A certain paper suggested that a normal distribution with mean 3,500 grams and a standard deviation of 560 grams is a reasonable model for birth weights of babies born in Canada.
One common medical definition of a large baby is any baby that weighs more than 4,000 grams at birth.
What is the probability that a randomly selected Canadian baby is a large baby?

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.

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.

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.

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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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.

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

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.

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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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

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.

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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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!

Answer:

Step-by-step explanation:

18 a)Since line DE is parallel to line BC, it means that triangle ADE is similar to triangle ABC. Therefore,

AB/AD = AC/AE = BC/DE

AC = AE + CE = 18 + 9

AC = 27

AB = AD + DB

AB = AD + 5

Therefore,

27/18 = (AD + 5)/AD

Cross multiplying, it becomes

27 × AD = 18(AD + 5)

27AD = 18AD + 90

27AD - 18AD = 90

9AD = 90

AD = 90/9

AD = 10

b) AC = AE + EC = 13 + 3

AC = 16

To find AD,

16/13 = 24/AD

16 × AD = 13 × 24

16AD = 312

AD = 312/16

AD = 19.5

DB = 24 - 19.5

DB = 4.5

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.).

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

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

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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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.

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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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.

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()

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.

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).

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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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]

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.

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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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.

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.

Answer:

rm(list=ls(all=TRUE))

set.seed(12345)

N=c(10,100,500)

Rate=0.2

B=1000

MN=SE=rep()

for(i in 1:length(N))

{

n=N[i]

X=rexp(n,rate=Rate)

EST=1/mean(X)

ESTh=rep()

for(j in 1:B)

{

Xh=rexp(n,rate=EST)

ESTh[j]=1/mean(Xh)

}

MN[i]=mean(ESTh)

SE[i]=sd(ESTh)

}

cbind(N,Rate,MN,SE)

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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Answer:

The confidence interval is 27.5 hg less than mu less than 29.1 hg

(A) Yes, because the confidence interval limits are not similar.

Step-by-step explanation:

Confidence interval is given as mean +/- margin of error (E)

mean = 28.3 hg

sd = 6.1 hg

n = 202

degree of freedom = n-1 = 202-1 = 201

confidence level (C) = 95% = 0.95

significance level = 1 - C = 1 - 0.95 = 0.05 = 5%

critical value corresponding to 201 degrees of freedom and 5% significance level is 1.97196

E = t×sd/√n = 1.97196×6.1/√202 = 0.8 hg

Lower limit = mean - E = 28.3 0.8 = 27.5 hg

Upper limit = mean + E = 28.3 + 0.8 = 29.1 hg

95% confidence interval is (27.5, 29.1)

When mean is 28.3, sd = 6.1 and n = 202, the confidence limits are 27.5 and 29.1 which is different from 27.8 and 29.6 which are the confidence limits when mean is 28.7, sd = 1.8 and n = 17

The confidence intervals for the two experiments have overlapping intervals, suggesting that the results are not very different.

The first experiment resulted in a 95% confidence interval of 3.070-3.164 g for the population mean weight of newborn girls. The second experiment had a 95% confidence interval of 3.035-3.127 g. Although the two confidence intervals are not identical, the mean for each experiment is within the confidence interval of the other experiment. This suggests that the results are not very different and there is an appreciable overlap between the two intervals. Therefore, the answer is C. No, because each confidence interval contains the mean of the other confidence interval.

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.

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

B. Evaluating Efficiency

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

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.

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

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 .

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.

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.

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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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%.

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.

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.

Answer:

Perimeter = 72

Step-by-step explanation:

Mathematics Puzzle

The area of a rectangle is the product of the base and the height

A=b.h

We know the base and the height are two consecutive prime numbers, not much of useful information so far.

We also know the area is composed only of the two smallest prime digits. Those digits are 2 and 3. If the number is reversed and it's not changed, then we only have two possible values for the area: 232 and 323.

We only need to find two consecutive prime numbers which product is one of the above. The number 232 has no prime factors: 232 = 8*29.

The number 323 is the product of 17 and 19, two consecutive prime numbers, thus the dimensions of the rectangle are 17 and 19.

The perimeter of that rectangle is 2*17+2*19= 72

Answer:

72 units

Step-by-step explanation:

72 units

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