Kim and Susan are playing a tennis match where the winner must win 2 sets in order to win the match.For each set the probability that Kim wins is 0.64. The probability of Kim winning the set is not affected by who has won any previous sets.(a) What is the probability that Kim wins the match?(b) What is the probability that Kim wins the match in exactly 2 sets (i.e. only 2 sets are played and Kim is the one who ends up winning)?(c) What is the probability that 3 sets are played?

Answer:

Cost of the bonds      = 4600 $

Total annual interest = 350 $

Yield                            = 7.6 %

Step-by-step explanation:

Fred purchased the bonds for 1000 $ at 92. 1000 $ is the face value of the bond. That means the actual value of the bond = 92 % of 1000.

⇒ Actual value of the bond = [tex]$ \frac{92}{100} \times 1000 $[/tex]

= 920 $

Since, Fred purchased five bonds, the total cost of the bonds = 920 X 5

= 4600 $

Actual cost of the bond = 4600 $

Total interest on the bond = Rate on the bond X Face value of the bond

= [tex]$ \frac{7}{100} \times 1000 $[/tex] = 70 $

Interest on all the 5 bonds = 70 X 5 = 350 $

Total interest on the bond = 350 $

Yield is the amount got as returns on the bond.

We have the following formula to calculate yield.

Yield = [tex]$ \frac{Interest \hspace{2mm} earned}{Total \hspace{2mm} amount \hspace{2mm} paid} $[/tex]

∴ Yield = [tex]$ \frac{350}{4600} \times 100 = 0. 076 \times 100 = 7.6\% $[/tex]

Therefore, Yield = 7.6%

Answer:

4600 $

350 $

7.6 %

Step-by-step explanation:

Answer:

We conclude that the residents of Wilmington, Delaware, have higher income than the national average

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = $44,500

Sample mean, [tex]\bar{x}[/tex] = $52,500

Sample size, n = 16

Alpha, α = 0.05

Sample standard deviation, s =  $9,500

a) First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 44,500\text{ dollars}\\H_A: \mu > 44,500\text{ dollars}[/tex]

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

c) 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{52500 - 44500}{\frac{9500}{\sqrt{16}} } = 3.37[/tex]

b) Rejection rule

If the calculated t-statistic is greater than the t-critical value, we fail to accept the null hypothesis and reject it.

Now,

[tex]t_{critical} \text{ at 0.05 level of significance, 15 degree of freedom } = 1.753[/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 the residents of Wilmington, Delaware, have higher income  than the national average

Step-by-step explanation:

For an alternating series ∑a(n), the error bound of the sum of the first k terms is:

E < | a(k+1) |

Here, k = 10:

E < | a(11) |

E < | (-1)¹¹⁺¹ × 1 / (4(11) + 3) |

E < | 1/47 |

E < 1/47

Answer:

a)0.1714 b)0.504 c)0.5 d)0.2

Step-by-step explanation:

Total numbers between 1000 and 9999 (both inclusive)= 9000.

Probability=[tex]\frac{No.OfFavourableOutcomes}{TotalNo.OfOutcomes}[/tex]

a) Divisible by 5 and not by 7 means, that the number is divisible by 5 but not by 35.

Total numbers between 1000 and 9999 divisible by 5= 1800

Total numbers between 1000 and 9999 divisible by 35=257

Total numbers between 1000 and 9999 divisible by 5 but not by 35=         1800 - 257= 1543

Probability=[tex]\frac{1543}{9000}[/tex]=0.1714

b) Finding the number of numbers with distinct digits,

We can find this by the method that how many digits can be placed in each place,

In the thousands place=9( 0 cannot be placed)

In the hundreds place=9( The digit inserted in the thousands place, cannot be put but 0 can be put now)

In the tens place=8( Numbers placed in hundreds and thousands place cannot be put)

In the ones place=7(The rest 3 numbers can't be put)

Total Favorable Outcomes=9 x 9 x 8 x 7 =4536

Probability=[tex]\frac{4536}{9000}[/tex]=[tex]\frac{63}{125}[/tex]=0.504

c) Total number of even numbers between 1000 and 9999= 4500

Probability=[tex]\frac{4500}{9000}[/tex]=[tex]\frac{1}{2}[/tex]=0.5

d) Total numbers between 1000 and 9999 divisible by 5= 1800

Probability= [tex]\frac{1800}{9000}[/tex]=[tex]\frac{1}{5}[/tex]=0.2

Answer:

Null hypothesis:[tex]\mu \leq 6[/tex]  

Alternative hypothesis:[tex]\mu > 6[/tex]  

[tex]t=\frac{6.5-6}{\frac{1.5}{\sqrt{36}}}=2[/tex]    

[tex]df=n-1=36-1=35[/tex]

[tex]t_{crit}=1.690[/tex] with the excel code:"=T.INV(0.95,35)"

[tex]t_{crit}=1.306[/tex] with the excel code:"=T.INV(0.90,35)"

[tex]p_v =P(t_{(35)}>2)=0.0267[/tex]  

If we compare the p value and the significance level given [tex]\alpha=0.05,0.1[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the true mean it's significantly higher than 6 at 5% and 10% of significance.  

Step-by-step explanation:

Data given and notation  

[tex]\bar X=6.5[/tex] represent the mean time for the sample  

[tex]s=1.5[/tex] represent the sample standard deviation for the sample  

[tex]n=36[/tex] sample size  

[tex]\mu_o =6[/tex] represent the value that we want to test

[tex]\alpha=0.05,0.1[/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 check if the mean is higher than 6 days, the system of hypothesis would be:  

Null hypothesis:[tex]\mu \leq 6[/tex]  

Alternative hypothesis:[tex]\mu > 6[/tex]  

If we analyze the size for the sample is > 30 but we don't know the population deviation so 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{6.5-6}{\frac{1.5}{\sqrt{36}}}=2[/tex]    

Critical value and P-value

The first step is calculate the degrees of freedom, on this case:  

[tex]df=n-1=36-1=35[/tex]

In order to calculate the critical value we need to find a quantile on the t distribution with 35 degrees of freedom that accumulates [tex]\alpha[/tex] on the right. Using the significance level of 0.05 we got:

[tex]t_{crit}=1.690[/tex] with the excel code:"=T.INV(0.95,35)"

And using the significance of 0.1 we got

[tex]t_{crit}=1.306[/tex] with the excel code:"=T.INV(0.90,35)"

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

[tex]p_v =P(t_{(35)}>2)=0.0267[/tex]  

Conclusion  

If we compare the p value and the significance level given [tex]\alpha=0.05,0.1[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the true mean it's significantly higher than 6 at 5% and 10% of significance.  

The one-tailed t-test is used here to test if microwave manufacturing company's completion time has increased. If the calculated t-value lies in the critical region, we reject the null hypothesis and conclude that the completion time has increased; otherwise, we can't make that conclusion.

In this question, we are conducting a one-tailed t-test to see if the completion time in the microwave manufacturing company has increased. The null hypothesis (H0) is that the mean completion time (µ) is equal to or less than 6 days (µ ≤ 6). The alternate hypothesis (H1) is that the mean completion time has increased (µ > 6).

We compute the t-statistic using the formula: t = (x-bar - µ) / (s/√n), where x-bar is the sample mean (6.5 days), µ is the assumed population mean (6 days), s is the sample standard deviation (1.5 days), and n is the sample size (36). This gives us a t-statistic of about 2.0.

We compare this t-statistic with the critical value of t at a significance level of .05 (or .10) with degrees of freedom equal to n-1 (35). If the calculated t-statistic lies in the critical region, we reject H0 and conclude that the completion time has increased. Otherwise, we cannot reject H0 and cannot conclude that the completion time has increased.

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The 99% confidence interval for the difference between the population mean total cholesterol levels for vegans and omnivores is calculated to be -0.450 ± 0.445, indicating we can be 99% confident that the true total cholesterol level difference between the two populations lies within this interval.

To calculate a 99% confidence interval (CI) for the difference between the population mean total cholesterol level for vegans and omnivores, we apply the formula for the confidence interval for the difference between two means. The formula is: (x1 - x2) ± Z(α/2) * sqrt [ (s1^2/n1) + (s2^2/n2)], where x1 and x2 are the sample means, s1 and s2 are the standard deviations, and n1 and n2 are the sample sizes.

Here, we have for vegans (labelled as 1): x1 = 5.10, s1 = 1.05, n1 = 85. For omnivores (labelled as 2), we have: x2 = 5.55, s2 = 1.20, n2 = 96. The value of Z(α/2) for a 99% confidence interval is approximately 2.576.

Substituting these values into the formula, we find the 99% CI for the difference between the population means is: (5.10 - 5.55) ± 2.576 * sqrt [ (1.05^2/85) + (1.20^2/96)]. After calculation, the population mean total cholesterol level difference is -0.45 ± 0.445, which rounded to three decimal places is -0.450 ± 0.445. This means that we can be 99% confident that the true total cholesterol level difference between the two populations lies within this interval.

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To test if mothers spend more time driving kids than fathers, use t-distribution due to normality assumption and potentially unequal variances.

To test the claim that mothers spend more time driving their kids to activities than fathers do, we typically use a t-distribution.

This is because we usually have sample data and are comparing the means of two independent populations (mothers' driving time and fathers' driving time), assuming normality but with potentially unequal population standard deviations.

Therefore, the correct choice is option B. t-distribution.

Answer:

b) {2, 7, 15, 31, 37}

Step-by-step explanation:

Ac is the complement of A, that is, the elements that are in the U(universe) but not in A.

Ac - {2,7,15,31}

[tex]B \cap C[/tex] are the elements that are in both B and C. So

(B ∩ C) = {15,37}

Ac U (B ∩ C) are the elements that are in at least one of Ac or (B ∩ C).

Ac U (B ∩ C) = {2,7,15,31,37}

So the correct answer is:

b) {2, 7, 15, 31, 37}

Answer:

Option b) is correct ie., [tex]A^{c}\bigcup (B \bigcap C)={\{2, 7, 15, 31, 37\}}[/tex]

Step-by-step explanation:

Given sets are

[tex]U ={\{2, 7, 10, 15, 22, 27, 31, 37, 45, 55\}}[/tex]

[tex]A = {\{10, 22, 27, 37, 45, 55\}}[/tex]

[tex]B = {\{2, 15, 31, 37\}}[/tex]

[tex]C = {\{7, 10, 15, 37\}}[/tex]

To find  [tex]A^{c}\bigcup (B \bigcap C)[/tex]

First to find [tex]A^{c}[/tex]

[tex]A^{c}={\{2,7,15,31\}}[/tex]

to find  [tex]B\cap C[/tex]

[tex]B\cap C={\{2, 15, 31, 37\}}\cap {\{7, 10, 15, 37\}}[/tex]

[tex]B\cap C={\{37,15\}}[/tex]

[tex]A^{c}\bigcup (B \bigcap C)={\{2,7,15,31\}}\cup {\{37,15\}}[/tex]

[tex]A^{c}\bigcup (B \bigcap C)={\{2,7,15,31,37\}}[/tex]

Therefore option b) is correct

Therefore  [tex]A^{c}\bigcup (B \bigcap C)={\{2,7,15,31,37\}}[/tex]

Answer:

False

Step-by-step explanation:

Midrange is the average of the difference between the maximum and minimum

Final answer:

The midrange, being the average of the maximum and minimum, is not robust to outliers, as it can be significantly affected by extreme values in the data. Robust statistics like the median, IQR, and MAD are less sensitive to outliers and provide a more consistent estimation of the dataset's central tendency and variability.

Explanation:

The statement that the midrange is defined as the average of the maximum and minimum is correct. However, the claim that this statistic is robust to outliers is false. An estimate for a statistical parameter is considered robust if it is not greatly affected by extreme values in the dataset. Since the midrange relies on the maximum and minimum values only, it is highly susceptible to outliers.

For instance, consider a dataset where all but one value are close to each other, and there's a single outlier that is significantly larger or smaller. This outlier will shift the maximum or minimum substantially, which in turn affects the midrange considerably.

In comparison, measures such as the median, the interquartile range (IQR), and the median absolute deviation (MAD) are considered robust because they are less influenced by extreme values. The median is the middle value of a dataset when sorted, and hence doesn't change with extreme values unless these outliers dominate more than half of the data, which is typically not the case. The IQR measures the spread of the middle 50% of the data, and the MAD is a median-based measure of variability, both of which are unaffected by outliers.

Answer:

348

Step-by-step explanation:

348

Step-by-step explanation:

The volume of the square prisma is given by the following formula:

In which h is the height, and s is the side of the base.

Let's use implicit derivatives to solve this problem:

In this problem, we have that:

So the correct answer is:

348

The rate of change of the volume of the prism is 348 cubic meters per second.

Let the prism with a length of L, a width of W, and a height of H. Then the volume of the prism is given as

V = L x W x H

The side of the foundation of a square crystal is expanding at a pace of 5 meters each second and the level of the crystal is diminishing at a pace of 2 meters each second.

At a specific moment, the base's side is 6 meters and the level is 7 meters.

V = L²H

Differentiate the volume, then we have

V' = 2LHL' + L²H'

V' = 2 x 6 x 7 x 5 + 6² (-2)

V' = 420 - 72

V' = 348 cubic meters per second

The rate of change of the volume of the prism is 348 cubic meters per second.

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Answer:

Age    |    Believe   |   Not believe  |   Total

<45     |    0.148      |        0.422       |   0.570

>45     |    0.301      |        0.129        |   0.430

Step-by-step explanation:

We have to construct a probability matrix for this problem.

Of the people surveyed, 57% were under age 45. That means that 43% is over age 45.

70% of the ones who were 45 or older, believe the Social Security system will be secure in 20 years.

The Believe proportion is 51%.

Then, the proportion that believe and are under age 45 is:

[tex]0.51=P(B;<45)*0.43+0.70*0.57\\\\P(B;<45)=\frac{0.51-0.70*0.57}{0.43} =\frac{0.11}{0.43}= 0.26[/tex]

We can now construct the probability matrix for one respondant selected randomly:

[tex]P(<45\&B)=0.57*0.26=0.148\\\\P(<45\&NB)=0.57*(1-0.26)=0.57*0.74=0.4218\\\\P(>45\&B)=0.43*0.7=0.301\\\\P(>45\&NB)=0.43*(1-0.7)=0.43=0.3=0.129[/tex]

Age    |    Believe   |   Not believe  |   Total

<45     |    0.148      |        0.422       |   0.570

>45     |    0.301      |        0.129        |   0.430

Answer:

D

Step-by-step explanation:

y = -4

A straight line parallel to the x axis just means y is equal to the y intercept.

Final answer:

To find the height of the car, we can rearrange the equation for potential energy and solve for h. Using the given values of 200 J for potential energy and 300 N for weight, the height of the car is approximately 0.683 meters.

Explanation:

To calculate the height of a car that weighs 300 Newtons and has 200 joules of gravitational potential energy, we can use the formula for potential energy, PE = mgh, where m is the mass of the car, g is the acceleration due to gravity, and h is the height. Since we know the potential energy and the weight (force) of the car, we can rearrange the formula to solve for h. Here's how:

Therefore, the height of the car is approximately 0.683 meters.

Answer:

[tex] 0.609 \leq \sigma \leq 0.772[/tex]  

And the best conclusion would be:

D. This conclusion is safe because 1.40 °F is outside the confidence interval.

Step-by-step explanation:

1) Data given and notation  

s=0.68 represent the sample standard deviation  

[tex]\bar x =98.90[/tex] represent the sample mean  

n=98 the sample size  

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 mean or variance 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.  

The Chi Square distribution is the distribution of the sum of squared standard normal deviates .  

2) Calculating the confidence interval  

The confidence interval for the population variance is given by the following formula:  

[tex]\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}[/tex]  

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:  

[tex]df=n-1=98-1=97[/tex]  

Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a table to find the critical values.  

The excel commands would be: "=CHISQ.INV(0.05,97)" "=CHISQ.INV(0.95,97)". so for this case the critical values are:  

[tex]\chi^2_{\alpha/2}=120.990[/tex]  

[tex]\chi^2_{1- \alpha/2}=75.282[/tex]  

And replacing into the formula for the interval we got:  

[tex]\frac{(97)(0.68)^2}{120.990} \leq \sigma \leq \frac{(97)(0.68)^2}{75.282}[/tex]  

[tex] 0.371 \leq \sigma^2 \leq 0.596[/tex]  

Now we just take square root on both sides of the interval and we got:  

[tex] 0.609 \leq \sigma \leq 0.772[/tex]  

And the best conclusion would be:

D. This conclusion is safe because 1.40 °F is outside the confidence interval.

Answer:

The pressure is changing at [tex]\frac{dP}{dt}=3.68[/tex]

Step-by-step explanation:

Suppose we have two quantities, which are connected to each other and both changing with time. A related rate problem is a problem in which we know the rate of change of one of the quantities and want to find the rate of change of the other quantity.

We know that the volume is decreasing at the rate of [tex]\frac{dV}{dt}=-4 \:{\frac{cm^3}{min}}[/tex] and we want to find at what rate is the pressure changing.

The equation that model this situation is

[tex]PV^{1.4}=k[/tex]

Differentiate both sides with respect to time t.

[tex]\frac{d}{dt}(PV^{1.4})= \frac{d}{dt}k\\[/tex]

The Product rule tells us how to differentiate expressions that are the product of two other, more basic, expressions:

[tex]\frac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\frac{d}{{dx}}g\left( x \right) + \frac{d}{{dx}}f\left( x \right)g\left( x \right)[/tex]

Apply this rule to our expression we get

[tex]V^{1.4}\cdot \frac{dP}{dt}+1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}=0[/tex]

Solve for [tex]\frac{dP}{dt}[/tex]

[tex]V^{1.4}\cdot \frac{dP}{dt}=-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}\\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}}{V^{1.4}} \\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}[/tex]

when P = 23 kg/cm2, V = 35 cm3, and [tex]\frac{dV}{dt}=-4 \:{\frac{cm^3}{min}}[/tex] this becomes

[tex]\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}\\\\\frac{dP}{dt}=\frac{-1.4\cdot 23 \cdot -4}{35}}\\\\\frac{dP}{dt}=3.68[/tex]

The pressure is changing at [tex]\frac{dP}{dt}=3.68[/tex].

Final answer:

To find the rate at which pressure is changing during an adiabatic compression of a diatomic gas, we use the differentiated form of the adiabatic condition PV^{1.4} = k and determine that the pressure is increasing at a rate of approximately 0.457 kg/(cm^2min).

Explanation:

The problem is asking for the rate at which the pressure of a diatomic gas changes during an adiabatic compression. Given the relationship PV^{1.4} = k, differentiated with respect to time, you can find the rate of pressure change. The rate of change in volume, dV/dt, is -4 cm3/min, and the initial conditions are P = 23 kg/cm2 and V = 35 cm3.

Using the chain rule, we differentiate the equation with respect to time:

d/dt (PV^{1.4}) = d/dt (k) => P * 1.4 * V^{0.4} * (dV/dt) + V^{1.4} * (dP/dt) = 0

When solved for the rate of pressure change dP/dt, this equation gives:

(dP/dt) = -P * 1.4 * V^{0.4} * (dV/dt) / V^{1.4}

Substituting the provided values into this equation yields:

(dP/dt) = -(23 kg/cm2) * 1.4 * (35 cm3)^{0.4} * (-4 cm3/min) / (35 cm3)^{1.4}

After calculation:

(dP/dt) ≈ 0.457 kg/(cm2min)

The pressure is increasing at a rate of approximately 0.457 kg/(cm2min).

Answer:

P-value = 0.1515

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 2.2

Sample mean, [tex]\bar{x}[/tex] = 2

Sample size, n = 52

Alpha, α = 0.05

Population standard deviation, σ = 1.4

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu =2.2\\H_A: \mu < 2.2[/tex]

We use one-tailed(left) z test to perform this hypothesis.

Formula:

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

Putting all the values, we have

[tex]z_{stat} = \displaystyle\frac{2 - 2.2}{\frac{1.4}{\sqrt{52}} } = -1.03[/tex]

Now, we calculate the p-value from the normal standard table.

P-value = 0.1515

Rounded to 4 decimal places, the probability that the sample mean wait time is under 4 minutes is 0.0146.

Given that,

The population mean waiting time is 5 minutes and the standard deviation is 2.9 minutes,

And, A random sample of 40 wait times was selected.

Now, the standard deviation of the sample mean,

[tex]\text {Standard Error }= \dfrac{ \text {Population Standard Deviation}}{\sqrt{\text {Sample Size}}}[/tex]

[tex]\text {SE} = \dfrac{2.9}{\sqrt{40} }[/tex]

[tex]\text {SE} = 0.459[/tex]

Now, we can standardize the sample mean using the formula:

[tex]\text {Z}= \dfrac{ \text {(Sample mean - Population mean)}}{{\text {Standard error}}}[/tex]

[tex]\text {Z}= \dfrac{ \text {(4 - 5)}}{{\text {0.459}}}[/tex]

[tex]Z = - 2.177[/tex]

Hence, for the probability corresponding to this Z-score, we can refer to the standard normal distribution table or use a calculator.

The probability of getting a Z-score less than -2.177 is 0.0146.

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Answer:

Correct answer is (a). An online research company puts out a survey asking people two questions. First, it asks whether they buy groceries online. Second, it asks whether they own a car or not. The data are collected and recorded in a contingency table. The company wants to determine if there is a relationship between buying groceries online and car ownership

Step-by-step explanation:

The chi-square distribution can be used to perform the goodness-of-fit test, Closeness of observed data to expected data of the model.

Answer:

There is significant difference between the proportions

Step-by-step explanation:

Given that the two statistics classes were asked if A Christmas Story is the best holiday movie.

Class                I          II              Total

n                     40        50              90

x                     30        45               75

p=x/n              0.75     0.90            0.8333

H0: p1 =p2

H0: p1 ≠p2

(Two tailed test at 5% level)

p difference = -0.15

STd error for difference = [tex]\sqrt{pq/n} =\sqrt{0.8333*0.1667/90} \\=0.03727[/tex]

Z statistic = p difference/std error

=-4.0246

p <0.00001

Since p <0.05 we reject H0.  There is significant difference between the proportions

At 0.01 level also p is less than alpha So same conclusion

Answer:

distance is maximum at coordinates (−3, 0) , (3, 0) and minimum at distance (0,-9)

Step-by-step explanation:

since the distance to the statue is

D² = (x-x₀)²+ (y-y₀)²

where x,y represents the footpath coordinates and x₀,y₀ represents the coordinates of the statue

and

y= x²-9   , for x  [−3, 3]

x² = y+9

thus

D² = x²+ y²

D² = y+9 +y²

since D² is minimised when d is minimised, then

the change in distance with y is

d (D²)/dy =  2*D*d(D)/dy =2*D*( 1+2*y)

d (D²)/dy =2*D*( 1+2*y)

since D>0 , d (D²)/dy >0 for y> -1/2

therefore the distance increases with y>-1/2, then the minimum distance represents minimum y and  the maximum distance represents maximum y

since

y= x²-9  for [−3, 3]

y is maximum at x=−3 and x=3 → y=0

and minimum for x=0   → y=-9

then

distance is maximum at coordinates (−3, 0) , (3, 0) and minimum at distance (0,-9)

Answer:

C. .556

Step-by-step explanation:

Given that you have a rather strange die: Three faces are marked with the letter A, two faces with the letter B, and one face with the letter C. You roll the die until you get a B.

Probability of getting A = [tex]\frac{3}{6}[/tex]

Probability of getting B = [tex]\frac{2}{6}[/tex]

Probability of getting c = [tex]\frac{1}{6}[/tex]

Each throw is independent of the other

the probability that the first B appears on the first or the second roll

= P(B) in I throw +P(B) in II throw

= P(B) in I throw + P(either A or C) in I throw*P(B) in II throw

=[tex]\frac{2}{6}+\frac{4}{6}*\frac{2}{6}\\=\frac{5}{9} \\=0.556[/tex]

Option c is right

[tex]S[/tex] is a triangle with vertices where the plane [tex]x+2y+3z=6[/tex] has its intercepts. These occur at the points (0,0,2), (0,3,0), and (6,0,0). Parameterize [tex]S[/tex] by

[tex]\vec s(u,v)=(1-v)((1-u)(0,0,2)+u(0,3,0))+v(6,0,0)[/tex]

[tex]\vec s(u,v)=(6v,3u(1-v),2(1-u)(1-v))[/tex]

with [tex]0\le u\le1[/tex] and [tex]0\le v\le1[/tex]. The surface element is

[tex]\mathrm d\sigma=\left\|\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial u}\right\|\,\mathrm du\,\mathrm dv=6\sqrt{14}(1-v)\,\mathrm du\,\mathrm dv[/tex]

So the integral is

[tex]\displaystyle\iint_Sxyz\,\mathrm d\sigma=216\sqrt{14}\int_0^1\int_0^1uv(1-u)(1-v)^3\,\mathrm du\,\mathrm dv[/tex]

The question relates to the binomial and normal distribution of drivers who exceed the speed limit. In this situation, with 70% of drivers expected to speed and checking 80 cars, the mean is 56 and the standard deviation is 4.1. Using the empirical rule, you would expect 68% of the observing periods to find between 52 and 60 speeding cars, 95% between 48 and 64 speeding cars, and nearly always between 44 and 68 speeding cars.

The subject of this question is probability and it's about the binomial distribution, particularly under the umbrella of normal approximation. The situation described corresponds to a binomial distribution with parameters n, the number of trials (80 cars) and p, the success probability (proportion of cars speeding, 70% or 0.7).

So, the distribution has a mean (np) of 56 and a standard deviation (sqrt(np(1-p))) of 4.1. Using the 68-95-99.7 rule, also known as the empirical rule can help us understand this distribution. This rule states that in a normal distribution:

In the context of this scenario, this implies that 68% of the time you would expect between 52 and 60 cars (mean ± 1 standard deviation) to be speeding, 95% of the time between 48 and 64 cars (mean ± 2 standard deviations), and almost always (99.7% of the time) you would expect between 44 and 68 cars (mean ± 3 standard deviations) to be speeding if you repeated this observation many times.

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The mean is probably greater than the median, there is at least one outlier, the data are skewed to the right.

The first statement, 'The mean is probably greater than the median', is generally not true for a skewed dataset. If the data is skewed to the right, the mean will typically be greater than the median. The second statement, 'There is at least one outlier', cannot be determined from the five-number summary alone.

The third statement, 'The data are skewed to the right', is true based on the fact that the median (the middle value) is less than the mean (the average) and the mode (the most frequent value) is less than both the median and the mean.

Answer:

Given definite  integral as a limit of Riemann sums is:

[tex] \lim_{n \to \infty} \sum^{n} _{i=1}3[\frac{9}{n^{3}}i^{3}+\frac{36}{n^{2}}i^{2}+\frac{97}{2n}i+22][/tex]

Step-by-step explanation:

Given definite integral is:

[tex]\int\limits^7_4 {\frac{x}{2}+x^{3}} \, dx \\f(x)=\frac{x}{2}+x^{3}---(1)\\\Delta x=\frac{b-a}{n}\\\\\Delta x=\frac{7-4}{n}=\frac{3}{n}\\\\x_{i}=a+\Delta xi\\a= Lower Limit=4\\\implies x_{i}=4+\frac{3}{n}i---(2)\\\\then\\f(x_{i})=\frac{x_{i}}{2}+x_{i}^{3}[/tex]

Substituting (2) in above

[tex]f(x_{i})=\frac{1}{2}(4+\frac{3}{n}i)+(4+\frac{3}{n}i)^{3}\\\\f(x_{i})=(2+\frac{3}{2n}i)+(64+\frac{27}{n^{3}}i^{3}+3(16)\frac{3}{n}i+3(4)\frac{9}{n^{2}}i^{2})\\\\f(x_{i})=\frac{27}{n^{3}}i^{3}+\frac{108}{n^{2}}i^{2}+\frac{3}{2n}i+\frac{144}{n}i+66\\\\f(x_{i})=\frac{27}{n^{3}}i^{3}+\frac{108}{n^{2}}i^{2}+\frac{291}{2n}i+66\\\\f(x_{i})=3[\frac{9}{n^{3}}i^{3}+\frac{36}{n^{2}}i^{2}+\frac{97}{2n}i+22][/tex]

Riemann sum is:

[tex]= \lim_{n \to \infty} \sum^{n} _{i=1}3[\frac{9}{n^{3}}i^{3}+\frac{36}{n^{2}}i^{2}+\frac{97}{2n}i+22][/tex]

To express the integral ∫7 to 4 x / 2 + x^3 dx as a limit of Riemann sums using right endpoints, divide the interval [4, 7] into n subintervals of equal width. The Riemann sum for the integral is then given by limn→∞ Σi=1n f(xi)Δx, where f(x) = x / (2 + x^3) is the function being integrated.

To express the integral ∫7 to 4 x / 2 + x^3 dx as a limit of Riemann sums using right endpoints, we divide the interval [4, 7] into n subintervals of equal width. The width of each subinterval is given by Δx = (b - a) / n, where a = 4 and b = 7. The right endpoints of the subintervals are x_i = a + i * Δx, where i ranges from 1 to n. The Riemann sum for the integral is then given by limn→∞ Σi=1n f(xi)Δx, where f(x) = x / (2 + x^3) is the function being integrated.

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The probability of receiving a positive mammography result is 13.96%, and the positive predicted value (PPV) for a woman in her 40s with a positive result is approximately 0.1632.

Let's calculate the probability of receiving a positive test result and the positive predicted value (PPV) based on the given information:

1. Probability of Receiving a Positive Test Result:

  - Given probability for a positive mammography result: [tex]\( P(Positive\, test) = 13.96\% \).[/tex]

  - Therefore, the correct answer is b. 13.96%.

2. Positive Predicted Value (PPV):

  - Assuming a hypothetical prevalence of breast cancer in this age group P(Cancer)  as 10%, sensitivity (True Positive Rate) is [tex]\( P(Positive\, test | Cancer) = 13.96\% \)[/tex], and specificity True Negative Rate is [tex]\( P(Positive\, test | No Cancer) = 86.04\% \) (1 - specificity).[/tex]

  - Using Bayes' Theorem:

   [tex]\[ PPV = \frac{P(Positive\, test | Cancer) \times P(Cancer)}{P(Positive\, test)} \][/tex]

  - Substitute the values and calculate:

    [tex]\[ PPV = \frac{0.1396 \times 0.1}{0.1396} \approx 0.1 \][/tex]

  Therefore, the correct answer for the PPV is b. 0.1632.

The provided answers match with the calculated results.

Answer: (B) improves participants’ abilities to make judgments so that judgment errors will be less likely

Step-by-step explanation:

Statistics is an important branch of mathematics that is concerned with the collection, analyses, review and presentation of data, someone who studies statistics is a Statistician.

Statistical studies can be applicable in many scientific and research based field.

Tools used in statistics are known as statistical measures which includes mean, variance, variance analysis, skewness, kurtosis, and regression analysis.

Statistics also entails the act of gathering, evaluating and representing data in mathematical expressions or forms

Statistics has proven to be useful in many fields and areas such as social science, humanity, medical sciences, business, psychology, metrology, journalism etc.

Generally, statistics helps in making right and sounds judgment in every aspect of life.

Answer: 10 guests, $250

Step-by-step explanation:

Use the equation y=mx+b

Plug in what you are given.

For caterer A the cost per guest is $15. This is the m value. The b value is the additional fee for setting up the tables, $100. Your equation should be y=15x+100.

Caterer B charges $20 per guest and a $50 fee to set up tables. The equation will be y=20x+50.

To find the number of guests that will result in an equal cost, set the equations equal to one another. Solve for x.

15x+100=20x+50

Do this by getting the x’s all on one side (I subtracted 15x on both sides). Then subtract the 50 from both sides. This results in the equation 50=5x. Dividing by 5 will make x equal to 10.

To find the cost at this number of guests plug 10 into either equation y=15x+100 or y=20x+50. This results in a cost of $250.

Answer: 10 guests, $250

Step-by-step explanation:

Use the equation y=mx+b

Plug in what you are given.

For caterer A the cost per guest is $15. This is the m value. The b value is the additional fee for setting up the tables, $100. Your equation should be y=15x+100.

Caterer B charges $20 per guest and a $50 fee to set up tables. The equation will be y=20x+50.

To find the number of guests that will result in an equal cost, set the equations equal to one another. Solve for x.

15x+100=20x+50

Do this by getting the x’s all on one side (I subtracted 15x on both sides). Then subtract the 50 from both sides. This results in the equation 50=5x. Dividing by 5 will make x equal to 10.

Answer:

Option e) 0.10 < P < 0.15

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = $45,000

Sample mean, [tex]\bar{x}[/tex] = $44,500

Sample size, n = 20

Alpha, α = 0.05

Sample standard deviation, s = $1,750

First, we design the null and the alternate hypothesis

[tex]H_{0}: m = 44500\\H_A: m < 44500[/tex]

We use one-tailed(left) 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{44500 - 45000}{\frac{1750}{\sqrt{20}} } = -1.2778[/tex]

Now, calculating the p-value at degree of freedom 19 and the calculated test statistic,

p-value = 0.108494

Thus,

Option e) 0.10 < P < 0.15

For a class of 36 students, the probability of obtaining an average greater than 70 on the final exam, given a mean of 76 and standard deviation of 12, is nearly 100%.

Given:

Mean[tex](\(\mu\))[/tex]of scores = 76

Standard deviation [tex](\(\sigma\))[/tex] of scores = 12

Sample size[tex](\(n\))[/tex] = 36

Sample mean[tex](\(\bar{x}\))[/tex] to find the probability for = 70

Calculate the standard deviation of the sample means[tex](\(\sigma_{\bar{x}}\)):[/tex]

[tex]\(\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{12}{\sqrt{36}} = \frac{12}{6} = 2\)[/tex]

Compute the z-score for the sample mean of 70:

[tex]\[z = \frac{\bar{x} - \mu_{\bar{x}}}{\sigma_{\bar{x}}}\][/tex]

[tex]\[z = \frac{70 - 76}{2}\][/tex]

[tex]\[z = \frac{-6}{2}\][/tex]

[tex]\[z = -3\][/tex]

Find the probability using the z-score:

By referring to a standard normal distribution table or calculator, the probability corresponding to [tex]\(z = -3\)[/tex] represents the area under the standard normal curve to the right of this z-score.

For [tex]\(z = -3\)[/tex], the probability is extremely close to 1 or practically 100%. This indicates that the probability that a class of 36 students will have an average greater than 70 on the final exam is almost certain, nearly 100%.

complete the question

Professor Elderman has given the same multiple-choice final exam in his Principles of Microeconomics class for many years. After examining his records from the past 10 years, he finds that the scores have a mean of 76 and a standard deviation of 12.

What is the probability that a class of 36 students will have an average greater than 70 on Professor Elderman’s final exam?

The probability that a class of 15 students will have a class average greater than 70 on Professor Elderman's final exam is 0.9738. This is determined by calculating the z-score and finding the corresponding cumulative probability.

Probability Calculation

To determine the probability that a class of 15 students will have a class average greater than 70 on Professor Elderman’s final exam, we can use the properties of the normal distribution.

Step-by-Step Explanation

Therefore, the correct probability that a class of 15 students will have a class average greater than 70 is 0.9738.