27. In constructing a confidence interval estimate of the population mean you decide to select 49 random observations to get your point estimate of the mean (sample mean). Your friend is also constructing a similar confidence interval estimate but decides to use a sample size of 36 random observations.
Which of the following is true?
a.) Your confidence interval estimate is narrower
b.) Your friend’s confidence interval estimate has a greater degree of confidence
c.) Your confidence interval estimate is wider
d.) Your confidence interval estimate has a greater degree of confidence
2.) The width of a confidence interval estimate for a proportion will be:
a.) Narrower for 99% confidence level than for a 95% confidence level
b.) Wider for a sample size of 100 than for a sample size of 75
c.) Narrower for 90% confidence level than for a 95% confidence level
d.) Narrower when the sample proportion is .50 than when the sample proportion is 20.

Answer:

D. The outlier will appear as a bar far from all of the other bars with a height that corresponds to a frequency of 1.

Step-by-step explanation:

An histogram measures how many times each value appears in the set we are studying. That is, it is a frequency measure.

Suppose we have the following set:

S = {1,1,1,1,1,1, 2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,3,3,3,4,4,4,4,4,4,4,4,5,5,5,5,5,5,100}

1 appears 6 times. That means that when the X axis is 1, the y axis is 6.

2 appears 8 times. The means that when the X axis is 2, the y axis is 8.

...

100 appears 1 time. This means that when the X axis is 100, the y axis is 1. The X is the outlier, and it is quite far from the other values.

So the correct answer is:

D. The outlier will appear as a bar far from all of the other bars with a height that corresponds to a frequency of 1.

Answer:

a) The sample range 25.9 [tex]ml\slash kg\slash \min[/tex]

b) The sample variance is 49.344 [tex]ml^2 \slash kg^2 \slash min^2[/tex]

c) The sample standard deviation 7.0245 [tex]ml\slash kg\slash \min[/tex]                  

Step-by-step explanation:

We are given the following data on oxygen consumption (mL/kg/min):

28.6, 49.4, 30.3, 28.2, 28.9, 26.4, 33.8, 29.9, 23.5, 30.2

a) The sample range

Range = Maximum - Minimum

[tex]\text{Range} = 49.4 - 23.5 = 25.9[/tex]

The sample range 25.9 [tex]ml\slash kg\slash \min[/tex]

b) The sample variance

[tex]\text{Variance} = \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{309.2}{10} = 30.92[/tex]

Sum of squares of differences =

5.3824 + 341.5104 + 0.3844 + 7.3984 + 4.0804 + 20.4304 + 8.2944 + 1.0404 + 55.0564 + 0.5184 = 444.096

[tex]s^2 = \dfrac{444.096}{9} = 49.344[/tex]

The sample variance is 49.344 [tex]ml^2 \slash kg^2 \slash min^2[/tex]

c)  The sample standard deviation

It is the square root of sample variance.

[tex]s = \sqrt{s^2} = \sqrt{49.344} = 7.0245[/tex]

The sample standard deviation 7.0245 [tex]ml\slash kg\slash \min[/tex]

A Simple Random Sample or a Stratified Random Sample of students across the CSU campuses would allow for reliable data collection on the average distance between student hometowns and their campus, taking into account that self-reporting may introduce errors.

When considering methods to sample the average distance between the hometowns of students and their California State University (CSU) campuses, there are several sampling techniques that can be considered:

Option (b) is incorrect as there are potential problems with self-reporting of distances, and option (d) is impractical for such a large population and not necessary for making inferences. Therefore, options (a), (c), and (e) are relevant to the question.

Answer:

Web searching, specialists consultations and comparisons.

Step-by-step explanation:

a. Does the Honda CR-V iclude traction control as a standard feature?

Research about the Honda CR-V on the internet, or reading an article about it.

b. How much money has our company´s philanthropic foundation donated to colleges and universities in each of the last three year?

Look over the company´s administrative records.

c.How does a 3D printer work?

Search on the web about the 3D printer function.

d. Could our Building 3 support a rooftop green space?

Consultation with an architect.

e. How can we determine whether we would save more money by switching to LED lighting in our corporate offices?

Search on the web about the LED lighting use of electricty and the use of electricty of the type of lighting that the company is already using and compare for the best one.

Answer:

0.9964 is the probability  that at least one of them meets specifications.

Step-by-step explanation:

We are given the following in the question:

B: Bolts meet the specification

N: Nails meet the specification

P(B) = 96% = 0.96

P(N) = 91% = 0.91

One bolt and one nail are chosen independently.

Thus, we can write

[tex]P(B\cap N) = P(B) \times P(N) = 0.96\times 0.91 = 0.8736[/tex]

We have to find the probability that at least one of them meets specifications.

[tex]P(B\cup N) = P(B) + P(N) -P(B\cap N)\\P(B\cup N) =0.96 + 0.91-0.8736\\P(B\cup N) =0.9964[/tex]

0.9964 is the probability  that at least one of them meets specifications.

Answer:option A is the correct answer.

Step-by-step explanation:

Ronald can read at a constant rate of p pages per minute.

If Roland can read a total number of N pages in minutes, then the equation representing the relationship between the number of pages, N and the time, m minutes would be

p pages = 1 minute

N pages = m minutes

Crossmultiplying, it becomes

p × m = N × 1

N = pm

Answer:

a) H0: μm = μf versus Ha: μm < μf

b) [tex]t=\frac{(2.68-2.7)-0}{\sqrt{\frac{0.65^2}{66}+\frac{0.5^2}{69}}}}=-0.200[/tex]

c) [tex]p_v =P(t_{133}<-0.200)=0.421[/tex]  

d) Fail to reject the claim that the average shower times are the same for male and female soldiers because the P-value is greater than 0.1.

Step-by-step explanation:

Data given and notation  

[tex]\bar X_{m}=2.68[/tex] represent the mean for the sample male

[tex]\bar X_{f}=2.7[/tex] represent the mean for the sample female

[tex]s_{m}=0.65[/tex] represent the sample standard deviation for the males

[tex]s_{f}=0.5[/tex] represent the sample standard deviation for the females  

[tex]n_{m}=66[/tex] sample size for the group male  

[tex]n_{f}=69[/tex] sample size for the group female  

t would represent the statistic (variable of interest)  

Part a

Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the average male soldier spends less time in the shower than the average female soldier, the system of hypothesis would be:  

Null hypothesis:[tex]\mu_{m}-\mu_{f}\geq 0[/tex]  

Alternative hypothesis:[tex]\mu_{m} - \mu_{f}< 0[/tex]  

Or equivalently:

Null hypothesis:[tex]\mu_{m}-\mu_{f}= 0[/tex]  

Alternative hypothesis:[tex]\mu_{m} - \mu_{f}< 0[/tex]  

And the best option is:

H0: μm = μf versus Ha: μm < μf

Part b

We don't have the population standard deviation, so for this case is better apply a t test to compare means, and the statistic is given by:  

[tex]t=\frac{(\bar X_{m}-\bar X_{f})-\Delta}{\sqrt{\frac{s^2_{m}}{n_{m}}+\frac{s^2_{f}}{n_{f}}}}[/tex] (1)

And the degrees of freedom are given by [tex]df=n_m +n_f -2=66+69-2=133[/tex]  

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

What is the test statistic?

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

[tex]t=\frac{(2.68-2.7)-0}{\sqrt{\frac{0.65^2}{66}+\frac{0.5^2}{69}}}}=-0.200[/tex]

Part c What is the p-value?

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

[tex]p_v =P(t_{133}<-0.200)=0.421[/tex]  

Part d

The significance level given is [tex] \alpha =0.1[/tex] since the p value is higher than the significance level we can conclude:

Fail to reject the claim that the average shower times are the same for male and female soldiers because the P-value is greater than 0.1.

The question relates to hypothesis testing in statistics, insights into average shower times for male and female soldiers. After formulating the hypotheses, we calculate the test statistic using the provided sample data, and then find the corresponding P-value. If the P-value is less than our significance level, we reject the null hypothesis and side with the alternative hypothesis.

The subject of this question falls under Mathematics, specifically it deals with hypothesis testing statistics.

a) The appropriate hypotheses for the engineer to consider would be: H0: μm = μf versus Ha: μm > μf

b) To calculate the test statistic, we use the formula for the test statistic in a independent two-sample t-test which incorporates the sample sizes, means, and standard deviations from the two groups. The formula is (avg(male soldiers)-avg(female soldiers))/sqrt(((sd(male soldiers))^2/number(male soldiers))+((sd(female soldiers))^2/number(female soldiers))). Plug in given values, we can obtain the test statistic.

c) The P-value can be obtained by looking up the test statistic in the T distribution table.

d) If the P-value is greater than the level of significance (0.1), we would fail to reject the null hypothesis. If the P-value is less than the level of significance, we would reject the null hypothesis. The conclusion is based on the specific P-value we computed.

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Answer:

84

Step-by-step explanation:

21x4=84

Answer:the total number of pieces of equipment in the bus is 84

Step-by-step explanation:

The total number of members of the soccer team on the bus is 21.

if each player carries on 4 pieces of equipment, then the total number of pieces of equipment in the bus would be

21 × 4 = 84

Answer:

[tex]y=2x+2[/tex]

Step-by-step explanation:

we know that

The equation of a linear function has no exponents higher than 1, and the graph of a linear function is a straight line.

Verify each case

case 1) we have

[tex]y=2x+2[/tex]

Is the equation of a line in slope intercept form

so

Is a straight line and has no exponents higher than 1

therefore

Is a linear equation

case 2) we have

[tex]y=2x^{2}+2[/tex]

Is a quadratic equation

Is a curved line and has at least one exponent higher than 1,

therefore

Is a non-linear equation

case 3) we have

[tex]y=2x^{3}+2[/tex]

Is a cubic equation

Is a curved line and has at least one exponent higher than 1,

therefore

Is a non-linear equation

case 4) we have

[tex]y=2x^{4}+2[/tex]

Is a quartic equation

Is a curved line and has at least one exponent higher than 1,

therefore

Is a non-linear equation

Answer:

[tex]\sum_{n=0}^9cos(\frac{\pi n}{2})=1[/tex]

[tex] \sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=0[/tex]

[tex] \sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})=\frac{1}{2}[/tex]

Step-by-step explanation:

[tex] \sum_{n=0}^9cos(\frac{\pi n}{2})=\frac{1}{2}(\sum_{n=0}^9 (e^{\frac{i\pi n}{2}}+ e^{\frac{i\pi n}{2}}))[/tex]

[tex]=\frac{1}{2}(\frac{1-e^{\frac{10i\pi}{2}}}{1-e^{\frac{i\pi}{2}}}+\frac{1-e^{-\frac{10i\pi}{2}}}{1-e^{-\frac{i\pi}{2}}})[/tex]

[tex]=\frac{1}{2}(\frac{1+1}{1-i}+\frac{1+1}{1+i})=1[/tex]

2nd

[tex]\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=\frac{1-e^{\frac{i2\pi N}{N}}}{1-e^{\frac{i2\pi}{N}}}[/tex]

[tex]=\frac{1-1}{1-e^{\frac{i2\pi}{N}}}=0[/tex]

3th

[tex] \sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})==\frac{1}{2}(\sum_{n=0}^\infty ((\frac{e^{\frac{i\pi n}{2}}}{2})^n+ (\frac{e^{-\frac{i\pi n}{2}}}{2})^n))[/tex]

[tex]=\frac{1}{2}(\frac{1-0}{1-i}+\frac{1-0}{1+i})=\frac{1}{2}[/tex]

What we use?

We use that

[tex] e^{i\pi n}=cos(\pi n)+i sin(\pi n)[/tex]

and

[tex]\sum_{n=0}^k r^k=\frac{1-r^{k+1}}{1-r}[/tex]

Geometric sum formulas are used to evaluate sums of a geometric series, with the result expressed in Cartesian form (a + bi) where a is the real part and bi is the imaginary part. The sum of a geometric series is calculated with the formula: Sum = a * (1 - r^n) / (1 - r), where a is the first term and r is the ratio. Please provide the specific sums for a detailed step-by-step calculation.

The problem at hand revolves around the usage of geometric sum formulas to evaluate sums and to express the result in Cartesian form. The critical point to remember is that a geometric series is a series with a constant ratio between successive terms. The sum of the first 'n' terms of a geometric sequence can be calculated using the formula:

Sum = a * (1 - rⁿ) / (1 - r)

Assuming 'a' represents the first term in the series and 'r' is the ratio.To convert a complex number into Cartesian form, you simply map the real and imaginary parts of the number 'a + bi', where 'a' is the real part, and 'bi' is the imaginary part.Unfortunately, without the specifics of the sums you're looking to evaluate, it's impossible to give a concrete step-by-step calculation. However, understanding the formulas and how they're applied should provide you with a good start.

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Answer:

a) 0.913

b) 0.397

c) 0.087

Step-by-step explanation:

We are given the following information:

We treat wearing tie too tight as a success.

P(Tight tie) = 15% = 0.15

Then the number of businessmen follows a binomial distribution, where

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

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 15

We have to evaluate:

a) at least one tie is too tight

[tex]P(x \geq 1) = P(x = 1) +....+ P(x = 15)\\=1 - P(x = 0)\\= 1 - \binom{15}{0}(0.15)^0(1-0.15)^{15}\\=1 - 0.087\\= 0.913[/tex]

b) more than two ties are too tight

[tex]P(x > 2) = P(x = 3) +....+ P(x = 15)\\=1 - P(x = 0) - P(x=1) - P(x=2)\\= 1 - \binom{15}{0}(0.15)^0(1-0.15)^{15}-\binom{15}{1}(0.15)^1(1-0.15)^{14}-\binom{15}{0}(0.15)^2(1-0.15)^{13}\\=1 - 0.087 - 0.231 - 0.285\\= 0.397[/tex]

c) no tie is too tight

[tex]P(x = 0)\\=\binom{15}{0}(0.15)^0(1-0.15)^{15}\\=0.087[/tex]

d) at least 18 ties are not too tight

This probability cannot be evaluated as the number of success or the failures exceeds the number of trials given which is 15.

The probability is asked for 18 failures which cannot be evaluated.

To calculate the probabilities, we can use the concept of binomial probability. Using appropriate formulas, we can find the probabilities (a) at least one tie is too tight (b) more than two ties are too tight (c) no tie is too tight (d) at least 18 ties are not too tight.

To calculate the probabilities, we can use the concept of binomial probability. In this case, the probability that a businessman wears the tie too tightly is 15% or 0.15. The total number of businessmen at the board meeting is 15.

(a) To calculate the probability that at least one tie is too tight, we need to find the complement of the probability that no ties are too tight. Using the formula 1 - P(no tight ties), we calculate 1 - (0.85^15) = 0.999 or 99.9%.

(b) To calculate the probability that more than two ties are too tight, we need to find the sum of probabilities of having 3 or more ties too tight. Using the binomial probability formula, we calculate P(X > 2) = 1 - (P(X=0) + P(X=1) + P(X=2)) = 1 - (0.85^15 + 15*0.15*(0.85^14) + 105*(0.15^2)*(0.85^13)) = 0.014 or 1.4%.

(c) To calculate the probability that no ties are too tight, we use the probability P(X=0) = 0.85^15 = 0.203 or 20.3%.

(d) To calculate the probability that at least 18 ties are not too tight, we need to find the sum of probabilities of having 18 or more ties not too tight. Using the binomial probability formula, we calculate P(X >= 18) = P(X=18) + P(X=19) + ... + P(X=15) = (15 choose 18)*(0.85^18)*(0.15^2) + (15 choose 19)*(0.85^19)*(0.15^1) + (15 choose 20)*(0.85^20)*(0.15^0) = 0.999 or 99.9%.

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Answer:

The complex numbers computed are:

A) [tex]q=0.8752+j0.4838=1e^{-j0.5049}[/tex]

B) [tex]r=-8-j8\sqrt{3} =16e^{j\pi \frac{4}{3}}[/tex]

The sketches are attached to this answer

Step-by-step explanation:

To compute these complex numbers you have to remember these rules:

[tex]Z=a+jb=(a^2+b^2)^{\frac{1}{2}}e^{jtan^{-1}(b/a)}[/tex] (a)

[tex]Z=|z|e^{j\alpha}=|z|cos(\alpha)+j|z|sin(\alpha)[/tex] (b)

Also for multiplication, division, and powers, if W and U are complex numbers and k is a real number:

[tex]{W}\cdot{U}={|W|e^{j\alpha}}{|U|e^{j\beta}}={|W|}{|U|}e^{j(\alpha+\beta)}[/tex]   (1)

[tex]\frac{W}{U}=\frac{|W|e^{j\alpha}}{|U|e^{j\beta}}=\frac{|W|}{|U|}e^{j(\alpha-\beta)}[/tex]                        (2)

[tex]W^{k}=|w|^{k}e^{j(\alpha\cdot k)}[/tex]                                      (3)

With these rules we will do the followings steps:

for A:

1) We solve first the divition, writing the 2 complex numbers exponential form (equation (a)).

2) With the rule (2) we solve the division.

3) with rule (3) we solve the power.

For B:

1)We write the numbers a, b, c, d, and f in exponential form (equation (a)).

2) We use the rule (1) for the product.

Answer:

Mean salary=$17818.68

Step-by-step explanation:

Salary($)          Employees(f)

5001-10,000     22

10,001-15,000    20

15,001-20,000   21

20,001-25,000  23

25,001-30,000  24

We know that company had 110 employees so ∑f should be equal to 110.

∑f=22+20+21+23+24=110

The mean salary can be computed as

[tex]xbar=\frac{sum(fx)}{sum(f)}[/tex]

The x be the midpoint can be calculated by taking the average of upper and lower class limit.

Class Interval Frequency(f)       x                fx

5001-10,000            22               7500.5     165011

10,001-15,000          20               12500.5     250010

15,001-20,000   21               17500.5     367510.5

20,001-25000   23              22500.5     517511.5

25,001-30,000 24      27500.5 660012

fx can be computed by multiplying each x value with frequency in the respective class.

∑fx=165011+250010+367510.5+517511.5+660012=1960055

[tex]xbar=\frac{1960055}{110}=17818.68[/tex]

Thus, the mean salary is $17818.68.

The mean salary is approximately $17,818.18.

To find the mean salary, we need to calculate the average of all the salaries. Here’s the step-by-step process:

Here's the detailed calculation:

Now, multiply each midpoint by the number of employees in that range:

Add these values together to get the total sum:

165,000 + 250,000 + 367,500 + 517,500 + 660,000 = 1,960,000

Now, divide by the total number of employees:

1,960,000 / 110 ≈ 17,818.18

Therefore, the mean salary is approximately $17,818.18.

The differential equation y' = y(y - 3) satisfies all the given conditions. It has equilibrium solutions at y = 0 and y = 3, y' > 0 for 0 < y < 3, and y' < 0 for -inf < y < 0 and 3 < y < inf.

The autonomous differential equation that meets the given conditions is

y' = y(y-3). To confirm this, we can check each condition:

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Answer:

Google it

Step-by-step explanation:

google is very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very useful

Answer:it is false.

Step-by-step explanation:

Let us assume that the regular price of the shirt is $x.

The shirt is on sale for 40% off the regular price. The amount that is taken off the shirt would be

40/100 × x = 0.4 × x = 0.4x

The new price of the shirt would be x - 0.4x = $0.6x

you have an additional 20% off coupon. The value of the coupon would be

20/100 × 0.6x = 0.2 × 0.6x = 0.12x

The cost of the shirt if the coupon is applied would be

0.6x - 0.12x = 0.48x

If you assumed that the shirt will ultimately be 60% off the original price, the cost of the shirt would be

x - 60/100 × x = x - 0.6x = 0.4x

Therefore, they are not equal and si, it is false.

Answer:

V = 20.2969 mm^3 @ t = 10

r = 1.692 mm @ t = 10

Step-by-step explanation:

The solution to the first order ordinary differential equation:

[tex]\frac{dV}{dt} = -kA[/tex]

Using Euler's method

[tex]\frac{dVi}{dt} = -k *4pi*r^2_{i} = -k *4pi*(\frac {3 V_{i} }{4pi})^(2/3)\\ V_{i+1} = V'_{i} *h + V_{i}    \\[/tex]

Where initial droplet volume is:

[tex]V(0) = \frac{4pi}{3} * r(0)^3 =  \frac{4pi}{3} * 2.5^3 = 65.45 mm^3[/tex]

Hence, the iterative solution will be as next:

[tex]V'_{i}  = -k *4pi*(\frac{3*65.45}{4pi})^(2/3)  = -6.283\\V_{i+1} = 65.45-6.283*0.25 = 63.88[/tex]

[tex]V'_{i}  = -k *4pi*(\frac{3*63.88}{4pi})^(2/3)  = -6.182\\V_{i+1} = 63.88-6.182*0.25 = 62.33[/tex]

[tex]V'_{i}  = -k *4pi*(\frac{3*62.33}{4pi})^(2/3)  = -6.082\\V_{i+1} = 62.33-6.082*0.25 = 60.813[/tex]

We compute the next iterations in MATLAB (see attachment)

Volume @ t = 10 is = 20.2969

The droplet radius at t=10 mins

[tex]r(10) = (\frac{3*20.2969}{4pi})^(2/3) = 1.692 mm\\[/tex]

The average change of droplet radius with time is:

Δr/Δt = [tex]\frac{r(10) - r(0)}{10-0} = \frac{1.692 - 2.5}{10} = -0.0808 mm/min[/tex]

The value of the evaporation rate is close the value of k = 0.08 mm/min

Hence, the results are accurate and consistent!

Using Euler's method, we approximated the volume and surface area of the object over a period of 10 minutes. Starting with an initial radius of 2.5 mm, and given a decay constant of 0.08 mm/min, we computed the final volume to be approximately 26.19 mm³ and the final surface area to be about 31.70 mm². This resulted in a final radius of approximately 2.06 mm.

To solve this problem using Euler's method, we'll use the following formulas:

dV/dt = -kA

A = 4πr²

V = (4/3)πr³

Given that the initial radius r₀ = 2.5 mm, we can compute the initial volume V₀ and surface area A₀. Then, we'll iterate through time steps using Euler's method:

Vₙ₊₁ = Vₙ - kA * Δt

Aₙ₊₁ = 4π(rₙ - (k/3) * Δt)²

Using k = 0.08 mm/min and a step size of Δt = 0.25 min, we perform the calculations:

V₀ = (4/3)π(2.5)³ ≈ 65.45 mm³

A₀ = 4π(2.5)² ≈ 98.17 mm²

Iterating:

t = 0.25 min:

V₁ ≈ 65.45 - 0.08 * 98.17 * 0.25 ≈ 61.44 mm³

A₁ ≈ 4π(2.44)² ≈ 59.17 mm²

t = 0.5 min:

V₂ ≈ 61.44 - 0.08 * 59.17 * 0.25 ≈ 59.58 mm³

A₂ ≈ 4π(2.42)² ≈ 58.01 mm²

Continuing this process until t = 10 min, we obtain:

V₄₀ ≈ 26.19 mm³

A₄₀ ≈ 31.70 mm²

Finally, calculating the radius r₄₀ corresponding to V₄₀:

r₄₀ = ((3V₄₀)/(4π))^(1/3) ≈ 2.06 mm

To learn more about Euler's method

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Answer:

a) [tex]P(X<3.0)=P(\frac{X-\mu}{\sigma}<\frac{3-\mu}{\sigma})=P(Z<\frac{3-4.5}{1.7})=P(z<-0.882)[/tex]

[tex]P(z<-0.882)=0.189[/tex]

b) [tex]P(X>7.0)=P(\frac{X-\mu}{\sigma}>\frac{7-\mu}{\sigma})=P(Z<\frac{7-4.5}{1.7})=P(z>1.47)[/tex]

[tex]P(z>1.47)=1-P(z<1.47) = 1-0.929=0.071[/tex]

c)  [tex]P(3<X<7)=P(\frac{3-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{7-\mu}{\sigma})=P(\frac{3-4.5}{1.7}<Z<\frac{7-4.5}{1.7})=P(-0.882<z<1.47)[/tex]

[tex]P(-0.882<z<1.47)=P(z<1.47)-P(z<-0.882)=0.929-0.189=0.740[/tex]

Step-by-step explanation:

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".  

Part a

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(4.5,1.7)[/tex]  

Where [tex]\mu=4.5[/tex] and [tex]\sigma=1.7[/tex]

We are interested on this probability

[tex]P(X<3.0)[/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}{\sigma}[/tex]

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

[tex]P(X<3.0)=P(\frac{X-\mu}{\sigma}<\frac{3-\mu}{\sigma})=P(Z<\frac{3-4.5}{1.7})=P(z<-0.882)[/tex]

And we can find this probability using excel or the normal standard table:

[tex]P(z<-0.882)=0.189[/tex]

Part b

We are interested on this probability

[tex]P(X>3.0)[/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}{\sigma}[/tex]

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

[tex]P(X>7.0)=P(\frac{X-\mu}{\sigma}>\frac{7-\mu}{\sigma})=P(Z<\frac{7-4.5}{1.7})=P(z>1.47)[/tex]

And we can find this probability using excel or the normal standard table:

[tex]P(z>1.47)=1-P(z<1.47) = 1-0.929=0.071[/tex]

Part c

[tex]P(3<X<7)=P(\frac{3-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{7-\mu}{\sigma})=P(\frac{3-4.5}{1.7}<Z<\frac{7-4.5}{1.7})=P(-0.882<z<1.47)[/tex]

And we can find this probability using excel or the normal standard table liek this:

[tex]P(-0.882<z<1.47)=P(z<1.47)-P(z<-0.882)=0.929-0.189=0.740[/tex]

Option:

A) The risk factor in the population of interest

B) The participants

C) The outcome in the population of interest

D) A & C

Answer:

D) A & C

Step-by-step explanation:

Answer: a. Parameter

Step-by-step explanation:

Parameter can be defined as a fact/characteristic about a whole population.

For example:

i. 20% of my class are boys

ii. 40% of Canadian senators are women.

Parameter usually deal with a small measurable population.

While statistic is a characteristic/fact about the sample.

Therefore the case above is a parameter because the students in the class is the population and the stated (Two thirds of all the students in this class are women) is a fact about the population.

Answer:

Parameter

Step-by-step explanation:

The given value is Parameter because it contains the measurement of population. In short, the given value 2/3 is the estimated value of population.

The population consists of all the students in the class. If the ratio of women from all the students in the class is calculated, then it is a measure of population. Thus, the given value is parameter.

Answer:  24

Step-by-step explanation:

Given : Choices for album covers = 3

Choices for television commercials = 4

Choices for album posters = 2

Now , the number of different combinations he needs in order to test each album cover, television commercial, and album poster = ( Choices for album covers ) x (Choices for television commercials) x (Choices for album posters)

= 3 x 4 x 2 = 24

Hence, the number of different combinations he needs in order to test each album cover, television commercial, and album poster is 24.

Answer:

D. $140.

Step-by-step explanation:

Given:

Cost of clothes = $76.17

Cost of books = $44.98

Cost of meal = $19.15

We need to find the best estimate of total cost of her shopping trip.

Solution:

First we will find the total cost of her shopping trip.

total cost of her shopping trip is equal to sum of Cost of clothes, Cost of books and Cost of meal.

framing in equation form we get;

total cost of her shopping trip = [tex]76.17+44.98+19.15 = \$140.3[/tex]

Now we can say that;

140.3 is close to 140

Hence Best estimate of total cost of shopping trip is $140.

Answer:  y=x+2

Step-by-step explanation:

Answer: 5.85 years

Therefore, an average associate is employed for 5.85 years.

Step-by-step explanation:

Given:

Rate of employment yearly = 20 associates per year

Total number of associates = 117 associates

Since the total number of associates remain constant the rate at which they employ new associates is equal to the rate at which associates leave = 20 per year

If 20 new associates are employed in a particular year it would take aan average of :

Average employment year A = total number of associates divided by the rate at which associates leave

A = 117/20

A = 5.85 years

Therefore, an average associate is employed for 5.85 years.

The average length of employment for an associate at the consulting company is calculated by dividing the total number of associates (117) by the annual turnover (20), yielding an average of approximately 5.85 years.

To calculate the average length of employment for associates at the consulting company, we can use the concept of employee turnover rate, which is the rate at which employees leave an organization and are replaced. Since the company must hire 20 new associates each year to replace those who have departed and the total number of associates is 117, we can use the formula for the average employment duration: Total Number of Associates / Annual Turnover = Average Length of Employment.

Using the numbers provided: 117 associates / 20 associates per year = 5.85 years.

This result signifies that the average associate is employed at the consulting company for approximately 5.85 years before they leave the company, although this is a simplification assuming a constant replacement and turnover rate.

Answer:

the probability that a student chosen at random plays football or cricket or both = [tex]\frac{1}{5} + \frac{2}{5} + \frac{4}{15} = \frac{13}{15}[/tex]

Step-by-step explanation:

i) 8 students play football and cricket

ii) 4 students do not play football or cricket

iii) total of 14 students play football.

iv) therefore the number of students who play only football is = 14 - 8 = 6

v) total of 20 students play cricket.

vi) therefore the number of students who play only cricket is = 20 - 8 = 12

vii) therefore the total number of students = 8 + 4 + 6 + 12 = 30

viii) the probability a student chosen at random plays football = [tex]\frac{6}{30} = \frac{1}{5}[/tex]

ix) the probability a student chosen at random plays cricket = [tex]\frac{12}{30} = \frac{2}{5}[/tex]

x) the probability a student chosen at random plays both football and cricket = [tex]\frac{8}{30} = \frac{4}{15}[/tex]

xi) therefore the probability that a student chosen at random plays football or cricket or both = [tex]\frac{1}{5} + \frac{2}{5} + \frac{4}{15} = \frac{13}{15}[/tex]

The probability that a student chosen at random plays football or cricket or both is [tex]\frac{13}{15}[/tex].

We have

Number of students play football and cricket = 8

Number of students do not play football or cricket = 4

Total Number of students play football = 14

 Therefore, the number of students who play only football

= 14 - 8

= 6

Total Number of students play cricket = 20

Therefore, the number of students who play only cricket

= 20 - 8

= 12

So, the total number of students

= 8 + 4 + 6 + 12

= 30

Now, the probability that a student chosen at random plays football

[tex]=\frac{6}{30} \\=\frac{1}{5}[/tex]

The probability that a student chosen at random plays cricket

[tex]=\frac{12}{30} \\=\frac{2}{5}[/tex]

The probability a student chosen at random plays both football and cricket  [tex]=\frac{8}{30} \\=\frac{4}{15}[/tex]

Therefore, the probability that a student chosen at random plays football or cricket or both

[tex]=\frac{1}{5} +\frac{2}{5}+\frac{4}{15}\\=\frac{3}{15} +\frac{6}{15}+\frac{4}{15}\\=\frac{13}{15}[/tex]

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Answer:B

Step-by-step explanation:

Answer: The expected value of the water depth is 4.5 m.

Step-by-step explanation:

Let x be a random variable which is uniformly distributed in interval [a,b] .

Then the mean of the distribution is ghiven by :-

[tex]E(x)=\dfrac{a+b}{2}[/tex]

Given : While conducting experiments, a marine biologist selects water depths from a uniformly distributed collection that vary between 2.00 m and 7.00 m.

Then, the expected value of the water depth = [tex]\dfrac{2+7}{2}=\dfrac{9}{2}=4.5[/tex]

Hence, the expected value of the water depth is 4.5 m.

Answer:

13112221

Step-by-step explanation:

Each sequence of numbers is a verbal representation of the sequence before it. Thus, starting with 1, the next sequence would be "one one," or "11." That sequence is followed by "two one," or "21," and so on and so forth.

This may also be a good explanation:

The first number is just ONE (amount) "1" (0-9 numeral). So if you say there's ONE "1" (seriously just say it aloud) the next number would be an 11. Then there are TWO "1's", creating 21. Then ONE "2" and ONE "1" which creates 1,211. Then ONE "1", ONE "2", and TWO "1's" creating 111,221 ... and so on.

The first number 1 is read as one one, so the second number is written as 11, this is read as two ones, so the next number is written as 21 ( two ones)

This continues throughout the sequence.

The last number written is 312211 which is read as one three, one one, two twos, two ones

This gets written as 13112221

Answer:

The true mean [tex]\mu[/tex] it probably could be larger, smaller, or equal to 22

Step-by-step explanation:

False.

By definition the sample mean is defined as:

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

For this case the value for the sample size is n =190 and the calculated sample mean is [tex] \bar X=22[/tex]. This value represent the sample and for this case we can't assume that this value represent at all the population as the population mean [tex]\mu[/tex] since we probably have variability from the data of the students at City College.

So we can conclude that the true mean [tex]\mu[/tex] it probably could be larger, smaller, or equal to 22

The true mean  it probably could be larger, smaller, or equal to 22.

Given that:

Total student, n = 190.

Average age of 190 students, [tex]\bar X = 22\\[/tex].

By definition the sample mean is defined as:

[tex]\bar X =\dfrac{{\sum_{i=1}^nX_i}}{n}[/tex]

For this case, the sample size n =190 and the calculated sample mean is [tex]\bar X = 22\\[/tex]. This value represent the sample and for this case can't assume that this value represent at all the population as the population mean  since probably have variability from the data of the students at city college.

Hence, conclude that the true mean  it probably could be larger, smaller, or equal to 22

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Answer:

The question is not clear, but here is a similar question with the same approach.

Integral from 0 to 1 Integral from 1 to 2 Integral from 2 to 3, { (x+y+z) } dx dy dz by changing the order of integration in an appropriate way.

Step-by-step explanation:

The approach is that of multiple integral where changing the order of integration is done appropriately

The step by step with detailed workings are shown in the attachment below.

Answer:

a) [tex] S= [x \in x \geq 0][/tex]

Because the weigth can't be negative.

b) AUB = "a weight exceeds 11 grams OR is less than or equal to 15 grams" and that is represent by all the sample space S.

c) A ∩ B ="a weight exceeds 11 grams AND is less than or equal to 15 grams" and that is represent by [tex] 11 < X \leq 15 [/tex] who is the same as [tex] 12 \leq X \leq 15[/tex] .

d) A' = "a weight NOT exceeds 11 grams" [tex] X \leq 11[/tex] that's the complement of the event A

e)  A ∪ B ∪ C = "represent all the possib;e values for the sample space or S"

f) (A ∪ C)'="for this case (AUC) represent the weigths that exceeds 11 gr OR are between 8 and less than 11, so on this case values [tex] X \geq 8[/tex], so then the complement (AUC)' woudl be all the values [tex] X <8[/tex]"

g) A ∩ B ∩ C =[tex]\emptyset[/tex] since we don't have a common interval for the 3 events at the same time

h) B' ∩ C = The complement of B are the [tex] X>15[/tex] and for C we have values [tex] 8 \leq X <12[/tex] and the intersection between these two events is the [tex]\emptyset[/tex].

i) A ∪ (B ∩ C) = For this case (B ∩ C)  represent the values between [tex] 8\leq X <12[/tex] and if we do the union A ∪ (B ∩ C) we got [tex] X \geq 8[/tex]

Step-by-step explanation:

For this case we have defined the following events, assuming that X represent the weight:

A= "a weight exceeds 11 grams" [tex]X>11[/tex]

B= " a weight is less than or equal to 15 grams" [tex] X \leq 15[/tex]

C= "a weight is greater than or equal to 8 grams and less than 12 grams" [tex] 8 \leq X < 12[/tex]

Part a

The sample space is given by:

[tex] S= [x \in x \geq 0][/tex]

Because the weigth can't be negative.

Part b

AUB = "a weight exceeds 11 grams OR is less than or equal to 15 grams" and that is represent by all the sample space S.

Part c

A ∩ B ="a weight exceeds 11 grams AND is less than or equal to 15 grams" and that is represent by [tex] 11 < X \leq 15 [/tex] who is the same as [tex] 12 \leq X \leq 15[/tex] .

Part d

A' = "a weight NOT exceeds 11 grams" [tex] X \leq 11[/tex] that's the complement of the event A

Part e

A ∪ B ∪ C = "represent all the possib;e values for the sample space or S"

Part f

(A ∪ C)'="for this case (AUC) represent the weigths that exceeds 11 gr OR are between 8 and less than 11, so on this case values [tex] X \geq 8[/tex], so then the complement (AUC)' woudl be all the values [tex] X <8[/tex]"

Part g

A ∩ B ∩ C =[tex]\emptyset[/tex] since we don't have a common interval for the 3 events at the same time

Part h

B' ∩ C = The complement of B are the values in[tex] X>15[/tex] and for C we have values [tex] 8 \leq X <12[/tex] and the intersection between these two events is the [tex]\emptyset[/tex].

Part i

A ∪ (B ∩ C) = For this case (B ∩ C)  represent the values between [tex] 8\leq X <12[/tex] and if we do the union A ∪ (B ∩ C) we got [tex] X \geq 8[/tex]

The question refers to the mathematical concept of events and their union and intersection. In this context, A, B, and C represent specific weight ranges. Various combinations of these, such as A ∪ B or A ∩ B ∩ C, refer to different sets of possible weights.

Sample space in this experiment would basically consist of all possible weights that the digital scale can measure to the nearest gram. In Principle, it's unlimited but practically it will depend on the maximum limit of the scale.

A is the event that a weight exceeds 11 grams. B is the event that a weight is less than or equal to 15 grams. C is the event that a weight is greater than or equal to 8 grams and less than 12 grams.

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