For the Data Set below, calculate the Variance to the nearest hundredth decimal place. (Do not use a coma in your answer) 175 349 234 512 638 549 500 611

Answer:

X = 87.5

Y = 137.5

Step-by-step explanation:

Let's X and Y be the xy-coordinates of the center warehouse.

We know that X is in between the x coordinates or the 2 plants:

50 < X < 150

Similarly Y is in between the y coordinates or the 2 plants:

100 < Y < 200

Using centroid method with the shipping units being weight we can have the following equations

250*50 + 150*150 = X*(250 + 150)

Hence X = (250*50 + 150*150)/(250+150) = 87.5

Similarly 250*100 + 150*200 = Y*(250 + 150)

Hence Y =  (250*100 + 150*200)/(250+150) = 137.5

Answer:

15

Step-by-step explanation:

We are given that

Number of alpha protein subunits=4

Number of beta protein subunits=2

Total number of protein sub-units=2+4=6

We have to find the number of different arrangements are there.

When r identical letters and y identical letters and total object are n then arrangements are

[tex]\frac{n!}{r!x!}[/tex]

n=6,r=2,x=4

By using the formula

Then, we get

Number of different arrangements =[tex]\frac{6!}{2!4!}[/tex]

Number of different arrangements=[tex]\frac{6\times 5\times 4!}{2\times 1\times 4!}[/tex]

Number of different arrangements=15

Hence, different arrangements are there= 15

Answer:

96.2% of slots meet these specifications.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 0.8750

Standard Deviation, σ = 0.0012

We are given that the distribution of width in inches of slots is a bell shaped distribution that is a normal distribution.

Formula:

[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]

P( widths between 0.8725 and 0.8775 inch)

[tex]P(0.8725 \leq x \leq 0.8775) = P(\displaystyle\frac{0.8725 - 0.8750}{0.0012} \leq z \leq \displaystyle\frac{0.8775-0.8750}{0.0012}) = P(-2.083 \leq z \leq 2.083)\\\\= P(z \leq 2.083) - P(z < -2.083)\\= 0.981 - 0.019 = 0.962 = 96.2\%[/tex]

[tex]P(0.8725 \leq x \leq 0.8775) = 96.2\%[/tex]

96.2% of slots meet these specifications.

The question asks for the proportion of slots meeting width specifications within a normal distribution with defined mean and standard deviation. We calculate the corresponding Z-scores for the lower and upper specification limits and then determine the probability of a slot falling within these limits.

The problem involves finding the proportion of slots that meet the specified width requirements in a normal distribution. In this case, the slot widths follow a normal distribution with a mean (μ) of 0.8750 inches, and a standard deviation (σ) of 0.0012 inches. The specifications require that slot widths be between 0.8725 inches and 0.8775 inches.

To find the proportion of slots that meet these specifications, we calculate the Z-scores for both the lower specification limit of 0.8725 and the upper specification limit of 0.8775. The Z-score formula is given by Z = (X - μ) / σ, where X is the value for which we want to find the Z-score.

For the lower limit, we have:

Z(lower) = (0.8725 - 0.8750) / 0.0012 = -2.083…

For the upper limit, we have:

Z(upper) = (0.8775 - 0.8750) / 0.0012 = 2.083…

Next, we use the standard normal distribution to find the probability corresponding to these Z-scores. The area under the curve between these two Z-scores represents the proportion of slots that are within the specifications. This can be found using standard normal distribution tables or a calculator with statistical functions.

Answer:

[tex] P(T<600)=P(Z< \frac{600-576}{144})=P(Z<0.167)=0.566[/tex]

Step-by-step explanation:

Previous concepts

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

[tex]P(X=x)=\lambda e^{-\lambda x}, x>0[/tex]

And 0 for other case. Let X the random variable that represent "The number of years a radio functions" and we know that the distribution is given by:

[tex]X \sim Exp(\lambda=\frac{1}{16})[/tex]

Or equivalently:

[tex]X \sim Exp(\mu=16)[/tex]

Solution to the problem

For this case we are interested in the total T, and we can find the mean and deviation for this like this:

[tex]\bar X =\frac{\sum_{i=1}^n X_i}{n}=\frac{T}{n}[/tex]

If we solve for T we got:

[tex] T= n\bar X[/tex]

And the expected value is given by:

[tex] E(T) = n E(\bar X)= n \mu= 16*36=576[/tex]

And we can find the variance like this:

[tex] Var(T) = Var(n\bar X)=n^2 Var(\bar X)= n^2 *\frac{\sigma^2}{n}=n \sigma^2[/tex]

And then the deviation is given by:

[tex]Sd(T)= \sqrt{n} \sigma=\sqrt{16} *36=144[/tex]

And the distribution for the total is:

[tex] T\sim N(n\mu, \sqrt{n}\sigma)[/tex]

And we want to find this probability:

[tex] P(T< 600)[/tex]

And we can use the z score formula given by:

[tex]z=\frac{T- \mu_T}{\sigma_T}[/tex]

And replacing we got this:

[tex] P(T<600)=P(Z< \frac{600-576}{144})=P(Z<0.167)=0.566[/tex]

Using the central limit theorem, the probability that the sum of the lifetimes of 16 light bulbs is less than 600 hours is found to be approximately 0.2514 after calculating the mean, standard deviation, and z-score for the sum.

To estimate the probability that the sum of the lifetimes of 16 light bulbs is less than 600 hours, we can use the central limit theorem. This theorem suggests that the sum (or average) of a large number of independent and identically distributed random variables will be approximately normally distributed, regardless of the original distribution of the variables. Here, each light bulb's lifetime is an exponential random variable with a mean of 36 hours.

First, we need to determine the mean (μ) and standard deviation (σ) of the sum of the lifetimes. For one light bulb, the mean is 36 hours, and since the standard deviation for an exponential distribution is equal to its mean, it is also 36 hours. For 16 light bulbs, the mean of the sum is 16 * 36 = 576 hours, and the standard deviation of the sum is √16 * 36 = 144 hours due to the square root rule for variances of independent sums.

To find the probability that the sum is less than 600 hours, we convert this to a standard normal distribution problem by calculating the z-score:

Z = (X - μ) / (σ/sqrt(n))
Z = (600 - 576) / (144/sqrt(16))
Z = 24 / 36
Z = 0.67

Now we look up the cumulative probability for a z-score of 0.67 using a standard normal distribution table or a calculator with normal distribution functions. The probability associated with a z-score of 0.67 is approximately 0.7486. Therefore, the probability that the sum of the lifetimes is less than 600 hours is 1 - 0.7486 = 0.2514.

Answer: a. −0.41

Step-by-step explanation:

The slope for the best-fitting linear equation is given by :-

[tex]b=\dfrac{SS_{xy}}{SS_x}[/tex]

where , [tex]SS_x[/tex] =sum of squared deviations from the mean of X.

[tex]SS_{xy}[/tex] = correlation between y and x in terms of the corrected sum of products.

As per given , we have

[tex]SS_x=10.00[/tex]

[tex]SS_{xy}=-16.32[/tex]

Then, the value of the slope for the best-fitting linear equation will be

[tex]b=\dfrac{-16.32}{40.00}=-0.408\approx -0.41[/tex]

Hence, the value of the slope for the best-fitting linear equation= -0.41

So the correct answer is a. −0.41 .

The value of the slope for the best-fitting linear equation is -0.41

The given parameters are:

[tex]SS_{xy} = -16.32[/tex] --- the correlation between y and x

[tex]SS_{x} = 40.00[/tex] --- the sum of squared deviations from the mean of X.

The slope (b) is calculated using the following formula

[tex]b = \frac{SS_{xy}}{SS_x}[/tex]

Substitute values for SSxy and SSx

[tex]b = \frac{-16.32}{40.00}[/tex]

Divide -16.32 by 40.00

[tex]b = -0.408[/tex]

Approximate

[tex]b = -0.41[/tex]

Hence, the value of the slope for the best-fitting linear equation is -0.41

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Answer:

B

Step-by-step explanation:

The retrospective or historic cohort story, is a longitudinal cohort story that considers a particular set of individuals that share the same exposure factor to ascertain its influence in the developments of an occurrence, which are compared with the other set or cohort which were not exposed to the same factors.

Retrospective studies have existed about the same time as prospective studies, hence their names.

Answer:

a. .8

Step-by-step explanation:

The point estimate of the difference between the means of Company 1 and Company 2 can be calculated as:

point estimate = mu1 - mu2 where

Therefore point estimate = $10.80- $10 =$ .8

Answer:

Convenience sampling. See explanation below.

Step-by-step explanation:

For this case they not use random sampling since all the individuals for the population are not included on the sampling frame, some individuals have probability of inclusion 0, because they are just using a charge for the call and some people would not answer the call.

Is not stratified sampling since we don't have strata clearly defined on this case, and other important thing is that in order to apply this method we need homogeneous strata groups and that's not satisfied on this case.

Is not systematic sampling since they not use a random number or a random starting point, and is not mentioned, they just use a call that is charge with 50 cents.

Is not cluster sampling since we don't have clusters clearly defined, and again in order to apply this method we need homogeneous characteristics on the clusters and an equal chance of being a part of the sample, and that's not satisfid again with the call charge used.

So then the only method that can be possible for this case is convenience sampling because they use a non probability sampling with some members of the potential population with probability of inclusion 0.

The sampling technique used in the given scenario is voluntary response sampling, where participants decide whether to take part in the survey. In this technique, participants chose to respond to the television survey by making a call. This method can be biased as the responses could lean towards those who hold strong views on the topic.

The sampling technique used in this scenario is referred as voluntary response sampling or self-selection sampling. In this method, participants themselves decide to participate or not, usually by responding to a call for participants. This often happens when surveys are disseminated widely such as through television or online. Since there was a call to answer "yes" or "no" for the question with a charge, individuals chose to participate by making a call. It is important to note that the main drawback of this technique is that it tends to be biased, as the sample could be skewed in favor of those who felt strongly about the topic.

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Answer:

Step-by-step explanation:

Hello!

You have two study variables

X₁: Time it takes to get to school by bus.

X₂: Time it takes to get to school by car.

Data:

Sample 1

Bus:(15,10,7,13,14,9,8,12,15,10,13,13,8,10,12,11,14,11,9,12)

n₁= 20

Mean X[bar]₁= 11.30

S₁= 2.39

Sample 2

Car:(5,8,7,6,9,12,11,10,9,6,8,10,13,12,9,11,10,7)

n₂= 18

Mean X[bar]₂= 9.06

S₂= 2.29

A.

To test if there is any difference between the times it takes to get to school using the bus or a car you need to compare the means of each population.

The condition needed to make a test for the difference between means is that both the independent population should have a normal distribution.

The sample sizes are too small to use an approximation with the CLT. You can test if the study variables have a normal distribution using different methods, and hypothesis test, using a QQ-plot or using the Box and Whiskers plot. The graphics are attached.

As you can see both samples show symmetric distribution, the boxes are proportioned, the second quantile (median) and the mean (black square) are similar and in the center of the boxes. The whiskers have the same length and there are no outliers. Both plots show symmetry centered in the mean consistent with a normal distribution. According to the plots you can assume both variables have a normal distribution.

The next step to select the statistic to test the population means is to check whether there is other population information available.

If the population variances are known, you can use the standard normal distribution.

If the population variances are unknown, the distribution to use is a Student's test.

If the unknown population variances are equal, you can use a t-test with a pooled sample variance.

If the unknown population variances are not equal, the t-test to use is the Welch approximation.

Using an F-test for variance homogeneity the p-value is 0.43 so at a 0.01 level, you can conclude that the population variances are equal.

The statistic to use is a pooled t-test.

B.

Degrees of freedom.

For each study variable, you can use a t-test with n-1 degrees of freedom.

For X₁ ⇒ n₁-1 = 20 - 1 = 19

For X₂ ⇒ n₂-1 = 18 = 17

For X₁ + X₂ ⇒ (n₁-1) + (n₂-1)= n₁ + n₂ - 2= 20 + 18 - 2= 36

C.

See above.

D.

The formula for the 99% confidence interval is:

(X[bar]₁ - X[bar]₂) ± [tex]t_{n_1+n_2-2; 1- \alpha /2}[/tex] * [tex]Sa\sqrt{\frac{1}{n_1} + \frac{1}{n_2} }[/tex]

[tex]Sa= \sqrt{\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2} }[/tex]

[tex]Sa= \sqrt{\frac{19*(2.39)^2+17*(2.29)^2}{36} }[/tex]

Sa= 2.34

[tex]t_{n_1+n_2-2; 1- \alpha /2}[/tex]

[tex]t_{36; 0.995}[/tex] = 2.72

(11.30 - 9.06) ± 2.72 * [tex]2.34\sqrt{\frac{1}{20} + \frac{1}{18} }[/tex]

[0.17;4.31]

With a 99% confidence level you'd expect that the difference between the population means of the time that takes to get to school by bus and car is contained in the interval [0.17;4.31].

E.

Couldn't find the original lesson to see what calculator is used.

F.

Same, no calculator available.

I hope it helps!

Answer:

this is nnot the answer i was looking for

Step-by-step explanation:

Answer:

The expected value is 1.8

Step-by-step explanation:

Consider the provided information.

Suppose there’s a 15%  chance of having 0 announcements, a 30% chance of having 1 announcement, a 25% chance of  having 2 announcements, a 20% chance of having 3 announcements, and a 10% chance of having 4  announcements.

[tex]\text{Expected Value}=a \cdot P(a) + b \cdot P(b) + c \cdot P(c) + \cdot\cdot[/tex]

Where a is the announcements and P(a) is the probability.

[tex]\text{Expected Value}=0\cdot 15\% + 1 \cdot 30\% + 2 \cdot 25\% + 3\cdot20\%+4\cdot10[/tex]

[tex]\text{Expected Value}=1 \cdot 0.30+2 \cdot 0.25 +3 \cdot 0.2 + 4\cdot 0.10[/tex]

[tex]\text{Expected Value}=1.8[/tex]

Hence, the expected value is 1.8

Answer:

Step-by-step explanation:

Final answer:

To find the nth Fibonacci number, dynamic programming stores previously calculated values in an array, which allows for efficient linear time computation by summing the two previous numbers to obtain the nth value.

Explanation:

The Fibonacci sequence is defined such that each number in the sequence is the sum of the two preceding ones, starting from 0 and 1. To calculate the nth Fibonacci number, denoted as Fib(n), we start by setting Fib(0) and Fib(1) equal to 0 and 1, respectively. For n ≥ 2, Fib(n) is defined recursively as Fib(n) = Fib(n - 1) + Fib(n - 2). A naive recursive algorithm could be inefficient due to repeated calculations. Using dynamic programming or memoization improves efficiency by storing intermediate results, thus avoiding unnecessary recalculations.

Computing Fibonacci Numbers Using Dynamic Programming

To compute the nth Fibonacci number using dynamic programming, we create an array or list to save previously computed Fibonacci numbers. The nth value, for instance Fib(8) = 21, is then easily found by summing up the n-1th and n-2th values from the array, which are already computed and stored. This approach leads to a time complexity that is linear, i.e., O(n), instead of exponential.

Answer:

Step-by-step explanation:

Feel free to ask if anything is unclear

Answer:

51.84%

Step-by-step explanation:

The percentage of data explained by regression line is assessed using R-square. Here, in the given scenario correlation coefficient r is given. We simply take square of correlation coefficient to get r-square. r-square=(-0.72)^2=0.5184=51.84%

Answer:

[tex]\cos (\frac{1}{t^3}-6)} + c[/tex]

Step-by-step explanation:

Given  function:

[tex]\int {\frac{3}{t^4}\sin (\frac{1}{t^3}-6)} \, dt[/tex]

Now,

let [tex]\frac{1}{t^3}-6[/tex] be 'x'

Therefore,

[tex]d(\frac{1}{t^3}-6)[/tex] = dx

or

[tex]\frac{-3}{t^4}dt[/tex] = dx

on substituting the above values in the equation, we get

⇒ ∫ - sin (x) . dx

or

⇒ cos (x) + c                      [ ∵ ∫sin (x) . dx = - cos (x)]

Here,

c is the integral constant

on substituting the value of 'x' in the equation, we get

[tex]\cos (\frac{1}{t^3}-6)} + c[/tex]

The matching is as follow:

a -> Statistic

b -> Data Sample

d -> Variable

e -> Parameter

f -> Population

a. Statistic: The proportion of the 380 observed people who buy the product

b. Data Sample: The 380 randomly selected people who are observed to see if they will buy the product

d. Variable: Purchase - Yes or No whether a person bought the product

e. Parameter: The proportion of all people in the company's region who buy the product

f. Population: All people in the marketing company's region

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The 380 randomly selected people are the 'Data Sample', the proportion of these who buy is a 'Statistic', all people in the region are the 'Population', the list of 380 Yes/No answers is the 'Variable', proportion of all people in the region who buy the product is 'Parameter', and yes/no answer for each person's purchase is also deemed a 'Variable'.

In this question, we are dealing with terms related to statistic studies. The 380 randomly selected people who are observed to see if they will buy the product represent the Data Sample. The proportion of the 380 observed people who buy the product is considered a Statistic. All people in the marketing company's region is the Population. The list of the 380 Yes or No answers to whether the person bought the product constitutes the Variable. The proportion of all people in the company's region who buy the product is an example of a Parameter. Lastly, the Purchase: Yes or No whether a person bought the product is the Variable.

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Answer:

0.23.

Step-by-step explanation:

We have been given that the weight of people on a college campus are normally distributed with mean 185 pounds and standard deviation 20 pounds.

First of all, we will find the z-score corresponding to sample score 200 using z-score formula.

[tex]z=\frac{x-\mu}{\sigma}[/tex], where,

[tex]z=[/tex] Z-score,

[tex]x=[/tex] Sample score,

[tex]\mu=[/tex] Mean,

[tex]\sigma=[/tex] Standard deviation.

[tex]z=\frac{200-185}{20}[/tex]

[tex]z=\frac{15}{20}[/tex]

[tex]z=0.75[/tex]

Now, we need to find [tex]P(z>0.75)[/tex]. Using formula  [tex]P(z>a)=1-P(z<a)[/tex], we will get:

[tex]P(z>0.75)=1-P(z<0.75)[/tex]

Using normal distribution table, we will get:

[tex]P(z>0.75)=1-0.77337 [/tex]

[tex]P(z>0.75)=0.22663 [/tex]

Round to nearest hundredth:

[tex]P(z>0.75)\approx 0.23[/tex]

Therefore, the probability that a person weighs more than 200 pounds is approximately 0.23.

Answer:the probability that a person weighs more than 200 pounds is 0.23

Step-by-step explanation:

Since the weight of people on a college campus are normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - u)/s

Where

x = weight of people on a college campus

u = mean weight

s = standard deviation

From the information given,

u = 185

s = 20

We want to find the probability that a person weighs more than 200 pounds. It is expressed as

P(x greater than 200) = P(x greater than 200) = 1 - P(x lesser than lesser than or equal to 200).

For x = 200,

z = (200 - 185)/20 = 0.75

Looking at the normal distribution table, the probability corresponding to the z score is 0.7735

P(x greater than 200) = 1 - 0.7735 = 0.23

Answer:

The procedure emphasizes the idea of the summation of one physical quantity. In this case, X.

Step-by-step explanation:

1. When we add fractions like these we do it simply by rewriting a new one, the summation of the numerators over the same denominator:

[tex]\frac{2}{3}X+\frac{4}{3})X=\frac{6}{3}X= 2X[/tex]

The procedure emphasizes the idea of the summation of one physical quantity, in this case, X.

2) This physical quantity x could be miles, oranges, gallons, etc.

Answer:

A. We can select a lower confidence level and increase the sample size.

Step-by-step explanation:

The length of a confidence interval is:

Direct proportional to the confidence interval. This means that the higher the confidence level, the higher the length of the interval is.

Inverse proportional to the size of the sample.This means that the higher the size of the sample, the lower, or narrower, the length of the interval is.

Which of the following could we do to produce higher precision in our estimates of the population proportion?

We want a narrower interval. So the correct answer is:

A. We can select a lower confidence level and increase the sample size.

Answer:

(a) Type II error

(b) Type I error

(c) It is analogous to choosing a lower value for a hypothesis test

(d) There will be more tendency of making type II error and less tendency of making type I error

Step-by-step explanation:

(a) The bank made a type II error because they accepted the null hypothesis when it is false

(b) The bank made a type I error because they rejected the null hypothesis when it is true

(c) By lowering the value for the hypothesis test, they give applicants who do not meet the initial cut-off point the benefit of doubt of repaying the loan thus increasing their chances of making more profit

(d) There will be more tendency of making type II error because the bank accepts the null hypothesis though they are not fully convinced the applicants will repay the loan and less tendency of making type I error because the bank rejects the null hypothesis knowing the applicants might not be able to repay the loan

In hypothesis testing, a person defaulting on a loan represents a Type I error, while missing an opportunity to make a loan to someone who would have repaid it represents a Type II error. Lowering the cut-off score is analogous to increasing the value in a hypothesis test, accepting more risk. This increases the likelihood of Type I errors but decreases the likelihood of Type II errors.

In the context of hypothesis testing in banking and the financial capital market, (a) when a person defaults on a loan, the bank made a Type I error: they lent money to an individual who failed to repay it. (b) If the bank does not lend money to someone who would have repaid it, it's a Type II error: they missed an opportunity to profit from interest because they incorrectly predicted the person would not pay back the loan. (c) Lowering the cut-off score from 250 points to 200 is analogous to choosing a higher value for a hypothesis test, which means the bank is willing to accept more risk. (d) Changing the cut-off value impacts the chance of each kind of error. By lowering the score, the bank is more likely to make Type I errors (lending to individuals who won't repay), but less likely to make Type II errors (not lending to individuals who would repay).

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Answer:

a) [tex]\alpha=0.01[/tex] is the significance level given

b) [tex]z=\frac{17.1-18}{\frac{5.1}{\sqrt{14}}}=-0.6603[/tex]    

c) Since is a one side left tailed test the p value would be:  

[tex]p_v =P(Z<-0.6603)=0.2545[/tex]  

Step-by-step explanation:

Data given and notation  

[tex]\bar X=17.1[/tex] represent the mean P/E ratio for the sample  

[tex]\sigma=5.1[/tex] represent the sample standard deviation for the population  

[tex]n=14[/tex] sample size  

[tex]\mu_o =18[/tex] represent the value that we want to test

[tex]\alpha=0.01[/tex] represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

[tex]p_v[/tex] represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean for the P/E ratio is less than 18, the system of hypothesis would be:  

Null hypothesis:[tex]\mu \geq 18[/tex]  

Alternative hypothesis:[tex]\mu < 18[/tex]  

If we analyze the size for the sample is < 30 but we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex]  (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

(a) What is the level of significance?

[tex]\alpha=0.01[/tex] is the significance level given

(b) What is the value of the sample test statistic?

We can replace in formula (1) the info given like this:  

[tex]z=\frac{17.1-18}{\frac{5.1}{\sqrt{14}}}=-0.6603[/tex]    

(c) Find (or estimate) the P-value. (Round your answer to four decimal places.)

Since is a one side left tailed test the p value would be:  

[tex]p_v =P(Z<-0.6603)=0.2545[/tex]  

Conclusion  

If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL reject the null hypothesis, so we can conclude that the true mean for the P/E ratio is not significantly less than 18.  

Answer:

a) [tex]v(t) = 3t^{2} - 16t + 2[/tex]

b) The velocity after 3 seconds is -3m/s.

c) [tex]t = 0.13s[/tex] and [tex]t = 5.2s[/tex].

Step-by-step explanation:

The position is given by the following equation.

[tex]s(t) = t^{3} - 8t^{2} + 2t[/tex]

(a) Find the velocity at time t.

The velocity is the derivative of position. So:

[tex]v(t) = s^{\prime}(t) = 3t^{2} - 16t + 2[/tex].

(b) What is the velocity after 3 seconds?

This is v(3).

[tex]v(t) = 3t^{2} - 16t + 2[/tex]

[tex]v(3) = 3*(3)^{2} - 16*(3) + 2 = -19[/tex]

The velocity after 3 seconds is -3m/s.

(c) When is the particle at rest?

This is when [tex]v(t) = 0[/tex].

So:

[tex]v(t) = 3t^{2} - 16t + 2[/tex]

[tex]3t^{2} - 16t + 2 = 0[/tex]

This is when [tex]t = 0.13s[/tex] and [tex]t = 5.2s[/tex].

Answer:

The volume of the solid is 256 cubic units.

Step-by-step explanation:

Given:

The solid lies between planes [tex]x=0\ and\ x=8[/tex]

The cross section of the solid is a square with diagonal length equal to the distance between the parabolas [tex]y=-2\sqrt{x}\ and\ y=2\sqrt{x}[/tex].

The distance between the parabolas is given as:

[tex]D=2\sqrt x-(-2\sqrt x)\\\\D=2\sqrt x+2\sqrt x\\\\D=4\sqrt x[/tex]

Now, we know that, area of a square with diagonal 'D' is given as:

[tex]A=\frac{D^2}{2}[/tex]

Plug in [tex]D=4\sqrt x[/tex]. This gives,

[tex]A=\frac{(4\sqrt x)^2}{2}\\\\A=\frac{16x}{2}\\\\A=8x[/tex]

Now, volume of the solid is equal to the product of area of cross section and length [tex]dx[/tex]. So, we integrate it over the length from [tex]x=0\ to\ x=8[/tex]. This gives,

[tex]V=\int\limits^8_0 {A} \, dx\\\\V=\int\limits^8_0 {(8x)} \, dx\\\\V=8\int\limits^8_0 {(x)} \, dx\\\\V=8(\frac{x^2}{2})_{0}^{8}\\\\V=4[8^2-0]\\\\V=4\times 64\\\\V=256\ units^3[/tex]

Therefore, the volume of the solid is 256 cubic units.

This question is about volume calculation using calculus. The solid between two planes at x=0 and x=8 has cross-sections which, when described by a function of x A(x), the volume of the object can be computed via integration of A(x) dx from x=0 to x=8.

The subject of this question falls under the field of Calculus, specifically, it's about Volume Calculation. The question describes a solid which is located between two planes at x=0 and x=8, perpendicular to the x-axis. Cross-sections perpendicular to the axis of this solid can be visualized like slices of the solid made along the x-axis.

If the area of these cross-sections can be represented by a function of x, A(x), then the volume of the entirety of the solid, V, can be calculated using the definite integral from x=0 to x=8 of A(x) dx. Essentially, this is summing up the volumes of the infinitesimal discs that make up the solid along the x-axis, from x=0 to x=8.

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Answer:

The point estimate for population mean is 8.129 Mpa.

Step-by-step explanation:

We are given the following in the question:

Data on flexural strength(MPa) for concrete beams of a certain type:

11.8, 7.7, 6.5, 6.8, 9.7, 6.8, 7.3, 7.9, 9.7, 8.7, 8.1, 8.5, 6.3, 7.0, 7.3, 7.4, 5.3, 9.0, 8.1, 11.3, 6.3, 7.2, 7.7, 7.8, 11.6, 10.7, 7.0

a) Point estimate of the mean value of strength for the conceptual population of all beams manufactured

We use the sample mean, [tex]\bar{x}[/tex] as the point estimate for population mean.

Formula:

[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]

[tex]\bar{x} = \dfrac{\sum x_i}{n} = \dfrac{219.5}{27} = 8.129[/tex]

Thus, the point estimate for population mean is 8.129 Mpa.

To estimate the mean flexural strength, the sum of strengths (219.5 MPa) is divided by the total number of beams measured (26), which yields a mean value of 8.442 MPa when rounded to three decimal places. The estimator used is the sample mean.

To calculate a point estimate of the mean value for flexural strength (MPa) for a conceptual population of concrete beams, we use the sum of all measured strengths and divide by the number of measurements. The sum of the flexural strengths is provided as Σxi = 219.5 MPa.

Given the dataset:

The number of measurements is the number of data points, which is 26. To find the mean:

mean = Sum of strengths / Number of measurements

mean = 219.5 MPa / 26

mean = 8.442 MPa (rounded to three decimal places)

The estimator used here is the sample mean (×).

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Answer: (143.07, 158.93)

Step-by-step explanation:

The formula to find the confidence interval is given by :-

[tex]\overline{x}\pm z^*\dfrac{\sigma}{\sqrt{n}}[/tex]

where n= sample size

[tex]\overline{x}[/tex] = Sample mean

z* = critical z-value (two tailed).                

[tex]\sigma[/tex] = Population standard deviation

We assume that the underlying population distribution is normal.

As per given , we have

n= 108

[tex]\overline{x}=151[/tex]

[tex]\sigma=32[/tex]

Critical value for 99% confidence level = 2.576  (By using z-table)

Then , the 99% confidence interval for the population mean hours spent watching television per month :-

[tex]151\pm (2.576)\dfrac{32}{\sqrt{108}}[/tex]

[tex]151\pm (2.576)\dfrac{32}{10.3923048454}[/tex]

[tex]151\pm (2.576)(3.07920143568)[/tex]

[tex]151\pm (7.93202289831)\approx151\pm7.93\\\\=(151-7.93,\ 151+7.93)\\\\=(143.07,\ 158.93 )[/tex]

Hence, the required 99% confidence interval for the population mean hours spent watching television per month. = (143.07, 158.93)

The 99% confidence interval for the average number of hours all Americans spend watching television per month, based on the given sample, is (143.76, 158.23). This is computed using the confidence interval formula with the given sample mean, standard deviation, and the z-score for a 99% confidence interval.

The question involves the concept of the confidence interval in statistics. Here we are given the sample size (n=108), the sample mean ([tex]\overline{X}[/tex] = 151), and the sample standard deviation (s=32). We are required to compute the 99% confidence interval.

To calculate a confidence interval, we apply this formula: [tex]\overline{X}[/tex] ± (z-value * (s/√n)) Where '[tex]\overline{X}[/tex]' is the sample mean, 'z-value' is the Z-score (which for a 99% confidence interval is 2.58), 's' is the standard deviation and 'n' is the sample size.

Substitute the given values into the formula: 151 ± (2.58 * (32/√108))

This results in: (143.76, 158.23)

So, we can say with 99% confidence that the average number of hours all Americans spend watching television per month is between 143.76 hours and 158.23 hours.

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Answer:

The expected number of vehicles owned to two decimal places is: 1.85.

Step-by-step explanation:

The table to the question is attached.

[tex]E(X) =[/tex]∑[tex]xp(x)[/tex]

Where:

E(X) = expected number of vehicles owned

∑ = Summation

x = number of vehicle owned

p(x) = probability of the vehicle owned

[tex]E(X) = (0 * 0.1) + (1 * 0.35) + (2 * 0.25) + (3 * 0.2) + (4 * 0.1)\\E(X) = 0 + 0.35 + 0.50 + 0.60 + 0.4\\E(X) = 1.85[/tex]

The expected number of vehicles owned is 1.85.

The expected number of vehicles owned, based on probability of ownership of 0 to 3 vehicles, is calculated by multiplying each possible number of vehicles by their corresponding probabilities and then summing up all the products. The calculated expected number is approximately 1.7 vehicles.

To find the expected number of vehicles owned, we first need to multiply each possible number of vehicles someone could own by the probability of them owning that many vehicles. Then, sum up all of these products.

For instance, if they could own up to 3 cars and the probability for owning 0, 1, 2, or 3 cars is 0.1, 0.3, 0.4, and 0.2 respectively:

For 0 cars: 0 * 0.1 = 0

For 1 car: 1 * 0.3 = 0.3

For 2 cars: 2 * 0.4 = 0.8

For 3 cars: 3 * 0.2 = 0.6    

Adding these together gives the expected number of cars:
0 + 0.3 + 0.8 + 0.6 = 1.7 (rounded to two decimal places).

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Answer:

test statistic is 4.27

Step-by-step explanation:

[tex]H_{0}[/tex] : mean waiting time in a residential district branch is the same as a commercial district branch

[tex]H_{a}[/tex] : mean waiting time in a residential district branch is more than a commercial district branch

commercial district branch:

mean waiting time:  [tex]\frac{4.14+5.66+3.04+5.34+4.82+2.69+3.32+3.41+4.42+6.01+0.15+5.11+6.59+6.43+3.72}{15} =4.32[/tex]

standard deviation:

mean squared differences from the mean = 1.63

residential district branch.

mean waiting time:  [tex]\frac{9.99+5.89+8.06+5.91+8.64+3.77+8.21+8.52+10.46+6.87+5.53+4.23+6.25+9.88+5.59}{15} =7.19[/tex]

standard deviation:

mean squared differences from the mean = 2.03

test statstic can be calculated using the formula:

[tex]z=\frac{X-Y}{\sqrt{\frac{s(x)^2}{N(x)}+\frac{s(y)^2}{N(y)}}}[/tex] where

[tex]z=\frac{7.19-4.32}{\sqrt{\frac{2.03^2}{15}+\frac{1.63^2}{15}}}[/tex] ≈4.27

Final answer:

The critical value for constructing a 99% confidence interval is 2.576.

Explanation:

To determine the critical value for constructing the 99% confidence interval, we need to find the Z-value that represents the level of confidence. For a 99% confidence interval, the alpha level (1 - confidence level) is 0.01. Since the data is approximately normally distributed and the sample size is greater than 30, we can use the Z-distribution. Using a Z-table or calculator, we find that the Z-value for a 0.01 alpha level is approximately 2.576.

Answer: d) 15 to 25

Step-by-step explanation:

Given : Sample size : n= 16

Degree of freedom = df =n-1 = 15

Sample mean : [tex]\overline{x}=20[/tex]

sample standard deviation : [tex]s= 12[/tex]

Significance level : [tex]\alpha= 1-0.90=0.10[/tex]

Since population standard deviation is unavailable , so the confidence interval for the population mean is given by:-

[tex]\overline{x}\pm t_{\alpha/2, df}\dfrac{s}{\sqrt{n}}[/tex]

Using t-distribution table , we have

Critical value = [tex]t_{\alpha/2, df}=t_{0.05 , 15}=1.7530[/tex]

90% confidence interval for the mean value will be :

[tex]20\pm (1.7530)\dfrac{12}{\sqrt{16}}[/tex]

[tex]20\pm (1.7530)\dfrac{12}{4}[/tex]

[tex]20\pm (1.7530)(3)[/tex]

[tex]20\pm (5.259)[/tex]

[tex](20-5.259,\ 20+5.259)[/tex]

[tex](14.741,\ 25.259)\approx(15,\ 25 )[/tex][Round to the nearest integer]

Hence, the 90% confidence interval (to the nearest integer) for the standard deviation of the random variable.= 15 to 25.

Final answer:

To obtain a 90% confidence interval for the standard deviation of a normally distributed random variable with a sample size of 16, sample mean of 20, and sample standard deviation of 12, use the chi-square distribution to calculate the lower and upper bounds. The 90% confidence interval is approximately 86 to 283.

Explanation:

To obtain a 90% confidence interval for the standard deviation of a normally distributed random variable, we can use the chi-square distribution. Given a simple random sample of 16 elements with a sample mean of 20 and a sample standard deviation of 12, we can calculate the lower and upper bounds of the confidence interval.

Step 1: Calculate the chi-square values for the lower and upper bounds using the following formulas:

Lower bound: (n-1)s² / X², where n is the sample size, s is the sample standard deviation, and X² is the chi-square value for a 90% confidence level with (n-1) degrees of freedom.

Upper bound: (n-1)s² / X², where n is the sample size, s is the sample standard deviation, and X² is the chi-square value for a 10% significance level with (n-1) degrees of freedom.

Substituting the values into the formulas, we get:

Lower bound: (15)(144) / 24.996 = 86.437

Upper bound: (15)(144) / 7.633 = 283.368

Rounding to the nearest integer, the 90% confidence interval for the standard deviation of the random variable is approximately 86 to 283.

Answer:

The number of cellphones to be produced per week is 500.

The cost of each cell phone is $250.

The maximum revenue is $1,25,000

Step-by-step explanation:

We are given the following information in the question:

The weekly​ price-demand equation:

[tex]p(x)=500-0.5x[/tex]

The cost equation:

[tex]C(x) = 25000+140x[/tex]

The revenue equation can be written as:

[tex]R(x) = p(x)\times x\\= (500-0.5x)x\\= 500x - 0.5x^2[/tex]

To find the maximum value of revenue, we first differentiate the revenue function:

[tex]\displaystyle\frac{dR(x)}{dx} = \frac{d}{dx}(500x - 0.5x^2) = 500-x[/tex]

Equating the first derivative to zero,

[tex]\displaystyle\frac{dR(x)}{dx} = 0\\\\500-x = 0\\x = 500[/tex]

Again differentiating the revenue function:

[tex]\displaystyle\frac{dR^2(x)}{dx^2} = \frac{d}{dx}(500 - x) = -1[/tex]

At x = 500,

[tex]\displaystyle\frac{dR^2(x)}{dx^2} < 0[/tex]

Thus, by double derivative test, R(x) has the maximum value at x = 500.

So, the number of cellphones to be produced per week is 500, in order to maximize the revenue.

Price of phone:

[tex]p(500)=500-0.5(500) = 250[/tex]

The cost of each cell phone is $250.

Maximum Revenue =

[tex]R(500) = 500(500) - 0.5(500)^2 = 125000[/tex]

Thus, the maximum revenue is $1,25,000