Speedy Oil provides a single-server automobile oil change and lubrication service. Customers provide an arrival rate of 2.5 cars per hour. The service rate is 5 cars per hour. Assume that arrivals follow a Poisson probability distribution and that service times follow an exponential probability distribution.
If required, round your answer to the nearest whole number.
(a) What is the average number of cars in the system?
(b) What is the average time that a car waits for the oil and lubrication service to begin?
(c) What is the average time a car spends in the system?
(d) What is the probability that an arrival has to wait for service?

Answers

Answer 1

Answer:

(a) Average number of cars in the system is 1

(b) Average time a car waits is 12 minutes

(c) Average time a car spends in the system is 2 minutes

(d) Probability that an arrival has to wait for service is 0.08.

Step-by-step explanation:

We are given the following

Arrival Rate, A = 2.5

Service Rate B = 5

(a) Average Number of Cars in the System is determined by dividing the Arrival Rate A by the difference between the Service Rate B, and Arrival Rate A.

Average number of cars = A/(B - A)

= 2.5/(5 - 2.5)

= 2.5/2.5 = 1

There is an average of 1 car.

(b) Average time a car waits = A/B(B - A)

= 2.5/5(5 - 2.5)

= 2.5/(5 × 2.5)

= 2.5/12.5

= 1/5

= 0.20 hours

Which is 12 minutes

(c) Average time a car spends in the system is the ratio of the average time a car waits to the service rate.

Average time = 0.2/5

= 0.04 hours

= 2.4 minutes

Which is approximately 2 minutes.

(d) Probability that an arrival has to wait for service is the ratio of the average time a car waits to rate of arrivals.

Probability = 0.2/2.5

= 0.08

Answer 2
Final answer:

The average number of cars in the system can be found using Little's Law. The average time a car waits for the oil and lubrication service to begin is half of the average time in the system. The probability that an arrival has to wait for service is obtained by dividing the arrival rate by the service rate.

Explanation:

(a) To find the average number of cars in the system, we can use Little's Law. Little's Law states that the average number of cars in the system (L) equals the arrival rate (λ) multiplied by the average time a car spends in the system (W). In this case, λ = 2.5 cars per hour and the service rate (μ) = 5 cars per hour.



So, L = λ * W. Rearranging the formula, W = L / λ. Substituting the values, W = (2.5 / 5) = 0.5 hours or 30 minutes.



(b) The average time that a car waits for the oil and lubrication service to begin is equal to half of the average time a car spends in the system, which is 30 minutes.



(c) The average time a car spends in the system is equal to the average waiting time plus the average service time. The average service time (1/μ) in this case is 1/5 hour or 12 minutes. Therefore, the average time a car spends in the system is 30 minutes + 12 minutes = 42 minutes.



(d) To find the probability that an arrival has to wait for service, we can use the formula P(wait) = λ / μ = 2.5 / 5 = 0.5 or 50%

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Related Questions

Solve the following problems (show your work). (a) (1 point). Give example of 2 x 2 non-zero matrices A, B, C such that AB = AC but B+C. (b) (1 point). Write 2 x 2 invertible matrices A and B such that A+B is not invertible. (c) (2 points). Write 3 x 3 singular matrices A and B such that A - B is non-singular.

Answers

Answer:

Step-by-step explanation: see attachment below

A corporation has 11 manufacturing plants. Of these, seven are domestic and four are outside the United States. Each year a performance evaluation is conducted for four randomly selected plants. What is the probability that a performance evaluation will include at least one plant outside the United States

Answers

Answer:

The probability that a performance evaluation will include at least one plant outside the United States is 0.836.

Step-by-step explanation:

Total plants = 11

Domestic plants = 7

Outside the US plants = 4

Suppose X is the number of plants outside the US which are selected for the performance evaluation. We need to compute the probability that at least 1 out of the 4 plants selected are outside the United States i.e. P(X≥1). To compute this, we will use the binomial distribution formula:

P(X=x) = ⁿCₓ pˣ qⁿ⁻ˣ

where n = total no. of trials

           x = no. of successful trials

           p = probability of success

           q = probability of failure

Here we have n=4, p=4/11 and q=7/11

P(X≥1) = 1 - P(X<1)

          = 1 - P(X=0)

          = 1 - ⁴C₀ * (4/11)⁰ * (7/11)⁴⁻⁰

          = 1 - 0.16399

P(X≥1) = 0.836

The probability that a performance evaluation will include at least one plant outside the United States is 0.836.

Circle your answer and justify it by showing your work. (a) T F: (b) T F: Let A be any square matrix, then ATA, AAT, and A + AT are all symmetric. If S is invertible, then ST is also invertible. If a row exchange is required to reduce matrix A into upper triangular form U, then (c) T F: A can not be factored as A = LU. (d) T F: Suppose A reduces to upper triangular U but U has a 0 in pivot position, then A has no LDU factorization. (e) T F: If A2 is not invertible, then A is not invertible. 10 10. [10points] (a) T F: All(x,y,z)∈R3 withx=y+z+1isasubspaceofR3 (b) T F: All(x,y,z)∈R3 withx+z=0isasubspaceofR3 (c) T F: All 2 × 2 symmetric matrices is a subspace of M22. (Here M22 is the vector space of all 2 × 2 matrices.) (d) T F: All polynomials of degree exactly 3 is a subspace of P5. (Here P5 is the vector space of all polynomials a5x5 + a4x4 + a3x3 + a2x2 + a1x + a0 of degree less than or equal to 5.) (e) T F: P3 is a subspace of P5. (Here Pi is the vector space of all polynomials aixi+ai−1xi−1+ ai−2x1−2 + ... + a2x2 + a1x + a0 of degree less than or equal to i.)a. TrueB. False.

Answers

Answer:

a-True

b-True

c-True

d- False

e-True

f-True

g-True

h- False

i-True

j- False

Step-by-step explanation:

See the attached pictures.

a triangle has side cequals6 and angles equals startfraction pi over 4 endfraction and equals startfraction pi over 3 endfraction . find the length a of the side opposite a. the length of side a is nothing.

Answers

Answer:

[TeX]a=\frac{12}{1+\sqrt{3}}[/Tex]

Step-by-step explanation:

c=6

Angle A= π/4

Angle B= π/3

Angle C = π - (π/4+π/3)= 5π/12

We want to determine side A.

Using Sine Rule

[TeX]\frac{a}{Sin A}=\frac{c}{Sin C}[/Tex]

[TeX]\frac{a}{Sin \frac{\pi}{4}}=\frac{6}{Sin \frac{5\pi}{12} }[/Tex]

[TeX]a=\frac{6}{Sin \frac{5\pi}{12} } X Sin \frac{\pi}{4}[/Tex]

[TeX]a=\frac{12}{1+\sqrt{3}}[/Tex]

a=4.39

Female III-4 is pregnant via male III-5. The owner of this breeding pair wants to know the probabilities of several possible outcomes for their offspring (IV-3). If you need help with how to approach calculating these probabilities, use the Hint.

Answers

Answer: the following are required for IV-3 to have condition

Explanation:

-II-4 passes an X b chromosome to III-4 (probability = 1/2).

-If III-4 has the genotype X B X b (accounted for by the above probability), then she passes an X b chromosome to IV-3 (probability = 1/2).

-III-5 passes a Y chromosome to IV-3 (probability = 1/2).

All of these requirements are needed in sequence, so you apply the product rule here, too (1/2 x 1/2 x 1/2 = 1/8).

Once the individual probabilities are known, the sum and/or product rules can be used for various combinations (both conditions, either condition, etc.).

1. You have an aluminum bar of dimensions 2cm*5cm*10 cm. You want to put it into electric circuit such a way that this bar will demonstrate the smallest possible resistance. You should connect your bar to the opposite faces with dimensions of:

Answers

Answer:

( 5 x 10 ) cm

Step-by-step explanation:

Given:

- The dimensions of the bar are:

                                  ( 2 x 5 x 10 ) cm

Find:

Which two faces with dimensions ( _x _ ) should be connected to get smallest possible resistance.

Solution:

- The electrical resistance R of any material with density ρ and corresponding dimensions is expressed as:

                                R = ρ*L / A

- Where, A: cross sectional Area

              L: The length in between the two faces.

- We need to minimize the electrical resistance of the bar. For that the Area must be maximized and Length should be minimized.

-                               A_max & L_min ---- > R_min

                               (5*10) & ( 2) ------> R_min

Hence, the electrical resistance is minimized by connecting the face with following dimensions ( 5 x 10 ) cm

QUESTION 7 A Randstad/Harris interactive survey reported that 25% of employees said their company is loyal to them. Suppose 9 employees are selected randomly and will be interviewed about company loyalty. What is the probability that none of the 9 employees will say their company is loyal to them? g

Answers

Answer:

7.51% probability that none of the 9 employees will say their company is loyal to them.

Step-by-step explanation:

For each employee, there are only two possible outcomes. Either they think that their company is loyal to them, or they do not think this. The probability of an employee thinking that their company is loyal to them is independent of other employees. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

25% of employees said their company is loyal to them.

This means that [tex]p = 0.25[/tex]

9 employees are selected randomly

This means that [tex]n = 9[/tex]

What is the probability that none of the 9 employees will say their company is loyal to them?

This is P(X = 0).

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{9,0}.(0.25)^{0}.(0.75)^{9} = 0.0751[/tex]

7.51% probability that none of the 9 employees will say their company is loyal to them.

The weights of packets of cookies produced by a certain manufacturer have a Normal distribution with a mean of 202 grams and a standard deviation of 3 grams. What is the weight that should be stamped on the packet so that only 1% of the packets are underweight?

Answers

Final answer:

The weight that should be stamped on the cookie packets to ensure that only 1% of packets are underweight, given a mean weight of 202 grams and a standard deviation of 3 grams, is approximately 195 grams.

Explanation:

The question asks for the weight to be inscribed on the packet so that only 1% of the packets are underweight. This is a statistics problem dealing with the Normal distribution, or bell curve. We know that the weight of the packets is Normally distributed with a mean of 202 grams and a standard deviation of 3 grams. We're asked to find out the weight at which only 1% of the packets would be considered underweight. The solution requires us to find the z-score that corresponds to the bottom 1% of the distribution. From the z-table, we know that a z-score of -2.33 corresponds roughly to the bottom 1%. The formula to calculate the X-value (the weight we’re looking for) is X = μ + Zσ. Substituting the given values we get, X = 202 + -2.33*3. This comes out to be approximately 195 grams. So, the weight that should be stamped on the packet to ensure that only 1% of packets are underweight is 195 grams.

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Which of the following are continuous variables, and which are discrete? (a) number of traffic fatalities per year in the state of Florida continuous discrete (b) distance a golf ball travels after being hit with a driver continuous discrete (c) time required to drive from home to college on any given day continuous discrete (d) number of ships in Pearl Harbor on any given day continuous discrete (e) your weight before breakfast each morning continuous discrete

Answers

Answer:

a) Discrete

b) Continuous

c) Continuous

d) Discrete

e) Continous

Step-by-step explanation:

Continuous:

Real numbers, can be integer, decimal, etc.

Discrete:

Only integer(countable values). So can be 0,1,2...

(a) number of traffic fatalities per year in the state of Florida

You cannot have half of a traffic fatality, for example. So this is discrete

(b) distance a golf ball travels after being hit with a driver

The ball can travel 10.25m, for example, which is a decimal number. So this is continuous.

(c) time required to drive from home to college on any given day

You can take 10.5 minutes, for example, which is a decimal value. So this is continuous.

(d) number of ships in Pearl Harbor on any given day

There is no half ship, for example. So this is discrete.

(e) your weight before breakfast each morning continuous discrete

You can weigh 80.4kg, for example, which is a decimal number. So this is continuous.

a) Discrete

b) Continuous

c) Continuous

d) Discrete

e) Continuous

Discrete data is the numerical type of data which includes whole, concrete numbers that has specific and fixed data values. therefore, they can be determined by counting.

example: Number of students in the class, ect.,

Continuous type of data includes complex numbers or the varying data values that are measured over a specific time interval.

example: Height of students in a school, etc.,

(a) Number of traffic fatalities per year in the state of Florida.

This is a Discrete data, as it is a countable data and will be always a whole be number. Number of traffic fatalities per year will be 75, 150, 200, etc.

(b) distance a golf ball travels after being hit with a driver.

This is a Discrete data, as the data can be measures and will be not always be a whole number. distance a golf ball travels after being hit can be 1.25 meters,  10.9 meters, 21.1 meters, etc.

(c) time required to drive from home to college on any given day.

This is a Discrete data, as the data can be measures and will be not always be a whole number. time required to drive from home to college can be 5minutes 30 seconds, 15 minutes 26 seconds.

(d) number of ships in Pearl Harbor on any given day.

This is a Discrete data, as it is a countable data and will be always a whole number. number of ships in Pearl Harbor on any given day will be always 1, 10, 8, etc.

(e) your weight before breakfast each morning.

This is a Discrete data, as the data can be measures and will be not always be a whole number. weight can be 44.56 kgs., 52.3 kgs, etc.

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What is the product?

Answers

Answer: the third option is correct

Step-by-step explanation:

The first matrix is a 2 × 3 matrix while the second matrix is a 3 × 2 matrix. To get the product of both matrices, we would multiply each term in each row by the terms in the corresponding column and add.

1) row 1, column 1

1×2 + 3×3 + 1×4 = 2 + 9 + 4 = 15

2) row 1, column 2

1×-2 + 3×5 + 1×1 = - 2 + 15 + 1 = 14

3) row 2, column 1

-2×-2 + 1×3 + 0×4 = - 4 + 3 + 0 = - 1

4) row 2, column 2

- 2×-2 + 1×5 + 0×1 = 4 + 5 = 9

The solution becomes

15 14

- 1 9

Suppose the coffee industry claimed that the average adult drinks 1.7 cups of coffee per day. To test this​ claim, a random sample of 40 adults was​ selected, and their average coffee consumption was found to be 1.9 cups per day. Assume the standard deviation of daily coffee consumption per day is 0.6 cups. Using alphaequals0.10​, complete parts a and b below. a. Is the coffee​ industry's claim supported by this​ sample? Determine the null and alternative hypotheses.

Answers

Answer:

(a) No, the coffee industry's claim is not supported by this sample.

(b) Null hypothesis: The average adult drinks 1.7 cups of coffee per day.

Alternate hypothesis: The average adult drinks more than 1.7 cups of coffee per day.

Step-by-step explanation:

(a) Test statistic (z) = (sample mean - population mean) ÷ sd/√n

sample mean = 1.9 cups per day

population mean = 1.7 cups per day

sd = 0.6 cups per day

n = 40

z = (1.9 - 1.7) ÷ 0.6/√40 = 0.2 ÷ 0.095 = 2.11

The test is a one-tailed test. Using alpha (significance level) = 0.1, the critical value is 2.326.

Conclusion:

Reject the null hypothesis because the test statistic 2.11 falls within the rejection region of the critical value 2.326.

The coffee industry's claim is contained in the null hypothesis, hence it is not supported by the sample because the null hypothesis is rejected.

(b) A null hypothesis is a statement from a population parameter which is either rejected or accepted (fail to reject) upon testing. It is expressed using the equality sign.

An alternate hypothesis is also a statement from the population parameter which negates the null hypothesis and is accepted if the null hypothesis is rejected. It is expressed using any of the inequality signs.

In a sample of 60 electric motors, the average efficiency (in percent) was 85 and the standard deviation was 2. Section 05.01 Exercise 12.a - Compute confidence interval; Find necessary sample size Find a 95% confidence interval for the mean efficiency. Round the answers to three decimal places.

Answers

Answer:

95% confidence interval for the mean efficiency is [84.483 , 85.517].

Step-by-step explanation:

We are given that in a sample of 60 electric motors, the average efficiency (in percent) was 85 and the standard deviation was 2.

So, the pivotal quantity for 95% confidence interval for the population mean efficiency is given by;

          P.Q. = [tex]\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]

where, [tex]\mu[/tex] = sample average efficiency = 85

            [tex]\sigma[/tex] = sample standard deviation = 2

            n = sample of motors = 60

            [tex]\mu[/tex] = population mean efficiency

So, 95% confidence interval for the mean efficiency, [tex]\mu[/tex] is ;

P(-2.0009 < [tex]t_5_9[/tex] < 2.0009) = 0.95

P(-2.0009  < [tex]\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }[/tex] < 2.0009 ) = 0.95

P( [tex]-2.0009 \times {\frac{s}{\sqrt{n} }[/tex] < [tex]{\bar X - \mu}[/tex] < [tex]2.0009 \times {\frac{s}{\sqrt{n} }[/tex] ) = 0.95

P( [tex]\bar X -2.0009 \times {\frac{s}{\sqrt{n} }[/tex] < [tex]\mu[/tex] < [tex]\bar X +2.0009 \times {\frac{s}{\sqrt{n} }[/tex] ) = 0.95

95% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X -2.0009 \times {\frac{s}{\sqrt{n} }[/tex] , [tex]\bar X +2.0009 \times {\frac{s}{\sqrt{n} }[/tex] ]

                                                 = [ [tex]85 -2.0009 \times {\frac{2}{\sqrt{60} }[/tex] , [tex]85 +2.0009 \times {\frac{2}{\sqrt{60} }[/tex] ]

                                                 = [84.483 , 85.517]

Therefore, 95% confidence interval for the population mean efficiency is [84.483 , 85.517].

Final answer:

The 95% confidence interval for the mean efficiency of electric motors, given a sample mean of 85, a standard deviation of 2, and a sample size of 60, is approximately (84.494, 85.506) when rounded to three decimal places.

Explanation:

To calculate the 95% confidence interval for the mean efficiency of electric motors, we use the sample mean, standard deviation, and the sample size along with the z-score for the 95% confidence level. Since the sample size is large (n > 30), we can use the z-distribution to approximate the sampling distribution of the sample mean.

The formula for a confidence interval is:

Confidence Interval = sample mean "+/-" (z-score * (standard deviation / sqrt(n)))

Given that the sample mean is 85, the standard deviation is 2, and the sample size (n) is 60, and using the z-score of approximately 1.96 for 95% confidence, we can compute the confidence interval.

Confidence Interval = 85 "+/-" (1.96 * (2 / √(60)))

After the calculation, we get the two ends of the interval:

Lower Limit = 85 - (1.96 * (2 / √(60))) = 84.494

Upper Limit = 85 + (1.96 * (2 / √(60))) = 85.506

Rounding these to three decimal places, the 95% confidence interval for the mean efficiency of electric motors is approximately (84.494, 85.506). This means we can be 95% confident that the true average efficiency of all electric motors is between 84.494% and 85.506%.

Consider the differential equation y '' − 2y ' + 37y = 0; ex cos(6x), ex sin(6x), (−[infinity], [infinity]). Verify that the given functions form a fundamental set of solutions of the differential equation on the indicated interval.The functions satisfy the differential equation and are linearly independent sincea) W(ex cos 6x, ex sin 6x) =

Answers

The question  is:

Consider the differential equation y'' - 2y' + 37y = 0;

[tex]e^x cos(6x), e^x sin(6x), (-\infty, \infty).[/tex]

Verify that the given functions form a fundamental set of solutions of the differential equation on the indicated interval.

The functions satisfy the differential equation and are linearly independent since

a) [tex]W(e^x cos 6x, e^x sin 6x) \neq 0[/tex]

Answer:

To verify if the given functions form a fundamental set of solutions to the differential equation, we find the Wronskian of the two functions.

The Wronskian of functions [tex]y_1 $and$ y_2[/tex] is given as[tex]W(y_1, y_2) = \left|\begin{array}{cc}y_1&y_2\\y_1'&y_2'\end{array}\right|\\\\y_1= e^x cos 6x \\y_2 = e^x sin 6x \\y_1' = -6e^x sin 6x \\y_2' = 6e^x cos 6x \\W\left(e^x cos 6x, e^x sin 6x \right) = \left|\begin{array}{cc}e^x cos 6x &e^x sin 6x \\ \\ -6e^x sin 6x&6e^x cos 6x \end{array}\right|\\ \\= 6e^{2x} cos^2 6x + e^{2x} sin^2 6x \\ \\ = 6e^{2x}\left( cos^2 6x + sin^2 6x\right)\\ \\$but $cos^2 6x + sin^2 6x = 1\\ \\W\left(y_1, y_2 \right) = 6e^{2x} \neq 0[/tex]

Because the Wronskian is not zero, we say the solutions are linearly independent

Final answer:

The two functions form a fundamental set of solutions to the differential equation if both satisfy the differential equation and their Wronskian is non-zero, indicating they are linearly independent. This can be confirmed by computing the determinant.

Explanation:

The given differential equation is a second-order homogeneous equation. To confirm if the given functions form a fundamental set of solutions, they should satisfy the differential equation and should be linearly independent.

To solve the equation, first find the Wronskian, which is a determinant used to test whether solutions are linearly independent. For this particular case:

W(ex cos 6x, ex sin 6x) = det[[ex cos(6x)]', [ex cos(6x)]'; [ex sin(6x)]', [ex cos(6x)]'].  

If the resulting value does not equal zero, the solutions are indeed linearly independent and they form a fundamental set of solutions for the given differential equation over the entire real line.

To conclude, if both solutions satisfy the differential equation and W does not equal zero, then ex cos(6x) and ex sin(6x) are linearly independent and they form a fundamental set of solutions for the given differential equation.

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Your body loses sodium when you sweat. Researchers sampled 38 random tennis players. The average sodium loss was 500 milligrams per pound and the standard deviation was 62 milligrams per pound. Construct and interpret a 99% confidence interval to estimate the mean loss in sodium in the population.

Answers

Answer:

The 99% confidence interval to estimate the mean loss in sodium in the population is between 474.10 milligrams and 525.90 milligrams. This means that we are 99% that the true mean loss in sodium in the population is between 474.10 milligrams and 525.90 milligrams.

Step-by-step explanation:

We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1-0.99}{2} = 0.005[/tex]

Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].

So it is z with a pvalue of [tex]1-0.005 = 0.995[/tex], so [tex]z = 2.575[/tex]

Now, find M as such

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

[tex]M = 2.575*\frac{62}{\sqrt{38}} = 25.90[/tex]

The lower end of the interval is the mean subtracted by M. So it is 500 - 25.90 = 474.10 milligrams.

The upper end of the interval is the mean added to M. So it is 500 + 25.90 = 525.90 milligrams

The 99% confidence interval to estimate the mean loss in sodium in the population is between 474.10 milligrams and 525.90 milligrams. This means that we are 99% that the true mean loss in sodium in the population is between 474.10 milligrams and 525.90 milligrams.

Answer:

Option A

Step-by-step explanation:

The 99% confidence interval is (472.69, 527.31). We are 99% confident that the true population mean of sodium loss for tennis players will be between 472.69 milligrams per pound and 527.31 milligrams per pound.

Air crew escape systems are powered by a solid propellant. The burning rate of this propellant is an important product characteristic. Specifications require that the mean burning rate must be 50 centimeters per second. The experimenter selects a random sample of and obtains an observed sample mean of the burning rate 51.3 centimeters per second and an observed sample standard deviation of the burning rate 2.0 centimeters per second. Based on a 5% type I error threshold, construct a hypothesis test and draw your conclusion.

Answers

Answer:

Step-by-step explanation:

Hello!

The objective is to test if the mean burning rate of a propellant follows the specificated 50 cm/s.

The variable of interest is X: burning rate of a propellant (cm/s)

A sample of n= 25 was taken and a sample mean X[bar]= 51.3 cm/s and a sample standard deviation S= 2.0 cm/s were obtained.

Assuming that the variable has a normal distribution, the parameter of interest is the population mean, and the statistic hypotheses are:

H₀: μ = 50

H₁: μ ≠ 50

α: 0.05

Since there is no information about the population variance and the variable has a normal distribution the statistic to choose is a one-sample t-test:

[tex]t_{H_0}= \frac{X[bar]-Mu}{\frac{S}{\sqrt{n} } } = \frac{51.3-50}{\frac{2.0}{\sqrt{25} } } = 3.25[/tex]

The p-value corresponding to this test is p-value 0.003402

The decision for deciding using the p-value is:

If p-value ≤ α ⇒ Reject the null hypothesis.

If p-value > α ⇒ Not reject the null hypothesis.

The p-value is less than the significance level, the decision is to reject the null hypothesis. Using a level of significance of 5% you can conclude that the population mean burning rate of propellant is not 50 cm/s.

I hope it helps!

Let X1,X2......X7 denote a random sample from a population having mean μ and variance σ. Consider the following estimators of μ:

θ1=X1+X2+......+X7 / 7
θ2= (2X1-X3+X5) / 2

a. Is either estimator unbiased?
b. Which estimator is best? In what sense is it best? Calculate the relative efficiency of the 2 estimtors.

Answers

Answer:

a) In order to check if an estimator is unbiased we need to check this condition:

[tex] E(\theta) = \mu[/tex]

And we can find the expected value of each estimator like this:

[tex] E(\theta_1 ) = \frac{1}{7} E(X_1 +X_2 +... +X_7) = \frac{1}{7} [E(X_1) +E(X_2) +....+E(X_7)]= \frac{1}{7} 7\mu= \mu[/tex]

So then we conclude that [tex] \theta_1 [/tex] is unbiased.

For the second estimator we have this:

[tex] E(\theta_2) = \frac{1}{2} [2E(X_1) -E(X_3) +E(X_5)]=\frac{1}{2} [2\mu -\mu +\mu] = \frac{1}{2} [2\mu]= \mu[/tex]

And then we conclude that [tex]\theta_2[/tex] is unbiaed too.

b) For this case first we need to find the variance of each estimator:

[tex] Var(\theta_1) = \frac{1}{49} (Var(X_1) +...+Var(X_7))= \frac{1}{49} (7\sigma^2) = \frac{\sigma^2}{7}[/tex]

And for the second estimator we have this:

[tex] Var(\theta_2) = \frac{1}{4} (4\sigma^2 -\sigma^2 +\sigma^2)= \frac{1}{4} (4\sigma^2)= \sigma^2[/tex]

And the relative efficiency is given by:

[tex] RE= \frac{Var(\theta_1)}{Var(\theta_2)}=\frac{\frac{\sigma^2}{7}}{\sigma^2}= \frac{1}{7}[/tex]

Step-by-step explanation:

For this case we assume that we have a random sample given by: [tex] X_1, X_2,....,X_7[/tex] and each [tex] X_i \sim N (\mu, \sigma)[/tex]

Part a

In order to check if an estimator is unbiased we need to check this condition:

[tex] E(\theta) = \mu[/tex]

And we can find the expected value of each estimator like this:

[tex] E(\theta_1 ) = \frac{1}{7} E(X_1 +X_2 +... +X_7) = \frac{1}{7} [E(X_1) +E(X_2) +....+E(X_7)]= \frac{1}{7} 7\mu= \mu[/tex]

So then we conclude that [tex] \theta_1 [/tex] is unbiased.

For the second estimator we have this:

[tex] E(\theta_2) = \frac{1}{2} [2E(X_1) -E(X_3) +E(X_5)]=\frac{1}{2} [2\mu -\mu +\mu] = \frac{1}{2} [2\mu]= \mu[/tex]

And then we conclude that [tex]\theta_2[/tex] is unbiaed too.

Part b

For this case first we need to find the variance of each estimator:

[tex] Var(\theta_1) = \frac{1}{49} (Var(X_1) +...+Var(X_7))= \frac{1}{49} (7\sigma^2) = \frac{\sigma^2}{7}[/tex]

And for the second estimator we have this:

[tex] Var(\theta_2) = \frac{1}{4} (4\sigma^2 -\sigma^2 +\sigma^2)= \frac{1}{4} (4\sigma^2)= \sigma^2[/tex]

And the relative efficiency is given by:

[tex] RE= \frac{Var(\theta_1)}{Var(\theta_2)}=\frac{\frac{\sigma^2}{7}}{\sigma^2}= \frac{1}{7}[/tex]

Both θ1 and θ2 are unbiased estimators of the population mean μ. However, θ1 is more efficient due to having a lower variance compared to θ2. The relative efficiency of θ1 compared to θ2 is 10.5.

To determine if the estimators [tex](\theta_1)[/tex] and [tex](\theta_2)[/tex] are unbiased and which one is best, we follow these steps:

A. Evaluating Unbiasedness

Estimator [tex](\theta_1):\theta_1=\frac{X_1+X_2+\\cdots+X_7}{7}[/tex][tex]\(E[\theta_1]=\frac{E[X_1+X_2+\cdots+X_7]}{7}=\frac{7\\mu}{7}=\mu\)[/tex]Thus, [tex](\theta_1)[/tex] is an unbiased estimator of \\(\\mu\\).Estimator [tex](\theta_2=\frac{2X_1-X_3+X_5}{2})[/tex][tex]\(E[\theta_2]=E[\frac{2X_1-X_3+X_5}{2}]=\frac{2E[X_1]-E[X_3]+E[X_5]}{2}=\frac{2\mu-\mu+\mu}{2}=\mu\)[/tex]Thus,[tex](\theta_2)[/tex] is also an unbiased estimator of [tex](\mu)[/tex].

B. Evaluating Efficiency

Variance of [tex](\theta_1):\theta_1=\frac{X_1+X_2+\cdots+X_7}{7}[/tex][tex]\(Var(\theta_1)=Var(\frac{X_1+X_2+\cdots+X_7}{7})=\frac{\sigma^2}{7}\)[/tex]Variance of[tex](\theta_2):\theta_2=\frac{2X_1-X_3+X_5}{2})[/tex][tex]\(Var(\theta_2)=Var(\frac{2X_1-X_3+X_5}{2})=\frac{4Var(X_1)+Var(X_3)+Var(X_5)}{4}=\frac{4\sigma^2+\sigma^2+\sigma^2}{4}=\frac{6\sigma^2}{4}=1.5\sigma^2)[/tex]Since[tex](Var(\theta_1)=\frac{\sigma^2}{7}) is less than \(Var(\theta_2)=1.5\sigma^2), (\theta_1)[/tex] is the more efficient estimator

The relative efficiency of [tex](\theta_1)[/tex] to [tex](\theta_2)[/tex] is:

[tex]\(RE=\frac{Var(\theta_2)}{Var(\theta_1)}=\frac{1.5\sigma^2}{\frac{\sigma^2}{7}}=10.5\)[/tex]

Both [tex](\theta_1)[/tex] and [tex](\theta_2)[/tex] are unbiased estimators of [tex](\mu)[/tex], but [tex](\theta_1)[/tex] is the best in terms of efficiency since it has a lower variance.

A client arrives in the emergency department in a crisis state. The client demonstrates signs of profound anxiety and is unable to focus on anything but the object of the crisis and the impact on self. The initial data collection would focus on which information

Answers

Answer:The client's physical condition

Step-by-step explanation: The first or initial data that should be collected when a person have a symptoms of profound anxiety and is unable to focus on anything but the object of the crisis and the impact on self, the initial data to be collected would focus on the physical condition of the clients as it will give the adequate Background data from where other actions or data analysis will be done in order to evaluate and handle such a situation, this initial data will also assist the health care provider to know what is the root cause of his condition and what other steps are required.

A researcher believes that 9% of females smoke cigarettes. If the researcher is correct, what is the probability that the proportion of smokers in a sample of 703 females would differ from the population proportion by more than 3%? Round your answer to four decimal places.

Answers

Final answer:

To calculate the probability that the proportion of smokers in a sample of 703 females would differ from the population proportion by more than 3%, we can use the normal approximation to the binomial distribution. The probability is approximately 0.0030.

Explanation:

To calculate the probability that the proportion of smokers in a sample of 703 females would differ from the population proportion by more than 3%, we can use the normal approximation to the binomial distribution. The first step is to calculate the standard error, which is the square root of the product of the population proportion (0.09) and its complement (0.91), divided by the sample size (703). The formula for the standard error is:

SE = sqrt((p * (1-p))/n)

Substituting the values into the formula, we get:

SE = sqrt((0.09 * 0.91)/703) ≈ 0.0096

Next, we need to calculate the z-score, which measures how many standard errors the observed proportion is from the population proportion. The formula for the z-score is:

z = (observed proportion - population proportion) / standard error

Substituting the values into the formula:

z = (observed proportion - population proportion) / SE

The observed proportion is unknown in this case, but we can calculate the range within which it can fall. To differ from the population proportion by more than 3%, the observed proportion can be either 3% greater than the population proportion or 3% less than the population proportion. So the range is from (0.09 - 0.03) to (0.09 + 0.03), which is (0.06 to 0.12).

Calculating the z-score for the lower limit of the range:

z = (0.06 - 0.09) / 0.0096 ≈ -3.125

Calculating the z-score for the upper limit of the range:

z = (0.12 - 0.09) / 0.0096 ≈ 3.125

To find the probability that the proportion differs from the population proportion by more than 3%, we need to find the area under the standard normal curve beyond these z-scores. We can use a standard normal table or a calculator to find the probabilities. However, since the z-scores are symmetric, we only need to find the probability for one tail and double it. Using a standard normal table or calculator, we find that the probability of a z-score less than -3.125 or greater than 3.125 is approximately 0.001521. Doubling this probability gives us the final answer:

Probability = 2 * 0.001521 ≈ 0.003042

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Your large corporation manufactures a certain popular brand of robot lawn mower. After manufacture, all mowers go through a standard quality control check. The Portland factory, which manufactures 30% of your mowers, has a probability of .8 that each mower will pass the quality control check. The Dallas factory, which manufactures 50% of your mowers, has a probability of .7 that each mower will pass the quality control check. For the Buffalo factory,
which manufactures the remaining 20% of your mowers, you have been unable to learn what the probability is that each mower will pass inspection. But, the corporation’s annual report claims that the overall probability that one of their manufactured mowers will pass the quality control check is .72

a) What do you conclude is the probability that a mower manufactured at the Buffalo factory will pass the quality control check?
b) A customer orders one of your mowers, and of course receives one that has passed the quality control check. What is the probability it was manufactured in Dallas?

Answers

Answer:

(a) The probability that a mower manufactured at the Buffalo factory will pass the quality control check is 0.65.

(b) The probability that a mower was manufactured in Dallas given that it passes the quality check is 0.4861.

Step-by-step explanation:

Denote the events as follows:

X = a mower is manufactured at the Portland factory

Y = a mower is manufactured at the Dallas factory

Z= a mower is manufactured at the Buffalo factory

A = a mower passes the quality check.

The information provided is:

[tex]P(X)=0.30\\P(A|X)=0.80\\P(Y)=0.50\\P(A|Y)=0.70\\P(Z)=0.20\\P(A)=0.72[/tex]

(a)

The probability that a mower manufactured at the Buffalo factory will pass the quality control check is:

P (A|Z)

Compute the value of P (A|Z) as follows:

[tex]P(A)=P(A\cap X)+P(A\cap Y) + P (A\cap Z)\\0.72=(0.80\times0.30)+(0.70\times0.50)+(0.20\times P(A|Z))\\0.20\times P(A|Z)=0.72-0.24-0.35\\P(A|Z)=\frac{0.13}{0.20}\\=0.65[/tex]

Thus, the probability that a mower manufactured at the Buffalo factory will pass the quality control check is 0.65.

(b)

Compute the value of P (Y|A) as follows:

[tex]P(Y|A)=\frac{P(A|Y)P(Y)}{P(A)}=\frac{0.70\times0.50}{0.72}=0.4861[/tex]

Thus, the probability that a mower was manufactured in Dallas given that it passes the quality check is 0.4861.

Final answer:

The probability that a mower from the Buffalo factory will pass the quality control check is 13%, and if a customer receives a mower that has passed the check, there is a 49% probability that it was manufactured in Dallas.

Explanation:

To find the probability that a robot lawn mower made at the Buffalo factory will pass the quality control check, we first understand that the total probability of a mower passing the check is a sum of all the probabilities from the three factories. This is given as 0.72.

The Portland factory which makes 30% of the mowers has a 0.8 probability of passing the check. Therefore, the contribution of Portland to the total probability is 0.3*0.8 = 0.24. Similarly, the Dallas factory makes 50% of the mowers and each has a passing probability of 0.7, so the Dallas contribution is 0.5*0.7 = 0.35.  

Knowing this, we can subtract the total contributions of Portland and Dallas from the overall probability of 0.72 to get Buffalo's contribution, which is the Buffalo passing probability. Therefore, the Buffalo passing probability becomes 0.72 - 0.24 - 0.35 = 0.13 or 13%.

In the second part, if the customer receives a mower that has passed the quality control check, the probability that it was manufactured in Dallas is the contribution of the Dallas factory to the passing mowers i.e., 0.35 ÷ 0.72 = 0.486 or approximately 49%.

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Professor stan der Deviation can take one of two routes on his way home from work. On the first route, there are four railroad crossings. The probability that he will be stopped by a train at any particulare on the the crossings is .1, and trains operate independently at the four crossings. The other route is longer but there are only two crossings, also independent of one another, with the same stoppage probability for each as on the first route. On a particular day, Professor Deviation has a meeting scheduled at home for a certain time. Whichever route he takes, he calculates that he will be late if he is stopped by trains at at least half of the crossings encountered.

A.) Which route should he take to minimize the probability of being late to the meeting?
B.) If he tosses a fair coin to decide on a route and is late, what is the probability he took the four crossing route?

Answers

Answer:

A) He should take the route with 4 crossings

B) The probability that he took the 4 crossing route is 0.2158

Step-by-step explanation:

Lets call X the number of crossings he encounters, A if he takes route 1 and B if he takes route 2.

Note that X given A is a binomial random variable with parameters n = 4 p = 0.1, and X given B has parameters n = 2, p = 0.1

The probability that the Professor is on time on route 1 is equal to

P(X|A = 0) + P(X|A = 1) = 0.9⁴ + 0.9³*0.1*4 = 0.9477

On the other hand, the probability that the professor is on time on route 2 is

P(X|B = 0) = 0.9² = 0.81

Hence, it is more likely for the professor to be late on route 2, thus he should take the route 1, the one with 4 crossings.

B) Lets call L the event 'The professor is late'. We know that

P(L|A) = 1-0.9477 = 0.0524

P(L|B) = 1-0.81 = 0.19

Also

P(A) = P(B) = 1/2 (this only depends on the result of the coin.

For the Bayes theorem we know, therefore that

[tex]P(A|L) = \frac{P(L|A) * P(A)}{P(L|A)*P(A) + P(L|B)*P(B)} = \frac{0.0523*0.5}{0.0523*0.5 + 0.19*0.5 } = 0.2158[/tex]

Hence, the probability that he took the 4 crossing route is 0.2158.

Your grandmother has been putting $ 3 comma 000$3,000 into a savings account on every birthday since your first​ (that is, when you turned​ 1). The account pays an interest rate of 7 %7%. How much money will be in the account on your 18th birthday immediately after your grandmother makes the deposit on that​ birthday?

Answers

Answer: there would be $102000 in the account.

Step-by-step explanation:

We would apply the future value which is expressed as

FV = C × [{(1 + r)^n - 1}/r]

Where

C represents the yearly payments on each birthday.

FV represents the amount of money

in your account at the end of 18 years.

r represents the annual rate.

n represents number of years or period.

From the information given,

r = 7% = 7/100 = 0.07

C = $3000

n = 18 years

Therefore,

FV = 3000 × [{(1 + 0.07)^18 - 1}/0.07]

FV = 3000 × [{3.38 - 1}/0.07]

FV = 3000 × 34

FV = $102000

A tennis club offers two payment options:
Option1: $35 monthly fee plus $4/hour for court rental
Option 2: No monthly fee but $6.50/hour for court rental.
Let x = hours per month of court rental time.
a) Write a mathematical model representing the total monthly cost, C, in terms of x for the following:
Option 1: C= _________________
Option 2: C=_________________
b) How many hours would you have to rent the court so that the monthly cost of option 1, is less than option 2. Set up an inequality and show your work algebraically using the information in part a.

Answers

Multiply the cost per hour by number of hours x and for option 1 you need to add the monthly fee:

A) option 1: C= 4x +35

Option2: C = 6.50x

B)

4x+35 < 6.50x

Subtract 4x from both sides:

35 < 2.50x

Divide both sides by 2.50 :

X < 14

You would need to rent more than 14 hours for option 1 to be cheaper.

According to the American Lung Association, 7% of the population has lung disease. Of those having lung disease, 90% are smokers; and of those not having lung disease, 25% are smokers. What is the probability that a smoker has lung disease?

Answers

Answer:

Probability that a smoker has lung disease = 0.2132

Step-by-step explanation:

Let L = event that % of population having lung disease, P(L) = 0.07

So,% of population not having lung disease, P(L') = 1 - P(L) = 1 - 0.07 = 0.93

S = event that person is smoker

% of population that are smokers given they are having lung disease, P(S/L) = 0.90

% of population that are smokers given they are not having lung disease, P(S/L') = 0.25

We know that, conditional probability formula is given by;

                        P(S/L) = [tex]\frac{P(S\bigcap L)}{P(L)}[/tex]  

                        [tex]P(S\bigcap L)[/tex] = P(S/L) * P(L)

                                      = 0.90 * 0.07 = 0.063

So,  [tex]P(S\bigcap L)[/tex] = 0.063 .

Now, probability that a smoker has lung disease is given by = P(L/S)

      P(L/S) = [tex]\frac{P(S\bigcap L)}{P(S)}[/tex]

P(S) = P(S/L) * P(L) + P(S/L') * P(L')

       = 0.90 * 0.07 + 0.25 * 0.93 = 0.2955

Therefore, P(L/S) = [tex]\frac{0.063}{0.2955}[/tex] = 0.2132

Hence, probability that a smoker has lung disease is 0.2132 .

1. Using R, construct time series (line) plots for both stock prices and return series. R functions ts.plot or plot can create the plot. Describe the patterns and compare the plots.

Answers

Answer:

The code in R for the time series is given as below.

Step-by-step explanation:

As the complete question is not presented thus a sample code for the variables is given as below

# Libraries

library(ggplot2)

library(dplyr)

library(plotly)

library(hrbrthemes)

# Load dataset

filepath=" Your dataset path here in csv format"

data <- read.table("filepath", header=T)

data$date <- as.Date(data$date)

# plot

data %>%

 ggplot( aes(x=date, y1=value1,x=date, y2=value2)) +

   geom_line(color="#69b3a2") +

   ylim(0,22000) +

   theme_ipsum()

Final answer:

In R, stock prices and return series can be visualized using the ts.plot() or plot() functions in a line graph. The time series graph will then help in identifying and understanding trends and patterns in the data over time. Comparing the plots of the stock prices and the return series can provide insights into market behavior.

Explanation:

To plot both stock prices and return series using R, one would use either the ts.plot or plot function. Firstly, you need to import or generate the stock prices and the return series data. The horizontal axis of the plot corresponds to the date or time increments, while the vertical axis corresponds to the values of the variable (i.e., stock prices or return series).

Load your time series data in R.Next, use the ts.plot() or plot() functions to visualize your data in a line graph.Pay attention to the patterns that emerge. Line graphs are an effective method to show the relationships between two changing variables such as time and stock price.Detail the trends and patterns you observe in your data. Are there seasonal trends, cyclical fluctuations or random variations? Is the trend increasing or decreasing over time? Comparatively analyze the plots for the stock prices and the return series.

Time series graphs facilitate spotting trends and are used to graphically represent the same variable values recorded over an extended period of time. In the context of stock prices and return series, these patterns over time can provide valuable insights into market behavior and investment opportunities.

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A balloon is 200 ft. off the ground and rising vertically at the constant rate of 15 ft/sec. An automobile passes beneath it traveling along a straight road at the constant rate of 45 mph or 66 ft/sec. How fast is the distance between them changing 1 sec. later?

Answers

After 1 second has passed, the balloon will be 215 ft above the ground, and the automobile will be 66 ft away from its initial position directly below the balloon, forming a right triangle whose hypotenuse has length [tex]\sqrt{215^2+66^2}=\sqrt{50,581}[/tex] ft.

The distance between the balloon and automobile [tex]d[/tex] is related to the height of the balloon [tex]b[/tex] and the distance [tex]a[/tex] the automobile has moved away from its starting position by the Pythagorean formula,

[tex]d^2=a^2+b^2[/tex]

Differentiate both sides with respect to time [tex]t[/tex]:

[tex]2d\dfrac{\mathrm dd}{\mathrm dt}=2a\dfrac{\mathrm da}{\mathrm dt}+2b\dfrac{\mathrm db}{\mathrm dt}[/tex]

Plug in everything you know and solve for the rate you want:

[tex]2(\sqrt{50,581}\,\mathrm{ft})\dfrac{\mathrm dd}{\mathrm dt}=2(66\,\mathrm{ft})\left(66\dfrac{\rm ft}{\rm s}\right)+2(215\,\mathrm{ft})\left(15\dfrac{\rm ft}{\rm s}\right)[/tex]

[tex]\implies\dfrac{\mathrm dd}{\mathrm dt}\approx33.71\dfrac{\rm ft}{\rm s}[/tex]

Final answer:

To find the rate at which the distance between the balloon and the car changes after 1 second, we can use the Pythagorean theorem and differentiate the equation with respect to time.

Explanation:

To find the rate at which the distance between the balloon and the car changes, we can use the Pythagorean theorem. Let's call the distance between them D. We have D^2 = (200 + 15t)^2 + (66t)^2, where t is the time in seconds. To find how fast D is changing after 1 second, we can differentiate the equation with respect to t and substitute t = 1. Finally, we can solve for the rate at which D is changing.

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The concentration C C (in mg/cc) of a particular drug in a person's bloodstream is related to the time t t (in hours) after injection in such a way that as t t changes from 0 0 to 8 8, C C changes from 0.9 0.9 to 0.1 0.1. Find the average rate of change of C C with respect to t t.

Answers

Final answer:

The average rate of change of C with respect to t is -0.1 mg/cc per hour.

Explanation:

The average rate of change of a function is calculated by finding the difference in the function values and dividing it by the difference in the independent variable values. In this case, we need to calculate the average rate of change of C with respect to t.

We are given that as t changes from 0 to 8, C changes from 0.9 to 0.1. So, the difference in C is 0.1 - 0.9 = -0.8 and the difference in t is 8 - 0 = 8.

Therefore, the average rate of change of C with respect to t is (-0.8)/8 = -0.1 mg/cc per hour.

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Can someone please help.

Answers

Step-by-step explanation:

Vectors are parallel if they point in the same or opposite directions.  In other words, they have the same slope.

Part A.

m₁ = 1 / √3

m₂ = -1 / -√3 = 1 / √3

The vectors are parallel.

Part B.

m₁ = 3 / 2

m₂ = -2 / -3 = 2 / 3

The vectors are not parallel.

12. Wendy sliced a loaf of bread into 12
equal slices. She used 4 of the slices to
make sandwiches. What fraction of the
loaf of bread was left?

Answers

Answer:2/3

Step-by-step explanation:

Answer: 2/3 the

loaf of bread was left.

Step-by-step explanation:

Wendy sliced a loaf of bread into 12

equal slices. This means that the portion or size of each slice is 1/12.

She used 4 of the slices to

make sandwiches. This means that the portion of the loaf of bread that she used to make the sandwiches is

4 × 1/12 = 4/12 = 1/3

Therefore, the fraction of the

loaf of bread that was left is

1 - 1/3 = 2/3

If two events are independent, then Group of answer choices the sum of their probabilities must be equal to one. they must be mutually exclusive. their intersection must be zero. None of the above.

Answers

Answer:

For this case we can define the following two events A and B.

In order to classify A and B as independent we needd to satisfy this condition:

[tex] P(A \cap B) = P(A) *P(B)[/tex]

None of the above.

True, because none of the options were correct.

See explanation below

Step-by-step explanation:

For this case we can define the following two events A and B.

In order to classify A and B as independent we need to satisfy this condition:

[tex] P(A \cap B) = P(A) *P(B)[/tex]

So let's analyze one by one the possible options:

the sum of their probabilities must be equal to one.

False, the sum of the probabilities can be <1 so this statement is not true

they must be mutually exclusive.

False when we talk about mutually exclusive events we are saying that:

[tex] P(A \cap B) =0[/tex]

But independence not always means that we have mutually exclusive events

their intersection must be zero.

False the intersection of the probabilities is 0 just if we have mutually exclusive events, not independent events

None of the above.

True, because none of the options were correct.

Final answer:

Independent events refer to the concept in which the outcome of one event doesn't influence the outcome of another. If two events are independent, the probability of both events taking place is the product of their individual probabilities.

Explanation:

In mathematics, when we say that events are independent, it means that the outcome of one event does not affect the outcome of the other. The correct statement is: 'If two events are independent, the probability of both events happening is the product of the probabilities of each event.' When the sum of their probabilities is equal to one, they are mutually exclusive, not independent. And their intersection being zero also refers to mutually exclusive events not independent ones.

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What is the age distribution of promotion-sensitive shoppers? A supermarket super shopper is defined as a shopper for whom at least 70% of the items purchased were on sale or purchased with a coupon.Age range, years 18-28 29-39 40-50 51-61 62 and overMidpoint x 23 34 45 56 67Percent of super shoppers 6% 47% 21% 12% 14%For the 62-and-over group, use the midpoint 67 years.(a) Using the age midpoints x and the percentage of super shoppers, do we have a valid probability distribution? Explain.(b) Use a histogram to graph the probability(c) Compute the expected age μ of a super shopper. (Round your answer to two decimal places.)μ =

Answers

Answer:

(a) The probability distribution is valid.  

(b) The histogram is attached below.

(c) The mean is 33.47.

Step-by-step explanation:

(a)

A probability distribution is valid if:

P (X = x) ≥ 0 for all values of x.∑ P (X = x) = 1

From the probability distribution table it can be seen that the probability for all x is greater than 0.

Compute the sum of all probabilities as follow:

[tex]\sum P(X=x)=0.06+0.47+0.21+0.12+0.14=1[/tex]

Thus, the probability distribution is valid.  

(b)

The histogram is attached below.

(c)

The formula of mean of a discrete probability distribution is:

[tex]\mu=\sum x.P(X=x)[/tex]

Compute the mean as follows:

[tex]\mu=\sum x.P(X=x)\\=1.38+15.989.45+6.725+9.38\\=33.474\\\approx 33.47[/tex]

Thus, the mean is 33.47.

Other Questions
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