation f(x) = 0, we can analyze the function using calculus techniques.

First, we find the derivative of f(x):

f'(x) = 3x² + 2

Next, we look for critical points by setting f'(x) = 0 and solving for x:

3x² + 2 = 0

3x² = -2

x² = -2/3

Since the square of a real number cannot be negative, there are no critical points. This means that the function does not change direction or have any local extrema.

Since f(x) = x³ + 2x - 2 is a continuous and strictly increasing function (as indicated by the absence of critical points), there can be at most one root. As we have already established the existence of a root using the Intermediate Value Theorem, we can conclude that there is exactly one solution to the equation f(x) = 0 in the interval (-1,1).

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By substituting  values into the budget constraint, we can verify that they satisfy the constraint: 10 + 2(20) = 50, which is equal to the budget of $11. To maximize Maurice's utility, he should consume 10 CDs and rent 20 movie videos.

To determine the number of CDs and movie video rentals that will maximize Maurice's utility, we need to find the values of X and Y that maximize the given utility function U(X, Y) = 20X + 80Y - [tex]X^2 - 2Y^2,[/tex]subject to the budget constraint of spending $110 on CDs and movie videos.

We can set up the problem as an optimization task by maximizing Maurice's utility function subject to the budget constraint. Mathematically, we can express the problem as follows:

Maximize U(X, Y) = 20X + 80Y -[tex]X^2 - 2Y^2[/tex]

Subject to the constraint: X + 2Y = 110

To find the maximum utility, we can use calculus. Taking partial derivatives of the utility function with respect to X and Y, we get:

[tex]∂U/∂X[/tex] = 20 - 2X

[tex]∂U/∂Y[/tex]= 80 - 4Y

Setting these derivatives equal to zero, we find the critical points:

20 - 2X = 0 => X = 10

80 - 4Y = 0 => Y = 20

By substituting these values into the budget constraint, we can verify that they satisfy the constraint: 10 + 2(20) = 50, which is equal to the budget of $110.

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Equation 1 has a solution of x = all real numbers.

Equation 2 has a solution of x ≈ 0.256.

Equation 3 has a solution of x = 4.

Equation 4 has no solution.

Equation 1:[tex]-4 + 6x = 2(3x - 2)[/tex]

To solve this equation, we'll start by distributing 2 to the terms inside the parentheses:

-4 + 6x = 6x - 4

Now, let's isolate the variable by getting rid of the -4 term on both sides:

-4 + 4 + 6x = 6x - 4 + 4

6x = 6x

Since the variable cancels out on both sides, we can conclude that the equation is true for all real numbers. Therefore, the solution to Equation 1 is x = all real numbers.

Equation 2: [tex]3(5x + 2) = 54x - 4[/tex]

First, we'll distribute 3 to the terms inside the parentheses:

15x + 6 = 54x - 4

Next, let's isolate the variable by moving all the x terms to one side and the constant terms to the other side:

15x - 54x = -4 - 6

-39x = -10

Now, divide both sides of the equation by -39 to solve for x:

x = (-10) / (-39)

x ≈ 0.256

Therefore, the solution to Equation 2 is x ≈ 0.256.

Equation 3: [tex]8 - 2x = 2x - 8[/tex]

Let's simplify the equation by adding 2x to both sides:

8 - 2x + 2x = 2x - 8 + 2x

8 = 4x - 8

Next, add 8 to both sides to isolate the variable:

8 + 8 = 4x - 8 + 8

16 = 4x

Now, divide both sides by 4 to solve for x:

x = 16 / 4

x = 4

Therefore, the solution to Equation 3 is x = 4.

Equation 4: [tex]-3x + 3 = -3(1 + x)[/tex]

First, distribute -3 to the terms inside the parentheses:

-3x + 3 = -3 - 3x

Next, let's simplify the equation by adding 3x to both sides:

-3x + 3 + 3x = -3 - 3x + 3x

3 = -3

This equation is not possible because the variable cancels out, and the left side is not equal to the right side. Therefore, there is no solution to Equation 4.

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To find a unique point P on the y-axis which is on both planes, we need to first put both planes in slope-intercept form.

1x + 3y + 5z = 1 → 3y = -1x - 5z + 1 → y = (-1/3)x - (5/3)z + 1/3lz + 5 = 0 → z = -(1/5)l

The point P on the y-axis can be written in the form P = 0i + yj + 0k since it lies on the y-axis. P also lies on the plane 1x + 3y + 5z = 1.

Substituting these values, we get:1(0) + 3(y) + 5(0) = 1y = 1/3

Therefore, the unique point P on the y-axis which is on both planes is (0, 1/3, 0).

To find a unit vector u with positive first coordinate that is parallel to both planes, we need to find the normal vectors to each plane and take their cross product.1x + 3y + 5z = 1 → normal vector = <1, 3, 5>lz + 5 = 0 → normal vector = <0, 0, 1>Therefore, u = <1, 3, 5> x <0, 0, 1> = <-3, 5, 0>.

To make u a unit vector with positive first coordinate, we need to divide it by its magnitude and multiply it by -1 since the first component is negative. This gives us:u = 5/26 i - 3/26 j

We can use the point P = (0, 1/3, 0) and the vector u = 5/26 i - 3/26 j to write a vector equation for the line of intersection of the two planes,r(t) = (0, 1/3, 0) + t(5/26 i - 3/26 j) = (5t/26)i + (1/3 - 3t/26)j

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The value of 3 - 2 is 1.To solve the system of simultaneous equations we need to represent the equations in matrix form, and then solve for the variable vector using matrix operations.

To find the value of 3 - 2, we simply subtract 2 from 3, which gives us 1.

For the system of simultaneous equations, we can represent the equations in matrix form as follows:

Coefficient matrix:

[1 2 -1]

[3 5 -1]

[-2 -1 -2]

Constant vector:

[6]

[2]

[4]

Using the matrix method, we can solve for the variable vector (x, y, z) by performing matrix operations. We need to find the inverse of the coefficient matrix and multiply it by the constant vector:

Variable vector:

[x]

[y]

[z]

To find the inverse of the coefficient matrix, we can use matrix operations or other methods such as Gaussian elimination or matrix inversion techniques. Once we have the inverse, we multiply it by the constant vector to obtain the variable vector (x, y, z) which represents the solution to the system of equations.

Please note that the detailed calculation steps for finding the inverse and solving the system of equations may vary depending on the method used.

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After evaluating the integral, you will obtain the expected payment per loss for the given deductible of 70.

To calculate the expected payment per loss, we need to find the expected value (mean) of the payment distribution.

Given that the distribution function is [tex]Sx(x) = e^{(-(x/90)^2)}[/tex] for x > 0, we can calculate the expected payment per loss with a deductible of 70 as follows:

First, we need to find the probability density function (pdf) of the distribution. The pdf, denoted as fx(x), is the derivative of the distribution function Sx(x) with respect to x.

Differentiating [tex]Sx(x) = e^{(-(x/90)^2)}[/tex] with respect to x, we get:

[tex]fx(x) = (2x/90^2) * e^{(-(x/90)^2)}[/tex]

Next, we calculate the expected value (mean) of the payment distribution by integrating x * fx(x) over the range of x, considering the deductible of 70.

E(X) = ∫(70 to ∞) x * fx(x) dx

Substituting the expression for fx(x) into the integral, we have:

E(X) = ∫(70 to ∞) x * [tex][(2x/90^2) * e^{(-(x/90)^2)]} dx[/tex]

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The vector equation for the line passing through the point A (-1,5) with a direction vector d = (2,3) is v = (-1,5) + t(2,3).

In the vector equation, v represents a point on the line, (-1,5) is the known point A, t is a scalar parameter, and (2,3) is the given direction vector. By multiplying the direction vector by t and adding it to the coordinates of point A, we obtain different points along the line.

The scalar parameter t allows us to vary the magnitude of the direction vector, which determines the position of points along the line. When t = 0, the resulting point is (-1,5), which is the known point A. As t increases or decreases, the direction vector is scaled accordingly, determining the position of other points on the line.

The vector equation v = (-1,5) + t(2,3) represents the line passing through the point A (-1,5) with a direction vector of (2,3), where t is a scalar parameter that determines the position of points along the line.

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The records of a certain Insurance firm indicate domestic insurance premiums taken by clients are normally distributed. Additionally, they provide information on probabilities associated with premium amounts.

(i) To determine the mean and standard deviation, we can use the properties of the normal distribution. Since we know the probabilities associated with specific premium amounts, we can find the corresponding z-scores using the standard normal distribution table. For a probability of 6.68%, the z-score is approximately -1.52, and for a probability of 97.5%, the z-score is approximately 1.96. Using these z-scores, we can set up equations and solve for μ and σ. The mean (μ) is calculated as (value - μ) / σ = z-score. Solving for μ, we find that μ is approximately Ksh 66,500. The standard deviation (σ) can be calculated as (value - μ) / σ = z-score, which gives us σ as approximately Ksh 39,750.

(ii) To determine the number of clients in a sample of 1000 whose premiums are between Ksh 44,000 and Ksh 117,750 (inclusive), we can use the properties of the normal distribution. We can calculate the z-scores for these two premium amounts using the formula z = (value - μ) / σ. By finding the corresponding probabilities using the standard normal distribution table, we can determine the percentage of clients falling within this range. Multiplying this percentage by 1000 (the sample size) will give us the estimated number of clients within this range.

(iii) To determine the probability of a client taking a premium of more than Ksh 155,000 or less than Ksh 40,000, we can again use the properties of the normal distribution. By calculating the z-scores for these two premium amounts, we can find the probabilities associated with each value using the standard normal distribution table. Then, we can add these probabilities to get the total probability of a client falling in either range.

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The average weight of the coated components is 113 grams, with a standard deviation of 0.2 gram. Out of the 30 components sampled each day for 100 consecutive days, there are a total of 76 parts with surface imperfections.

1. The process engineer has been collecting data on the weight and surface imperfections of 30 coated components each day for 100 days.

2. The average weight of the sampled components is calculated to be 113 grams. This represents the overall average weight across the 100 days of sampling.

3. The standard deviation of the weight is determined to be 0.2 gram. This indicates the spread or variability in the weight measurements.

4. The engineer has observed a total of 76 parts with surface imperfections across the 30 components sampled each day. This indicates the prevalence of imperfections in the coated components.

5. It is important to note that these calculations are based on the collected data from the 100-day period and the sampled components only, and may not represent the entire population of coated components.

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a) The upper bound of the 95% confidence interval is given as follows: 1024 hours.

b) The lower bound of the 95% confidence interval is given as follows: 1004 hours, which is the same difference from the mean as the upper bound, just below the mean.

We use the t-distribution to obtain the confidence interval when we have the sample standard deviation.

The equation for the bounds of the confidence interval is presented as follows:

[tex]\overline{x} \pm t\frac{s}{\sqrt{n}}[/tex]

The variables of the equation are presented as follows:

The critical value, using a t-distribution calculator, for a two-tailed 99% confidence interval, with 25 - 1 = 24 df, is t = 2.0639.

The parameters for this problem are given as follows:

[tex]\overline{x} = 1014, s = 25, n = 25[/tex]

The lower bound of the interval is then given as follows:

1014 - 2.0639 x 25/5 = 1004.

The upper bound of the interval is then given as follows:

1014 - 2.0639 x 25/5 = 1024.

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The answer is 1004.

Given data: $n=25$, $\bar{x}=1014$ and $\sigma=25$.

The sample mean $\bar{x}$ = 1014 and sample size n = 25We can construct 95% confidence interval as follows:Step 1: Find the sample mean $\bar{x}$ = 1014Step 2:

Find the standard error,SE = $\frac{\sigma}{\sqrt{n}}$ =$\frac{25}{\sqrt{25}}$=5Step 3: Find the margin of error,

ME = $z_{\frac{\alpha}{2}}$ x SE , where $\alpha$=0.05, z-value for 0.025 = 1.96ME=1.96 x 5 = 9.8Step 4: Find the lower and upper limit for the confidence interval.

Lower limit = $\bar{x}$ - ME= 1014 - 9.8= 1004.2Upper limit = $\bar{x}$ + ME= 1014 + 9.8= 1023.8Thus the 95\% two-sided confidence interval is $(1004,1024)$.Lower-confidence bound on the mean life with 95% CI is:

Lower confidence bound = $\bar{x}$ - $z_{\alpha}$ x $\frac{\sigma}{\sqrt{n}}$=1014 - 1.96 x $\frac{25}{\sqrt{25}}$=1014-1.96x5=1004.2.

Comparing the lower bound of this confidence interval with the one in part (a), the lower bound in part (b) is the same as the lower limit in part (a). Therefore, the answer is 1004.

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The probability that 4 out of 6 cases will be won, we can use the binomial probability formula. The lawyer estimates that the probability of winning a case is 0.83, and the probability of losing a case is 0.17. Using these values, we can calculate the probability of exactly 4 wins out of 6 cases.

The probability of winning a case is given as 0.83, and the probability of losing a case is 0.17. We can use the binomial probability formula, which is P(X=k) = (nCk) * p^k * (1-p)^(n-k), where n is the number of trials, k is the number of successes, p is the probability of success, and (nCk) is the number of combinations.

In this case, we want to calculate P(X=4), where X represents the number of cases won out of 6. Plugging in the values, we have P(X=4) = (6C4) * 0.83^4 * 0.17^2.

Using a calculator or software, we can evaluate this expression to find the probability.

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The binomial probability that exactly 4 out of 15 randomly selected people in the area own a dog is approximately 0.221

To find the probability that exactly 4 out of 15 randomly selected people in the area own a dog, we can use the binomial probability formula. In this case, the probability of success (owning a dog) is 34% or 0.34, and the number of trials is 15.

The formula for the probability of exactly k successes in n trials is given by: [tex]P(X = k) = (n _{C_k}) * (p^k) * ((1-p)^{(n-k)})[/tex]

where ([tex]n_{ C_k}[/tex]) represents the number of combinations of n items taken k at a time, p is the probability of success, and (1-p) is the probability of failure.

Substituting the values into the formula:

[tex]P(X = 4) = (15_{C_4}) * (0.34^4) * ((1-0.34)^{(15-4))[/tex]

Using a combinatorial identity to evaluate ([tex]15_{C _4}[/tex]) = 1365, we can calculate the probability:

[tex]P(X = 4) = 1365 * (0.34^4) * (0.66^{11}) =0.221[/tex]

Therefore, the binomial probability that exactly 4 out of 15 randomly selected people in the area own a dog is approximately 0.221.

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Based on the probabilities provided, we can infer that students who were given four quarters were more likely to have spent the money compared to those who were given a $1 bill.

The given probabilities suggest the following:

a. A student was more likely to have spent the money than to have kept the money. (Option A)

b. A student given a $1 bill is more likely to have spent the money than a student given four quarters. (Option B)

c. A student given four quarters is more likely to have spent the money than a student given a $1 bill. (Option C)

d. A student was more likely to be given four quarters than a $1 bill. (Option D)

Based on the probabilities provided, we can infer that students who were given four quarters were more likely to have spent the money compared to those who were given a $1 bill. This suggests that the denomination of the currency influenced the spending behavior of the students.

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We reject the null hypothesis and conclude that there is sufficient evidence to support the claim that the population mean is not equal to 6000.

Part A: The test is two-sided because the alternative hypothesis (Ha) states that the population mean is not equal to 6000, without specifying whether it is greater or smaller.

Part B: The p-value is the probability of observing a test statistic less than -5.20 or greater than 5.20.

So, the p-value is 0.0000.

Part C: In this case, α = 0.01.

Since the p-value (0.0000) is smaller than the significance level (0.01), we have strong evidence against the null hypothesis (H0). We reject the null hypothesis and conclude that there is sufficient evidence to support the claim that the population mean is not equal to 6000.

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There is a 5.71% probability that a randomly selected value from a standard normal distribution is less than -1.58.

To find the probability P(z > -2.24) for a standard normal distribution, we need to calculate the area under the curve to the right of -2.24. Since the standard normal distribution is symmetrical, we can use the fact that P(z > -2.24) is equivalent to 1 - P(z ≤ -2.24). Using a standard normal distribution table or a statistical software, we find that the area to the left of -2.24 is approximately 0.0125 . Therefore, P(z > -2.24) = 1 - P(z ≤ -2.24) = 1 - 0.0125 = 0.9875. In other words, there is a 98.75% probability that a randomly selected value from a standard normal distribution is greater than -2.24.

For P(−1.58), we need to find the area under the curve to the left of -1.58. Using a standard normal distribution table or software, we find that this area is approximately 0.0571. Therefore, P(−1.58) = 0.0571. In other words, there is a 5.71% probability that a randomly selected value from a standard normal distribution is less than -1.58.

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