A researcher wants to estimate the mean weekly family expenditure on clothing purchases and maintenance. First, she needs to determine the number of families that must be sampled, in order to estimate the mean weekly expenditure on clothing to within $15 at alpha = 0.01. From prior similar studies she is confident that the standard deviation of weekly family expenditures on clothing purchases and maintenance is $125. The required sample size to be taken is___________.

A. 21
B. 22
C. 461
D. 460
E. 460.46

Answer:

a) [tex]P(t) = 2e^{0.275t}[/tex]

b) 54.225 square centimeters.

c) 2.52 hours

Step-by-step explanation:

The population growth law is:

[tex]P(t) = P_{0}e^{rt}[/tex]

In which P(t) is the population after t hours, [tex]P_{0}[/tex] is the initial population and r is the growth rate, as a decimal.

In this problem, we have that:

The colony occupied 2 square centimeters initially, so [tex]P_{0} = 2[/tex]

The colony triples every 4 hours. So

[tex]P(4) = 3P_{0} = 6[/tex]

(a) An expression for the size P(t) of the colony at any time t.

We have to find the value of r. We can do this by using the P(4) equation.

[tex]P(t) = P_{0}e^{rt}[/tex]

[tex]6 = 2e^{4r}[/tex]

[tex]e^{4r} = 3[/tex]

Applying ln to both sides, we get:

[tex]4r = 1.1[/tex]

[tex]r = 0.275[/tex]

So

[tex]P(t) = 2e^{0.275t}[/tex]

(b) The area occupied by the colony after 12 hours.

[tex]P(t) = 2e^{0.275t}[/tex]

[tex]P(12) = 2e^{0.275*12}[/tex]

[tex]P(12) = 54.225[/tex]

(c) The doubling time for the colony?

t when [tex]P(t) = 2P_{0} = 2*2 = 4[/tex].

[tex]P(t) = 2e^{0.275t}[/tex]

[tex]4 = 2e^{0.275t}[/tex]

[tex]e^{0.275t} = 2[/tex]

Applying ln to both sides

[tex]0.275t = 0.6931[/tex]

[tex]t = 2.52[/tex]

Answer:

Step-by-step explanation:

The distance of each lap around pavia park is 1 7/8 miles. Converting

1 7/8 miles to improper fraction, it becomes 15/8 miles.

Ellen rode her bike for 3 1/2 laps before leaving the park. Converting

3 1/2 laps to improper fraction, it becomes 7/2 laps.

The total number of miles that Ellen rode her bike in pavia park would be the product of the distance of each lap and the number of laps that he covered. It becomes

15/8 × 7/2 = 105/16 = 6.5626 miles

Answer:

Step-by-step explanation:

If a number is evenly divisible by another number, it means that the number divides it completely without a remainder.

6) 44 is evenly divisible by 2 and 4. 44 divided by 3 and 6 would have remainders.

7) 38 is evenly divisible by 2. There would be remainders if 38 divides 3, 4 or 6

8) 726 is evenly divisible by 2, 3 and 6. It is not evenly divisible by 4

9) 2112 is evenly divisible by 2, 3 4 and 6

10) 1221 is evenly divisible by 3. It is not evenly divisible by 2 4 and 6

Answer:

[tex]t=\frac{188-200}{\frac{28}{\sqrt{40}}}=-2.7105[/tex]    

[tex]p_v =2*P(t_{39}<-2.7105)=0.004967[/tex]    

Step-by-step explanation:

Data given and notation    

[tex]\bar X=188[/tex] represent the sample mean    

[tex]s=28[/tex] represent the sample standard deviation    

[tex]n=40[/tex] sample size    

[tex]\mu_o =2-0[/tex] represent the value that we want to test    

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

t 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 apply a two tailed tailed test.    

What are H0 and Ha for this study?    

Null hypothesis:  [tex]\mu = 200[/tex]    

Alternative hypothesis :[tex]\mu \neq 200[/tex]    

Compute the test statistic  

The statistic for this case is given by:    

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

t-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".    

Calculate the statistic    

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

[tex]t=\frac{188-200}{\frac{28}{\sqrt{40}}}=-2.7105[/tex]    

Give the appropriate conclusion for the test  

First we need to find the degrees of freedom given by:

[tex]df=n-1=40-1=39[/tex]

Since is a two tailed test the p value would be:    

[tex]p_v =2*P(t_{39}<-2.7105)=0.004967[/tex]    

Conclusion    

If we compare the p value and a significance level assumed for example [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can concldue that the true mean is significantly different from 200 minutes at 5% of significance.    

Answer:

a) [tex]\bar X= 19.2[/tex] represent the sample mean. And that represent the best estimator for the population mean since [tex]\hat \mu =\bar X=19.2[/tex]  b) The 90% confidence interval is given by (17.3;21.1)  

c) No, because 18 is above the lower limit of the confidence interval.

d) Because the parent population is assumed to be normally distributed.

The reason of this is because the t distribution is an special case of the normal distribution when the degrees of freedom increase.

Step-by-step explanation:

1) Notation and definitions  

n=17 represent the sample size

Part a  

[tex]\bar X= 19.2[/tex] represent the sample mean. And that represent the best estimator for the population mean since [tex]\hat \mu =\bar X=19.2[/tex]  Part b

[tex]s=4.4[/tex] represent the sample standard deviation  

m represent the margin of error  

Confidence =90% or 0.90

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

2) Calculate the critical value tc  

In order to find the critical value is important to mention that we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 90% of confidence, our significance level would be given by [tex]\alpha=1-0.90=0.1[/tex] and [tex]\alpha/2 =0.05[/tex]. The degrees of freedom are given by:  

[tex]df=n-1=17-1=16[/tex]  

We can find the critical values in excel using the following formulas:  

"=T.INV(0.05,16)" for [tex]t_{\alpha/2}=-1.75[/tex]  

"=T.INV(1-0.05,16)" for [tex]t_{1-\alpha/2}=1.75[/tex]  

The critical value [tex]tc=\pm 1.75[/tex]  

3) Calculate the margin of error (m)  

The margin of error for the sample mean is given by this formula:  

[tex]m=t_c \frac{s}{\sqrt{n}}[/tex]  

[tex]m=1.75 \frac{4.4}{\sqrt{17}}=1.868[/tex]  

4) Calculate the confidence interval  

The interval for the mean is given by this formula:  

[tex]\bar X \pm t_{c} \frac{s}{\sqrt{n}}[/tex]  

And calculating the limits we got:  

[tex]19.2 - 1.75 \frac{4.4}{\sqrt{17}}=17.332[/tex]  

[tex]19.2 + 1.75 \frac{4.4}{\sqrt{17}}=21.068[/tex]  

The 90% confidence interval is given by (17.332;21.068)  and rounded would be:  (17.3;21.1)

Part c

No, because 18 is above the lower limit of the confidence interval.

Part d

Because the parent population is assumed to be normally distributed.

The reason of this is because the t distribution is an special case of the normal distribution when the degrees of freedom increase.

Answer:

0.1325

Step-by-step explanation:

Weight of the first stock (w1) = .25

Weight of the second stock (w2) = .75

Expected return for the first stock (E(R1)) = .08

Expected return for the second stock (E(R2)) = .15

The expected return of the portfolio is given by the weighted average of the expected return of each stock:

[tex]E(R_p)=w_1*E(R_1)+w_2*E(R_2)\\E(R_p)=0.25*.08 +0.75*.15\\E(R_p)=0.1325[/tex]

The portfolio expected return, E(Rp), is 0.1325

Answer:

C. There is not enough information to answer this question

Step-by-step explanation:

Conclusion cannot be made whether to retain or reject the null hypothesis because the information given is not sufficient

Answer:

D. The variable is qualitative because it is an attribute characteristic.

Step-by-step explanation:

In an address, the street name is an attribute of the address.

An attribute is a qualitative variable.

So the correct answer is:

D. The variable is qualitative because it is an attribute characteristic.

The variable is qualitative because it is an attribute characteristic.

Option D is correct

A qualitative variable is a variable whose values are varied by attributes or characteristics. Examples are hair color, course done in school, gender, etc.

A quantitative variable is a variable whose values are varied by actual measurement. Examples are number of odd numbers, number of students in a class, the population of a country, etc.

The description of the given variable is:

Street name of address

This description represents the attribute or characteristic of a location. Therefore, it is a qualitative variable

Learn more on qualitative and quantitative variables here: https://brainly.com/question/14037311

Answer: She should blend 98 lbs of high-quality beans.

She should blend 72 lbs of cheaper beans

Step-by-step explanation:

Let x represent the number of pounds of high quality beans that she should blend.

Let y represent the number of pounds of cheaper beans that she should blend.

She needs to prepare 170 lbs of blended coffee beans. This means that

x + y = 170

She plans to do this by blending together a high-quality bean costing $4.75 per pound and a cheaper bean at $2.00 per pound. The blend would sell for $3.59 per pound. This means that the total cost of the blend would be 3.59×170 = $610.3. This means that

4.75x + 2y = 610.3 - - - - - - - - - -1

Substituting x = 170 - y into equation 1, it becomes

4.75(170 - y) + 2y = 610.3

807.5 - 4.75y + 2y = 610.3

- 4.75y + 2y = 610.3 - 807.5

- 2.75y = - 197.2

y = - 197.2/-2.75 = 71.9

y = 72 pounds

x = 170 - y = 170 - 71.9

x = 98.1

x = 98 pounds

Answer:

We conclude that children in district are brighter, on average, than the general population.

Step-by-step explanation:

We are given the following data set:

105, 109, 115, 112, 124, 115, 103, 110, 125, 99

Formula:

[tex]\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}[/tex]  

where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.  

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

[tex]Mean =\displaystyle\frac{1117}{10} = 111.7[/tex]

Sum of squares of differences = 642.1

[tex]S.D = \sqrt{\frac{642.1}{49}} = 8.44[/tex]

We are given the following in the question:  

Population mean, μ = 106

Sample mean, [tex]\bar{x}[/tex] = 111.7

Sample size, n = 10

Alpha, α = 0.05

Sample standard deviation, s = 8.44

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 106\\H_A: \mu > 106[/tex]

We use one-tailed(right) t test to perform this hypothesis.

Formula:

[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex]

Putting all the values, we have

[tex]t_{stat} = \displaystyle\frac{111.7 - 106}{\frac{8.44}{\sqrt{10}} } = 2.135[/tex]

Now,

[tex]t_{critical} \text{ at 0.05 level of significance, 9 degree of freedom } = 1.833[/tex]

Since,                  

[tex]t_{stat} > t_{critical}[/tex]

We fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

We conclude that children in district are brighter, on average, than the general population.

Answer:

There is not enough evidence to support the claim that union membership increased.

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 400

p = 11.3% = 0.113

Alpha, α = 0.05

Number of women belonging to union , x = 52

First, we design the null and the alternate hypothesis

[tex]H_{0}: p = 0.113\\H_A: p > 0.113[/tex]

The null hypothesis sates that 11.3% of U.S. workers belong to union and the alternate hypothesis states that there is a increase in union membership.

Formula:

[tex]\hat{p} = \dfrac{x}{n} = \dfrac{52}{400} = 0.13[/tex]

[tex]z = \dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}[/tex]

Putting the values, we get,

[tex]z = \displaystyle\frac{0.13-0.113}{\sqrt{\frac{0.113(1-0.113)}{400}}} = 1.073[/tex]

now, we calculate the p-value from the table.

P-value = 0.141636

Since the p-value is greater than the significance level, we fail to reject the null hypothesis and accept the null hypothesis.

Thus, there is not enough evidence to support the claim that union membership increased.

The evidence isn't sufficient enough to support the claim that union membership increased.

This is a statistical measurement used to validate a hypothesis against observed data.

Sample size, n = 400

p = 11.3% = 0.113

Alpha, α = 0.05

Number of women belonging to union   = 52

H₀ : p = 0.113

Hₐ : p  > 0.113

This means 11.3% of U.S. workers belong to union and there was an increase.

p = x / n

z = p - p /(( √p(1-p) /n

 = 53/400 = 0.13

z = p - p /(( √p(1 - p) /n)).

Substitute the values into the equation.

z = 0.13 - 0.113 / ((√0.113(1-0.113)/400)) = 1.073

P-value = 0.141636 from the table which is greater than the significance level, hence we accept the null hypothesis.

The evidence is therefore not sufficient enough to support the claim that union membership increased.

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Answer: b. (.54, .71)

Step-by-step explanation:

Confidence interval for population proportion is given by :-

[tex]\hat{p}\pm z\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]

,where [tex]\hat{p}[/tex] = sample proportion

z= Critical z-value

n= sample size.

Let p be the proportion of people who support additional handgun control.

As per given , we have

n= 80

[tex]\hat{p}=\dfrac{50}{80}=0.625[/tex]

Critical z-value for 90% confidence interval is 1.645

Now , a 90% confidence interval for the population proportion of those who support additional handgun control will become:

[tex]0.625\pm (1.645)\sqrt{\dfrac{0.625(1-0.625)}{80}}[/tex]

[tex]=0.625\pm (1.645)\sqrt{0.0029296875}[/tex]

[tex]=0.625\pm 0.089\\\\=(0.625-0.089, 0.625+0.089)\\\\=(0.536,\ 0.714)\approx(0.54,\ 0.71)[/tex]

So the correct answer is : b. (.54, .71)

Answer:

b. 1.71

[tex]z=\frac{0.5296 -0.5}{\sqrt{\frac{0.5(1-0.5)}{827}}}=1.71[/tex]  

Step-by-step explanation:

1) Data given and notation

n=827 represent the random sample taken

X=438 represent the  people that indicate that they would like to see the new show in the lineup

[tex]\hat p=\frac{438}{827}=0.5296[/tex] estimated proportion of people that indicate that they would like to see the new show in the lineup

[tex]p_o=0.5[/tex] is the value that we want to test

[tex]\alpha[/tex] represent the significance level

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is higher than 0.5:  

Null hypothesis:[tex]p \leq 0.5[/tex]  

Alternative hypothesis:[tex]p > 0.5[/tex]  

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)  

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

3) Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:  

[tex]z=\frac{0.5296 -0.5}{\sqrt{\frac{0.5(1-0.5)}{827}}}=1.71[/tex]  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level assumed is [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.  

Since is a right tailed test the p value would be:  

[tex]p_v =P(Z>1.71)=0.044[/tex]  

If we compare the p value obtained and the significance level assumed [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the true proportion is higher than 0.5.  

Answer:

Step-by-step explanation:

Given that a sample of 161children was selected from fourth and fifth graders at elementary schools in Philadelphia. In addition to recording the grade level, the researchers determined whether each child had a previously undetected reading disability

a) The two qualitative variables are disability and not having disability and secondly the grades of children

b) Contingency table:

Grade                  4                5                      Total

Normal read.       32              34                        66

Not normal read.  23                6                        29  

Total                      55             40                         95

H0: Reading disability is independent of grade.

Ha: There is association between the two

c)  4 5 Total

Nor read 38.21052632 27.78947368 66

Not norm 16.78947368 12.21052632 29

Expected cells are obtained using the formula

row total*col total/grand total

Answer:

[tex]t=\frac{23500-20000}{\frac{3900}{\sqrt{100}}}=8.974[/tex]      

[tex]p_v =P(t_{99}>8.974)=9.43x10^{-15}[/tex]  

If we compare the p value and the significance level given for example [tex]\alpha=0.05[/tex] we see that [tex]p_v<<\alpha[/tex] so we can conclude that we to reject the null hypothesis, and the actual mean is significantly higher than 20000.      

Step-by-step explanation:

1) Data given and notation      

[tex]\bar X=23500[/tex] represent the sample mean  

[tex]s=3900[/tex] represent the sample standard deviation      

[tex]n=100[/tex] sample size      

[tex]\mu_o =2000[/tex] represent the value that we want to test    

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

t 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 determine if the true mean is higher than 20000, the system of hypothesis would be:      

Null hypothesis:[tex]\mu \leq 2000[/tex]      

Alternative hypothesis:[tex]\mu > 2000[/tex]      

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:      

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

t-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".  

Calculate the statistic      

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

[tex]t=\frac{23500-20000}{\frac{3900}{\sqrt{100}}}=8.974[/tex]      

Calculate the P-value      

First we need to calculate the degrees of freedom given by:  

[tex]df=n-1=100-1=99[/tex]  

Since is a one-side upper test the p value would be:      

[tex]p_v =P(t_{99}>8.974)=9.43x10^{-15}[/tex]  

Conclusion      

If we compare the p value and the significance level given for example [tex]\alpha=0.05[/tex] we see that [tex]p_v<<\alpha[/tex] so we can conclude that we to reject the null hypothesis, and the actual mean is significantly higher than 20000.      

To evaluate the claim that automobiles are driven on average more than 20,000 kilometers per year, a one-sample t-test has to be conducted using the sample data. Based on the outcome of the test and the P-value generated, we can either reject or fail to reject the null hypothesis, hence determining whether to agree with the claim.

To determine whether you'd agree with the claim that automobiles are driven on average more than 20,000 kilometers per year based on the sample data, you would need to perform a hypothesis test. In this case, the null hypothesis (H0) would be that the average distance driven is equal to 20,000 kilometers. The alternative hypothesis (H1) would be that the average distance driven is more than 20,000 kilometers.

Your sample data indicates an average of 23,500 kilometers with a standard deviation of 3900 kilometers. To conduct the hypothesis test, you can use the formulas for a one-sample t-test because the population standard deviation is unknown. The P-value generated from the test will tell you whether to reject the null hypothesis.

If the P-value is less than the chosen significance level (commonly 0.05), you would reject the null hypothesis and conclude that automobiles are driven, on average, more than 20,000 kilometers per year. Otherwise, you would fail to reject the null hypothesis and could not conclusively agree with the claim. Without knowing the exact P-value derived from the test, it would be impossible to definitively agree or disagree with the claim based on the provided sample data.

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Answer:

The cost per gallon is US$ 1.87

Step-by-step explanation:

1. Let's review the information provided to us to answer the question correctly:

Number of gallons of gas = 13

Cost of the gallons of gas = US$ 24.31

2. What is the cost per gallon?

Cost per gallon = Cost of the gallons of gas/Number of gallons of gas

Replacing with the real values, we have:

Cost per gallon = 24.31/13

Cost per gallon = 1.87

The cost per gallon is US$ 1.87

Answer: A. 14.02 to 15.98

Step-by-step explanation:

Let [tex]\mu[/tex] denotes the mean waiting time for population.

Given : Sample size : n= 64

Sample mean : [tex]\overline{x}=15[/tex]   (minutes)

Population standard deviation = [tex]\sigma= 4[/tex]

Confidence level : 95%

By z-table , the critical values for 95% confidence = z*=1.96

Confidence interval for population mean : [tex]\overline{x}\pm z^* \dfrac{\sigma}{\sqrt{n}}[/tex]

The 95% confidence interval for the population mean of waiting times will be :

[tex]15\pm (1.96)\dfrac{4}{\sqrt{64}}[/tex]

[tex]15\pm (1.96)\dfrac{4}{8}[/tex]

[tex]15\pm (1.96)(0.5)[/tex]

[tex]15\pm 0.98[/tex]

[tex](15-0.98,\ 15+0.98)=(14.02,\ 15.98)[/tex]

Hence, the 5% confidence interval for the population mean of waiting times is 14.02 to 15.98.

Thus , the correct answer is Option A.

Answer:

[tex]P(\bar X<3.83)=0.06117[/tex]

Step-by-step explanation:

1) Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Let X the random variable that represent the thickness of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(4,1.1)[/tex]  

Where [tex]\mu=4[/tex] and [tex]\sigma=1.1[/tex]

And let [tex]\bar X[/tex] represent the sample mean, the distribution for the sample mean is given by:

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

On this case  [tex]\bar X \sim N(4,\frac{1.1}{\sqrt{100}})[/tex]

2) Solution to the problem

We are interested on this probability

[tex]P(\bar X<3.83)[/tex]

And the best way to solve this problem is using the normal standard distribution and the z score given by:

[tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

If we apply this formula to our probability we got this:

[tex]P(\bar X<3.83)=P(\frac{X-\mu}{\frac{\sigma}{\sqrt{n}}}<\frac{3.83-\mu}{\frac{\sigma}{\sqrt{n}}})[/tex]

[tex]=P(Z<\frac{3.83-4}{\frac{1.1}{\sqrt{100}}})=P(Z<-1.545)[/tex]

And in order to find this probability we can find tables for the normal standard distribution, excel or a calculator.  

[tex]P(Z<-1.545)=0.06117[/tex]

And the excel formula to calculate it would be:

"=NORM.DIST(-1.545,0,1,TRUE)"

The probability is 49.20%

The z score shows by how many standard deviations the raw score is above or below the mean. The z score is given by:

[tex]z=\frac{x-\mu}{\sigma} \\\\where\ x=raw\ score,\mu=mean,\sigma=standard\ deviation.\\\\For\ a\ sample\ size\ n:\\\\z=\frac{x-\mu}{\sigma/\sqrt{n} }[/tex]

Given that n = 100, μ = 4 mm, σ = 1.1 mm

For x < 3.83 mm:

[tex]z=\frac{x-\mu}{\sigma/\sqrt{n} } \\\\z=\frac{3.83-4}{1.1/\sqrt{100} } =-0.0187[/tex]

P(x < 3.83) = P(z < -0.0187) = 0.4920 = 49.20%

From the normal distribution table, the probability that the average thickness of the 100 sheets is less than 3.83 mm is 49.20%

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Answer:

a) during the first 10 hours of growth the weight will increase 271.8% relative to the initial state ( or 2.718 g)

b) during the first 20 hours of growth the weight will increase  467.1 % relative to the initial state ( or 4.671 g)

Step-by-step explanation:

since the growing rate law is

W'(t) = 0.1 gr/hour * e^(0.1gr/hour* t)  , W'(t) [gr/hour]

and following mathematical conventions: W'(t)= dW/dt

then

dW/dt=0.1 e^(0.1t)

∫dW =∫0.1 e^(0.1t) dt

W = e^(0.1t) + C

at the beginning, (time t=0) the weight is W=2 grams .Therefore

2 g = e^(0.1 g/h*0) + C  → 2 g = 1 g + C → C = 1 g

then

W = e^(0.1t) + 1 g

at t= 10 hours

W = e^(0.1 g/h*10h) + 1 g = 3.718 g/h

therefore the weight will increase

ΔW = 3.718 g -  1 g = 2.718 g  or 271.8% relative to the initial state

for t=20 hours

W = e^(0.1 g/h*20h) + 1 g = 8.389 g/h

thus, the from t= 10 hours to t= 20 hours the weight will increase

ΔW =  8.389 g/h -  3.718 g = 4.671 g  or 467.1 %relative to the initial state

Final answer:

To find the growth of the yeast culture, we integrate the growth rate function over the respective intervals. For the first 10 hours, the yeast weight increases by roughly 1.718 grams. From the 10th to the 20th hour, it increases by approximately 4.671 grams.

Explanation:

To do this, we integrate the provided growth rate function over the given time intervals. Initially, the yeast culture weighs 2 grams and grows at the rate of W'(t) = 0.1e0.1t grams per hour.

Finding the Weight Increase During the First 10 Hours

To find the total growth over the first 10 hours, we calculate the integral of the growth rate from t = 0 to t = 10.

[tex]\[ \int_{0}^{10} 0.1e^{0.1t} dt = \left[ \frac{0.1}{0.1}e^{0.1t} \right]_{0}^{10} = \left[ e^{0.1t} \right]_{0}^{10} = e^{1} - e^{0} = e - 1 \][/tex]

Since e is approximately 2.718, this value becomes approximately 2.718 - 1, which equals 1.718 grams. Hence, during the first 10 hours, the yeast culture will increase by about 1.718 grams.

Weight Increase from the End of the 10th to the End of the 20th Hour

Again, we integrate the growth rate, but this time from t = 10 to t = 20.

[tex]\[ \int_{10}^{20} 0.1e^{0.1t} dt = \left[ e^{0.1t} \right]_{10}^{20} = e^{2} - e^{1} \][/tex]

Similar to the previous calculation, this yields e² - e which is approximately 7.389 - 2.718, resulting in an increase of approximately 4.671 grams.

We can see here that at the 0.01 significance level, we can actually conclude that the single-earner couples on average spend more time watching television together.

To determine whether we can conclude that single-earner couples spend more time watching television together on average than dual-earner couples, we can perform a hypothesis test.

The null hypothesis (H₀) assumes that there is no difference in the mean time spent watching television between the two groups, while the alternative hypothesis (H₁) suggests that single-earner couples spend more time together watching television.

Let's set up the hypotheses:

Null Hypothesis (H₀): μ₁ ≤ μ₂ (The mean time spent together watching television for single-earner couples is less than or equal to the mean time for dual-earner couples.)

Alternative Hypothesis (H₁): μ₁ > μ₂ (The mean time spent together watching television for single-earner couples is greater than the mean time for dual-earner couples.)

Where:

μ₁ = population mean time spent watching television for single-earner couples

μ₂ = population mean time spent watching television for dual-earner couples

Next, we will use a two-sample t-test to test the hypotheses. Since we are trying to determine if single-earner couples spend more time watching television, this will be a one-tailed t-test.

Given the sample means, sample standard deviations, and sample sizes, we can calculate the t-statistic and compare it to the critical t-value at the 0.01 significance level (α = 0.01) with degrees of freedom d f = n₁ + n₂ - 2, where n₁ and n₂ are the sample sizes of single-earner and dual-earner couples, respectively.

Let's assume the sample sizes are n₁ = n₂ = 30 (the actual sample sizes from the study are not given in the question, but this is just for demonstration purposes).

Now, we can calculate the t-statistic:

t = (x₁ - x₂) / √((s₁²/n₁) + ([tex]s_{2}[/tex]² /n₂))

where:

x₁ = sample mean time for single-earner couples

x₂ = sample mean time for dual-earner couples

s₁ = sample standard deviation for single-earner couples

[tex]s_{2}[/tex] = sample standard deviation for dual-earner couples

n₁ = sample size for single-earner couples

n₂ = sample size for dual-earner couples

Using the provided values:

x₁ = 61 minutes

x₂ = 48.4 minutes

s₁ = 15.5 minutes

= 18.1 minutes

n₁ = n₂ = 30 (sample sizes assumed for demonstration)

Calculating the t-statistic:

t = (61 - 48.4) / √((15.5²/30) + (18.1²/30))

t ≈ 4.083

Next, we need to find the critical t-value from the t-distribution table at α = 0.01 significance level and df = 30 + 30 - 2 = 58 (degrees of freedom).

The critical t-value at α = 0.01 with d  f = 58 is approximately 2.660.

Since the calculated t-statistic (4.083) is greater than the critical t-value (2.660), we reject the null hypothesis (H₀).

Therefore, at the 0.01 significance level, we can conclude that single-earner couples, on average, spend more time watching television together than dual-earner couples based on the data provided in the study.

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To determine whether single-earner couples spend more time watching television together than dual-earner couples, an independent samples t-test must be conducted at the 0.01 significance level. Using the provided means and standard deviations, we would calculate a t-statistic and compare it against critical values to either reject the null or fail to reject it.

The question asks whether single-earner couples spend more time watching television together than dual-earner couples, based on a study with provided mean values and standard deviations for both groups. To determine if there is a statistically significant difference between the two means, we would conduct a hypothesis test, specifically an independent samples t-test, at the 0.01 significance level. The null hypothesis (0) would state that there is no difference in the mean television watching times between the groups, while the alternative hypothesis (A) would claim that there is a difference, specifically that single-earner couples watch more television.

Given that the mean time spent watching television for single-earner couples is 61 minutes with a standard deviation of 15.5 minutes, and for dual-earner couples it is 48.4 minutes with a standard deviation of 18.1 minutes, we would calculate the t-statistic and compare it against the t-distribution critical values for the given degrees of freedom. If the calculated t-statistic exceeds the critical value for a one-tailed test at the 0.01 level, we would reject the null hypothesis and conclude that there is a significant difference, supporting the claim that single-earner couples spend more time watching television together.

Answer:

Option A)  k = 2

Step-by-step explanation:

Multimonial Experiment

A multimonial experiment is an experiment with n repeated trials and each trial has a discrete number of possible outcomes.

Binomial Experiment

Binomial experiment is an experiment with n repeated trials and each trial has only two possible outcomes.

Thus, if k  represents the number of possible outcomes, then for k = 2, a multimonial experiment will become a binomial experiment.

Option A)  k = 2

The particle's position, represented by s(t), is found from its velocity v(t) using integration. The function of the velocity is rewritten, integrated, and a constant of integration is found using the given initial condition.

The problem given involves a particle moving with a certain velocity function, v(t) = 1.5√t, and an initial position at t = 4, that is, s(4) = 14. The problem asks for the position of the particle, which is often represented by a displacement or position function, denoted commonly as s(t).

To solve this problem, we need to use the fundamental relationship between velocity and position, which states that velocity is the rate of change of position with respect to time. This relationship implies that to find the position function from the velocity function, we need to find an antiderivative or integral of the velocity function.

By following this step-by-step method, we can determine the position of the particle based on the given information.

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The position of the particle is given by the function s(t) = t^(3/2) + 6.

To find the position of the particle, we need to integrate the given velocity function, [tex]v(t)=1.5\sqrt{t}[/tex].

First, let's find the indefinite integral of v(t):

[tex]\int\limits {v(t)} \, dt = \int\limits {1.5\sqrt{t} } \, dt[/tex]

Rewrite the square root as a power:

[tex]=\int\limits {1.5t^{\frac{1}{2} } \, dt[/tex]

Apply the power rule of integration:

[tex]=\frac{1.5t^{(\frac{1}{2}+1 )} }{\frac{1}{2} +1} + C[/tex]

[tex]= 1.5*(\frac{2}{3} )t^{\frac{3}{2}}+C[/tex]

[tex]=t^\frac{3}{2} +C[/tex]

Now we use the initial condition s(4) = 14 to solve for C:

Therefore:

So the position function is:

[tex]s(t)=t^\frac{3}{2} +6[/tex]

Answer:x = 9

Step-by-step explanation:

The attached photo is that of the given diagram. b represents the angle adjacent 75 degrees.

If line m is parallel to line n, it means that angle b degrees and angle (10x + 15) are corresponding angles. Corresponding angles are equal.

Therefore,

b = 10x + 15

The sum of angles on a straight line is 180 degrees. It means that

b + 75 = 180

b = 180 - 75 = 105

Therefore

10x + 15 = 105

10x = 105 - 15 = 90

x = 90/10 = 9

Answer:

x = 9°

Step-by-step explanation:

105° must be equal to 10x + 15 ° for lines to be parallel.

> 105° = 10x + 15°

> 10x = 90°

> x = 9°

Answer:

Step-by-step explanation:

Given that Student scores on exams given by a certain instruc-tor have mean 74 and standard deviation 14.

                                      Group I  X             Group II Y

Sample mean                     74                          74

n                                         25                          64

Std error (14/sqrtn)             2.8                        1.75

a) P(X>80) =[tex]1-0.9839\\= 0.0161[/tex]

b) P(Y>80) = [tex]1-0.9997\\=0.0003[/tex]

c) X-Y is Normal with mean = 0 and std deviation = [tex]\sqrt{2.8^2+1.75^2} \\=3.302[/tex]

P(X-Y>2.2) = [tex]1-0.8411\\=0.1589[/tex]

d) [tex]P(\bar x -\bar Y>2.2) = 0.1589[/tex]

Answer:

Step-by-step explanation:

Since the human body temperatures are normally distributed, the formula for normal distribution is expressed as

z = (x - u)/s

Where

x = human body temperatures

u = mean body temperature

s = standard deviation

From the information given,

u = 98.20oF

s = 0.62oF

We want to find the probability that their mean body temperature will be less than 98.50oF. It is expressed as

P(x lesser than 98.50)

For x = 98.50,

z = (98.50 - 98.20)/0.62 = 0.48

Looking at the normal distribution table, the corresponding probability to the z score is 0.6844

P(x lesser than 98.50) = 0.6844

Answer:

Step-by-step explanation:

Assuming the number of tickets sales from Mondays is normally distributed. the formula for normal distribution would be applied. It is expressed as

z = (x - u)/s

Where

x = ticket sales from monday

u = mean amount of ticket

s = standard deviation

From the information given,

u = 500 tickets

s = 50 tickets

We want to find the probability that the mean will be greater than 510. It is expressed as

P(x greater than 510) = 1 - P(x lesser than or equal to 510)

For x = 510

z = (510 - 500)/50 = 0.2

Looking at the normal distribution table, the probability corresponding to the z score is 0.9773

P(x greater than 510) = 1 - 0.9773 = 0.0227

Answer:

the correct answer is 0.1366

Step-by-step explanation:

Answer:

Step-by-step explanation:

move constant to the right and change the sign

X=12-8

X=4

Answer:

II. This finding is significant for a two-tailed test at .01.

III. This finding is significant for a one-tailed test at .01.

d. II and III only

Step-by-step explanation:

1) Data given and notation    

[tex]\bar X=19.2[/tex] represent the battery life sample mean    

[tex]\sigma=2.5[/tex] represent the population standard deviation    

[tex]n=25[/tex] sample size    

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

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

t would represent the statistic (variable of interest)    

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

2) State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to check if the mean battery life is equal to 18 or not for parta I and II:    

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

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

And for part III we have a one tailed test with the following hypothesis:

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

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

Since we know the population deviation, 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".    

3) Calculate the statistic    

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

[tex]z=\frac{19.2-18}{\frac{2.5}{\sqrt{25}}}=2.4[/tex]    

4) P-value    

First we need to calculate the degrees of freedom given by:  

[tex]df=n-1=25-1=24[/tex]  

Since is a two tailed test for parts I and II, the p value would be:    

[tex]p_v =2*P(t_{(24)}>2.4)=0.0245[/tex]

And for part III since we have a one right tailed test the p value is:

[tex]p_v =P(t_{(24)}>2.4)=0.0122[/tex]

5) Conclusion    

I. This finding is significant for a two-tailed test at .05.

Since the [tex]p_v <\alpha[/tex]. We reject the null hypothesis so we don't have a significant result. FALSE

II. This finding is significant for a two-tailed test at .01.

Since the [tex]p_v >\alpha[/tex]. We FAIL to reject the null hypothesis so we have a significant result. TRUE.

III. This finding is significant for a one-tailed test at .01.

Since the [tex]p_v >\alpha[/tex]. We FAIL to reject the null hypothesis so we have a significant result. TRUE.

So then the correct options is:

d. II and III only

Answer:

E. I and III only

Step-by-step explanation:

I. .05

III. one-tailed at .01

Answer:

-a discrete random variable

Step-by-step explanation:

The number of customers that enter a store can be 0,1,2,...,100,1000,etc...

If cannot be a decimal number, for example, 0.5. So it is a discrete random variable.

The correct answer is:

-a discrete random variable

The integral that gives the volume of the solid is

∫[from -11 to 11] 4(121 - y²) dy, which equals 21395 cubic units.

We have,

To find the volume of the given solid using the general slicing method, we need to integrate the areas of the individual slices perpendicular to the base.

Each cross-section perpendicular to the base and parallel to the diameter is a square.

Let's set up the integral to calculate the volume:

First, let's consider a vertical slice at a distance "y" from the x-axis.

This slice would be a square with a side length "2x," where "x" is the horizontal distance from the y-axis to the rightmost edge of the square.

Since the diameter of the semicircle is on the x-axis and the semicircle is centred on the y-axis, we have a right triangle formed by the radius (11), the distance from the y-axis (x), and the distance from the x-axis (y).

Using the Pythagorean theorem: x² + y² = 11²

Solving for "x": x² = 11² - y²,

so x = √(121 - y²)

Now, the area of the square slice is (side length)² = (2x)² = 4x².

Since the limits of integration are determined by the range of y values, which go from -11 to 11 (the radius of the semicircle), the integral for the volume is:

V = ∫[from -11 to 11] 4x² dy

Substitute the expression for "x" in terms of "y":

V = ∫[from -11 to 11] 4(121 - y²) dy

Simplify:

V = 4 ∫[from -11 to 11] (484 - 4y²) dy

Integrate:

V = 4 [484y - (4/3)y³] | from -11 to 11

V = 4 [484(11) - (4/3)(11)³] - [484(-11) - (4/3)(-11)³]

V = 4 [5344 - (4/3) * 1331] + [5344 + (4/3) * 1331]

V = 14276 + 7119

   = 21395 cubic units

Thus,

The integral that gives the volume of the solid is

∫[from -11 to 11] 4(121 - y²) dy, which equals 21395 cubic units.

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The volume of the solid is found by integrating the area of the square cross sections, which are determined by the y-coordinate on the semicircle, along the x-axis. The equation for the volume integral is V = ∫ from -11 to 11 of 4(11² - x²) dx.

To find the volume of the solid, we need to integrate the area of the square cross sections along the length of the semicircle. The side length of each square is equal to the y-coordinate at that point (which ranges from -11 to 11). The general form of our semicircle in this position is [tex]y = \sqrt (11^2 - x^2),[/tex] where -11 ≤ x ≤ 11. The equation for the area of each square, then, is  [tex]A(x) = (2y)^2 = 4y^2 = 4(11^2 - x^2).[/tex]

Following the disk method for volumes of revolution, we need to integrate the cross-sectional area along the x-axis, from -11 to 11. So the total volume V is given by V = ∫ from -11 to 11 of A(x) dx = ∫ from -11 to 11 of 4(11² - x²) dx.

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