Three airlines serve a small town in Ohio. Airline A has 50% of all the scheduled flights, airline B has 30%, and Airline C has 20%. Their on-time rates are 80%, 65%, and 40%, respectively. A plane has just left on-time. what is the overall probability of leaving on-time

Answer:

Step-by-step explanation:

1) The sample and the population of this study is the friends who replied his email which includes in his contact list. then, the number of the replied to his email are 9 friends.

population: the whole friends include in his contact list.

3) Type I error occurs when one incorrectly rejects the null hypothesis

Here there is possibility of type I error

Answer:

she would have $370.20

Step-by-step explanation:

24.68*15=370.2

The answer would be 320.2. 24.8 x 15= 320.5

Step-by-step explanation:

If the conditions for statistical inference for testing a difference in population proportions been met then we can say that  -(E )All conditions for making statistical inference have been met.

Step-by-step explanation:

We can test the claim and the assumptions  about the population proportion  under the following conditions

Thus we can say that by studying the question we can say that the above mentioned condition have been met.Hence

If the conditions for statistical inference for testing a difference in population proportions been met then we can say that  -(E )All conditions for making statistical inference have been met.

The conditions for statistical inference for testing a difference in population proportions have not been met.

In order to test for a difference in population proportions, certain conditions need to be met. Firstly, the condition for independence must be met, which means that random samples need to be selected. Secondly, the distribution of the difference in sample proportions needs to be approximately normal. This is determined by checking whether nO(pC) is greater than or equal to 10 and nO(1-pC) is greater than or equal to 10. In this case, the conditions for statistical inference have not been met because random samples were not selected (option A) and nN(pC) is not greater than or equal to 10 (option C).

Answer:

We accept H₀ we don´t have evidence of differences between the information from the sample and the population mean

Step-by-step explanation:

From data and excel (or any statistics calculator) we get:

X = 249,6 ml         and       s  1,26 ml

Sample mean and sample standard deviation respectively.

Population mean  μ₀  = 250 ml

We have a normal distribution but we dont know the standard deviation of the population. Furthermore we have a two tails test since we are finding if the sample give us evidence of differences ( in both senses ) when we compare them with the amount of water spec ( 250 ml )

Our test hypothesis are: null hypothesis       H₀    X = μ₀

Alternative Hypothesis                                 Hₐ     X ≠ μ₀

We also know that sample size is 8  therefore df  =  8 - 1   df = 7 , with this value and the fact that we are required to test at α = 0,05 ( two tails test)

t = 2,365

Then we evaluate our interval:

X ± t* (s/√n)   ⇒   249,6 ±  2,365 * ( 1,26/√8 )

 249,6 ±  2,365 * (1,26/2,83)  ⇒ 249,6 ±  2,365 *0,45

249,6 ± 1,052

P [ 250,652 ; 248,548]

Then the population mean 250 is inside the interval, therefore we must accept that the bottles have being  fill withing the spec. We accept H₀

Answer:

Because the p-value of 0.4304 is greater than the significance level of 0.05, we fail to reject the null hypothesis. We conclude the data provide convincing evidence that the mean amount of water in all the bottles filled that day does not differ from the target value of 250 milliliters.

Step-by-step explanation:

Answer:

  600

Step-by-step explanation:

As t gets very large, the exponential term goes to zero, and the expression nears the value ...

  P(∞) = 600/(1 +5·0) = 600

The maximum number of squirrels the park can support is modeled as being 600.

Answer:

20 feet

Step-by-step explanation:

Final answer:

The probability of more than 192 confidence intervals containing the true proportion when constructing 95% confidence intervals across 200 towns, which follows a binomial distribution.

Explanation:

When constructing confidence intervals for a proportion, if we state that we have 95% confidence, this means that if we were to take many samples and build a confidence interval from each sample, we would expect 95% of those intervals to contain the population proportion. In the case with 200 towns, 95% confidence suggests that approximately 190 out of 200 intervals would contain the true population proportion, as 95% of 200 is 190.

To find the probability that more than 192 of these intervals contain the true proportion, we would use the binomial distribution, where each interval has a 0.95 probability of containing the true proportion (success), and we are looking for the sum of the probabilities of 193, 194, ..., 200 successes out of 200 trials.

Answer:

a) The 95% of confidence intervals for the average spending

(23.872 , 32.128)

b) The calculated value t= 1< 1.711( single tailed test) at 0.05 level of significance with 24 degrees of freedom.

The null hypothesis is accepted

A survey of 25 young professionals statistically that the population mean is less than $30

Step-by-step explanation:

Step:-(i)

Given data a survey of 25 young professionals fond that they spend an average of $28 when dining out, with a standard deviation of $10

The sample size 'n' = 25

The mean of the sample   x⁻  = $28

The standard deviation of the sample (S) = $10.

Level of significance ∝=0.05

The degrees of freedom γ =n-1 =25-1=24

tabulated value t₀.₀₅ = 2.064

Step 2:-

The 95% of confidence intervals for the average spending

([tex](x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } ,x^{-} + t_{\alpha }\frac{S}{\sqrt{n} } )[/tex]

[tex](28 - 2.064 \frac{10}{\sqrt{25} } ,28 + 2.064\frac{10}{\sqrt{25} } )[/tex]

( 28 - 4.128 , 28 + 4.128)

(23.872 , 32.128)

a) The 95% of confidence intervals for the average spending

(23.872 , 32.128)

b)

Null hypothesis: H₀:μ<30

Alternative Hypothesis: H₁: μ>30

level of significance ∝ = 0.05

The test statistic

[tex]t = \frac{x^{-}-mean }{\frac{S}{\sqrt{n} } }[/tex]

[tex]t = \frac{28-30 }{\frac{10}{\sqrt{25} } }[/tex]

t = |-1|

The calculated value t= 1< 1.711( single tailed test) at 0.05 level of significance with 24 degrees of freedom.

The null hypothesis is accepted

Conclusion:-

The null hypothesis is accepted

A survey of 25 young professionals statistically that the population mean is less than $30

Answer:

109/10 or 10 9/10

Step-by-step explanation:

Dividing something by 10 puts it one place after the decimal point.

so 9/10 becomes 0.9.

Answer:

10 9/10

Step-by-step explanation:

Answer:

He would need 45,000 for his down payment. He would only have to come up with 9,000.

Step-by-step explanation:

10% times 450,000 is 45,000.

80% times 45,000 is 36,000. 45,000 minus 3600 is 9,000.

Answer:

$9,000

Step-by-step explanation:

10% of 450000

10/100 × 450000

= 45000

He has 80%, needs to arrange for 100-80 = 20% of the down payment

20/100 × 45000 = 9,000

Answer:

E. The entire population is measured in both cases, so the actual difference in means can be computed and a confidence interval should not be used.

Step-by-step explanation:

The entire population of Greek and Egyptian mathematicians was already measured by Amara. She can measure the actual difference in the mean. Therefore, she needn't use a confidence interval. It is mostly common for a researcher to be more interested in the difference between means than in the specific values of the means. The difference in sample means is used to compute the difference in population means.

Amara's sample is a random selection of Greek and Egyptian mathematicians. It is a smaller group drawn from the population that has the characteristics of the entire population. The observations and conclusions made against the sample data are attributed to the population. In this case, the entire position is measured.

A t-test is a type of inferential statistic used to determine if there is a significant difference between the means of two groups (Greek and Egyptian mathematicians) which may be related in certain features. It is mostly used when the data sets, like the data set recorded as the outcome from rolling a die 50 times, would follow a normal distribution and may have unknown variances. A t-test is used as a hypothesis testing tool, which allows testing of an assumption applicable to a population. Once the actual difference is known, a confidence interval should not be used.

Answer:

a) X ~ exp ( 10 )

b) E(X) = 0.1 , Var (X) = 0.01

c)  P ( X < 0.07 ) = 0.00698

Step-by-step explanation:

Solution:-

- The spike train, used to study neural activity, the given time in between consecutive spikes (ISI) where the firing rate = 10 neurons per seconds.

- Denote a random variable "X"represent a single interspike interval (ISI) having an exponential distribution.

- Where X follows exponential distribution defined by event rate parameter i.e λ.

                               X ~ Exp ( λ )

- The event rate (λ) is the number of times an event occurs per unit time. Since we are studying a single interspike interval (ISI) - which corresponds to the firing rate. So, event rate (λ) = firing rate = 10 neurons per second. Hence, the distribution is:

                               X ~ Exp ( 10 )

- The expected value E(X) denotes the amount of time in which a single an event occurs; hence, the time taken for a single neuron.

                               E(X) = 1 / λ

                               E(X) = 1 / 10

                               E(X) = 0.1 s per neuron.

- The variance is the variation in the time taken by a single neuron to be emitted. It is defined as:

                               Var (X) = 1 / λ^2

                               Var (X) = 1 / 10^2

                               Var (X) = 0.01 s^2

- The probability that ISI is less than t = 0.07 seconds: P ( X < t = 0.07 s):

- The cumulative distribution function for exponential variate "X" is:

                                P ( X < t ) = 1 - e^(-λ*t)

- Plug the values and the determine:

                                P ( X < 0.07 ) = 1 - e^(-0.1*0.07)

                                                      = 1 - 0.99302

                                                      = 0.00698

Answer:

a) (9/2)

b) (9/4)

Step-by-step explanation:

I = ∫¹₀ (x³ - 5x) dx = -(9/4)

a) ∫¹₀ (10x - 2x³) dx = -2 ∫¹₀ (x³ - 5x) dx

∫¹₀ (x³ - 5x) dx = -(9/4) from the given value for I

-2 ∫¹₀ (x³ - 5x) dx = -2 × (-9/4) = (9/2)

b) ∫¹₀ (5x - x³) dx = -1 ∫¹₀ (x³ - 5x) dx

∫¹₀ (x³ - 5x) dx = -(9/4) from the given value for I

-1 ∫¹₀ (x³ - 5x) dx = -1 × (-9/4) = (9/4)

Hope this Helps!!!

Answer:

a.

[tex]\int\limits_{0}^{1} 10x - 2x^3 \,dx = -2(\int\limits_{0}^{1} x^3 -5x\,dx) = -2*(-9/4) = 9/2[/tex]

b.

[tex]\int\limits_{0}^{1} 5x - x^3 \,dx = -1*(\int\limits_{0}^{1} x^3 -5x\,dx) = -1*(-9/4) = 9/4[/tex]

Step-by-step explanation:

According to the information given.

[tex]\int\limits_{0}^{1} x^3 - 5x \,dx = -9/4\\[/tex]

Now.

a.

[tex]\int\limits_{0}^{1} 10x - 2x^3 \,dx = -2(\int\limits_{0}^{1} x^3 -5x\,dx) = -2*(-9/4) = 9/2[/tex]

b.

[tex]\int\limits_{0}^{1} 5x - x^3 \,dx = -1*(\int\limits_{0}^{1} x^3 -5x\,dx) = -1*(-9/4) = 9/4[/tex]

The entire peach pie contains 2240 calories.

To find the number of calories in the entire peach pie, we need to multiply the number of calories in each slice by the total number of slices. In this case, there are 8 equal slices in the peach pie, and each slice has 280 calories. Therefore, the total number of calories in the entire peach pie is 8 multiplied by 280, which equals 2240 calories.

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Answer:

See explaination

Step-by-step explanation:

public class GaussElim{

private static final double eps = 1e-10; % set epsilon value

public static doublic[] fun(double[][] A,double[] b){

int n=b.length; %calculate length of vector b.

for( int j=0;j<n;j++){

int max=j; %find and swap pivot row.

for (int i=j+1;i<n;i++){

if(Math.abs(A[i][j])>Math.abs(A[max][j])){

max=i;

}

}

double[] t1= A[j]; %swap

A[j]=A[max];

A[max]=t1;

double t= b[j]; %swap

b[j]=b[max];

b[max]=t;

if(Math.abs(A[j][j])<=eps){

throw new ArithmeticException("Matrix is singular."); % if matrix A is a singular matrix then throw error.

}

for(int i=j+1;i<n;i++){

double alpha= A[i][j]/A[j][j];

b[i]=b[i]-alpha*b[j];

for(int k=j;k<n;k++){

A[i][k]=A[i][k]-alpha*A[j][k];

}

}

}

double[] x=new double[n]; % back substitution starts here

for(int i=n-1;i>=0;i--){

double sum=0.0;

for(int j=i+1;j<n;j++){

sum=sum+A[i][j]*x[j];

}

x[i]=(b[i]-sum)/A[i][i];

}

return x;

}

public static void main(String[] args){

int n=3;

double[][] A={{1,2,1},{4,2,0},{-1,5,-3}};

double[] b={5,3,21};

double[] x=fun(A,b);

for(int i=0;i<n;i++){

StdOut.println(x[i]);

}

}

}

Answer:

infinite solutions

Step-by-step explanation:

5(- 3x - 2) - (x - 3) = -4(4x + 5) + 13

Ditribute

-15x -10 -x +3 = -16x - 20+13

Combine like term

-16x -7 = -16x -7

Add 16x to each side

-16x+16x -7 = -16x+16x-7

-7 =-7

This is always true so we have infinite solutions.

answers

All real numbers are solutions

step by step

5(−3x−2)−(x−3)=−4(4x+5)+13

Step 1: Simplify both sides of the equation.

5(−3x−2)−(x−3)=−4(4x+5)+13

5(−3x−2)+−1(x−3)=−4(4x+5)+13(Distribute the Negative Sign)

5(−3x−2)+−1x+(−1)(−3)=−4(4x+5)+13

5(−3x−2)+−x+3=−4(4x+5)+13

(5)(−3x)+(5)(−2)+−x+3=(−4)(4x)+(−4)(5)+13(Distribute)

−15x+−10+−x+3=−16x+−20+13

(−15x+−x)+(−10+3)=(−16x)+(−20+13)(Combine Like Terms)

−16x+−7=−16x+−7

−16x−7=−16x−7

Step 2: Add 16x to both sides.

−16x−7+16x=−16x−7+16x

−7=−7

Step 3: Add 7 to both sides.

−7+7=−7+7

0=0

Answer:

(a) P([tex]\bar X[/tex] [tex]\geq[/tex] 76) = 0.2327

(b) P(73 < [tex]\bar X[/tex] < 75) = 0.5035

(c) P([tex]\bar X[/tex] < 74.8) = 0.77035

Step-by-step explanation:

We are given that the mean of a population is 74 and the standard deviation is 16.

Assuming the data follows normal distribution.

Let [tex]\bar X[/tex] = sample mean

The z-score probability distribution for sample mean is given by;

                         Z = [tex]\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)

where, [tex]\mu[/tex] = population mean = 74

            [tex]\sigma[/tex] = standard deviation = 16

            n = sample size

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

(a) Probability that a random sample of size 34 yielding a sample mean of 76 or more is given by = P([tex]\bar X[/tex] [tex]\geq[/tex] 76)

   P([tex]\bar X[/tex] [tex]\geq[/tex] 76) = P( [tex]\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] [tex]\geq[/tex] [tex]\frac{76-74}{\frac{16}{\sqrt{34} } }[/tex] ) = P(Z [tex]\geq[/tex] 0.73) = 1 - P(Z < 0.73)

                                                 = 1 - 0.7673 = 0.2327

The above probability is calculated by looking at the value of x = 0.73 in the z table which has an area of 0.7673.

(b) Probability that a random sample of size 120 yielding a sample mean of between 73 and 75 is given by = P(73 < [tex]\bar X[/tex] < 75) = P([tex]\bar X[/tex] < 75) - P([tex]\bar X[/tex] [tex]\leq[/tex] 73)

   P([tex]\bar X[/tex] < 75) = P( [tex]\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\frac{75-74}{\frac{16}{\sqrt{120} } }[/tex] ) = P(Z < 0.68) = 0.75175

   P([tex]\bar X[/tex] [tex]\leq[/tex] 73) = P( [tex]\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] [tex]\leq[/tex] [tex]\frac{73-74}{\frac{16}{\sqrt{120} } }[/tex] ) = P(Z [tex]\leq[/tex] -0.68) =1 - P(Z < 0.68)

                                                 = 1 - 0.75175 = 0.24825

Therefore, P(73 < [tex]\bar X[/tex] < 75) = 0.75175 - 0.24825 = 0.5035

The above probability is calculated by looking at the value of x = 0.68 in the z table which has an area of 0.75175.

(c) Probability that a random sample of size 218 yielding a sample mean of less than 74.8 is given by = P([tex]\bar X[/tex] < 74.8)

   P([tex]\bar X[/tex] < 74.8) = P( [tex]\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\frac{74.8-74}{\frac{16}{\sqrt{218} } }[/tex] ) = P(Z < 0.74) = 0.77035

The above probability is calculated by looking at the value of x = 0.74 in the z table which has an area of 0.77035.

Answer:

(2x100,000) 200,000,00

Step-by-step explanation:

200,000

Answer: 2 hundred thousands

Explanation: To determine what the digit 2 means in 7,239,103, if we put 7,239,103 into the place value chart, we can recognize that the digit 2 is in the hundreds column of the thousands period.

So in this problem, 2 refers to 2 hundred thousands.

Place value chart is attached

in the image provided.

Answer:

[tex]z=\frac{0.843-0.653}{\sqrt{0.748(1-0.748)(\frac{1}{300}+\frac{1}{300})}}=5.358[/tex]    

[tex]p_v =2*P(Z>5.358) = 4.2x10^{-8}[/tex]  

Comparing the p value with the significance level given [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, and we can say that we have singificantly differences between the two proportions.  

Step-by-step explanation:

Data given and notation  

[tex]X_{1}=253[/tex] represent the number with no defects in sample 1

[tex]X_{2}=196[/tex] represent the number with no defects in sample 1

[tex]n_{1}=300[/tex] sample 1

[tex]n_{2}=300[/tex] sample 2

[tex]p_{1}=\frac{253}{300}=0.843[/tex] represent the proportion of number with no defects in sample 1

[tex]p_{2}=\frac{196}{300}=0.653[/tex] represent the proportion of number with no defects in sample 2

z would represent the statistic (variable of interest)  

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

[tex]\alpha=0.05[/tex] significance level given

Concepts and formulas to use  

We need to conduct a hypothesis in order to check if is there is a difference in the the two proportions, the system of hypothesis would be:  

Null hypothesis:[tex]p_{1} - p_2}=0[/tex]  

Alternative hypothesis:[tex]p_{1} - p_{2} \neq 0[/tex]  

We need to apply a z test to compare proportions, and the statistic is given by:  

[tex]z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}[/tex]   (1)  

Where [tex]\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{253+196}{300+300}=0.748[/tex]  

z-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.  

Calculate the statistic  

Replacing in formula (1) the values obtained we got this:  

[tex]z=\frac{0.843-0.653}{\sqrt{0.748(1-0.748)(\frac{1}{300}+\frac{1}{300})}}=5.358[/tex]    

Statistical decision

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

[tex]p_v =2*P(Z>5.358) = 4.2x10^{-8}[/tex]  

Comparing the p value with the significance level given [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, and we can say that we have singificantly differences between the two proportions.  

Answer: Angle Bisector

Step-by-step explanation: You divided into two equal angles, you bisected. You turned F, I, G and I, G, H  into two separate triangles.

Answer:

257/2 degrees or 128.5 degrees

Step-by-step explanation:

radians to degrees is x radians * 180/π

[tex]\frac{257\pi }{360} * \frac{180}{\pi }= \frac{257}{2}[/tex]

it is 257/2 degrees or 128.5 degrees

Answer:

H₀: p = 0.20.

Hₐ: p ≠ 0.20.

Step-by-step explanation:

The question is:

The data in Nutrition Study include information on nutrition and health habits of a sample of 315 people. One of the variables is Smoke, indicating whether a person smokes or not (yes or no). Use technology to test whether the data provide evidence that the proportion of smokers is different from 20% given that 43 identify themselves as smokers. Clearly state the null and alternative hypotheses

In this case we need to test whether the proportion of smokers is different from 20%.

A one-proportion z-test can be used to determine the conclusion for this test.

The hypothesis defined as:

H₀: The proportion of smokers is 20%, i.e. p = 0.20.

Hₐ: The proportion of smokers is different from 20%, i.e. p ≠ 0.20.

The information provided is:

n = 315

X = number of people who identified themselves as smokers = 43

Compute the sample proportion of smokers as follows:

[tex]\hat p=\frac{X}{n}=\frac{43}{315}=0.137[/tex]

Compute the test statistic as follows:

[tex]z=\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}}}=\frac{0.137-0.20}{\sqrt{\frac{0.20(1-0.20)}{315}}}=-2.80[/tex]

The test statistic is -2.80.

Compute the p-value as follows:

[tex]p-value=2\times P(Z<-2.80)\\=2\times [1-P(Z<2.80)]\\=2\times [1-0.99744]\\=0.00512[/tex]

*Use a z-table.

The p-value is 0.00512.

The p-value is quite small. So, the null hypothesis will be rejected at any significance level.

Thus, it can be concluded that the  proportion of smokers is different from 20%.

The null hypothesis is that the proportion of smokers is 20% and the alternative hypothesis is that the proportion of smokers is not 20%.

The null hypothesis simply means that there's no effect or relationship between the variables while the alternative hypothesis simply states that the prediction of the research has an effect

In this case, the null hypothesis is that the proportion of smokers is 20% and the alternative hypothesis is that the proportion of smokers is not 20%.

Learn more about hypothesis on:

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Step-by-step explanation:

you are going to have to be more specific... please let us know which ones are the legs

Answer:

4 hours

Step-by-step explanation:

Answer:

We conclude that the average attention span for teenage boys is less than 5 minutes.

Step-by-step explanation:

We are given that the mean time to distraction for teenage boys working on an independent task was 4 minutes.

Suppose that this mean was based on a random sample of 50 teenage Australian boys and that the sample standard deviation was 1.4 minutes.

Let [tex]\mu[/tex] = average attention span for teenage boys

SO, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu \geq[/tex] 5 minutes    {means that the average attention span for teenage boys is more than or equal to 5 minutes}

Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] < 5 minutes    {means that the average attention span for teenage boys is less than 5 minutes}

The test statistics that will be used here is One-sample t test statistics as we don't know about the population standard deviation;

                    T.S.  = [tex]\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }[/tex]  ~ [tex]t_n_-_1[/tex]

where, [tex]\bar X[/tex] = sample mean attention time span for teenage boys = 4 min

             s = sample standard deviation = 1.4 min

             n = sample of teenage boys = 50

So, the test statistics  =  [tex]\frac{4-5}{\frac{1.4}{\sqrt{50} } }[/tex]  ~  [tex]t_4_9[/tex]

                                     =  -5.051

Now at 0.01 significance level, the t table gives critical value of -2.405 at 49 degree of freedom for left-tailed test. Since our test statistics is less than the critical value of t as -2.405 > -5.051, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the average attention span for teenage boys is less than 5 minutes.

Answer:

D - Can't be determined

Step-by-step explanation:

10 * 2 *4 * 16 = 1280

1280/2 = 640

Answer:

Answer:

The first box plot.

Step-by-step explanation:

Step-by-step explanation:

In a box plot, the lower quartile is the right hand side of the box.  In the first plot, this is 3.

The upper quartile is the left hand side of the box.  In the first plot, this is 8.

The interquartile range is the difference between these two values:

8-3 = 5.

Answer:

a) N(A∩B) = 0.9

b) N(A∩B) = 4.9

c) N(A or B) = 4.2

Step-by-step explanation:

Given that A and B each randomly, and independently, choose 3 of 10 objects;

P(A) = P(B) = 3/10 = 0.3

P(A') = P(B') = 1 - 0.3 = 0.7

a) chosen by both;

Probability of being chosen by both;

P(A∩B) = 0.3 × 0.3 = 0.09

Expected Number of objects being chosen by both;

N(A∩B) = P(A∩B) × N(total) = 0.09×10

N(A∩B) = 0.9

b) not chosen by either A or B;

Probability of not being chosen by either A or B;

P(A'∩B') = 0.7 × 0.7 = 0.49

Expected Number of objects being chosen by both;

N(A'∩B') = P(A'∩B') × N(total) = 0.49×10

N(A∩B) = 4.9

c) chosen by exactly one of A and B.

Probability of being chosen by exactly one of A and B

P(A∩B') + P(A'∩B) = 0.3×0.7 + 0.7 × 0.3 = 0.42

Expected Number of objects being chosen by both;

N(A or B) = 0.42 × 10

N(A or B) = 4.2

The expected number of objects chosen by both A and B is 0.9. The expected number of objects not chosen by either A or B is 9.1. The expected number of objects chosen by exactly one of A and B is 4.2.

To find the expected number of objects chosen by both A and B, we can use the multiplication rule. Since A and B each independently choose 3 objects from a set of 10, the probability of choosing a specific object is 3/10 for both A and B. Therefore, the expected number of objects chosen by both A and B is (3/10)(3/10)(10) = 0.9.

To find the expected number of objects not chosen by either A or B, we can subtract the expected number of objects chosen by both A and B from the total number of objects. So, the expected number of objects not chosen by either A or B is 10 - 0.9 = 9.1.

To find the expected number of objects chosen by exactly one of A and B, we can use the addition rule. Since A and B each independently choose 3 objects, the probability of choosing a specific object is 3/10 for both A and B. Therefore, the expected number of objects chosen by exactly one of A and B is 2(3/10)(7/10)(10) = 4.2.

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Answer:

0.0046 = 0.46% probability that the sample mean would differ from the population mean by more than 3.1 gallons

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

[tex]\mu = 86, \sigma = 7, n = 41, s = \frac{7}{\sqrt{41}} = 1.0932[/tex]

If 41 racing cars are randomly selected, what is the probability that the sample mean would differ from the population mean by more than 3.1 gallons?

Less than 86 - 3.1 = 82.9 or more than 86 + 3.1 = 89.1. Since the normal distribution is symmetric, these probabilities are equal, which means that we can find one of them and multiply by 2.

Less than 82.9

pvalue of Z when X = 82.9. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{82.9 - 86}{1.0932}[/tex]

[tex]Z = -2.84[/tex]

[tex]Z = -2.84[/tex] has a pvalue of 0.0023

2*0.0023 = 0.0046

0.0046 = 0.46% probability that the sample mean would differ from the population mean by more than 3.1 gallons

Final answer:

The probability that the sample mean of 41 racing cars would differ from the population mean by more than 3.1 gallons is 0.0046 or 0.46%.

Explanation:

To calculate the probability that the sample mean of 41 racing cars would differ from the population mean by more than 3.1 gallons, you use the Central Limit Theorem, which states that the sampling distribution of the sample mean will be normally distributed if the sample size is large enough. In this case, 41 is usually considered large enough. The formula to calculate the z-score for a given sample mean difference is z = (X - µ) / (σ/√n), where X is the sample mean, µ is the population mean, σ is the standard deviation, and n is the sample size.

First, we calculate the standard error of the mean (SEM): SEM = σ/√n = 7/√41 ≈ 1.0934. Next, we calculate the z-score for a difference of 3.1 gallons: z = 3.1 / 1.0934 ≈ 2.8354.

Now, we look up the z-score in standard normal distribution tables or use a calculator to find the probability corresponding to a z-score of 2.8354, which gives us the probability of a sample mean being 3.1 gallons or more away from the population mean on one side. However, since the question asks for the probability of the sample mean differing by more than 3.1 gallons on either side, we have to multiply this probability by 2 to get the final answer.

To obtain the final probability, we use statistical software or tables, which typically provide the probability for the region beyond a certain z-score. If P(z > 2.8354) is the probability of being beyond 2.8354 standard deviations from the mean on one side, the probability of being more than 3.1 gallons away from the mean on either side is 2 * P(z > 2.8354). Assuming P(z > 2.8354) = 0.0023 (from a standard normal distribution table), the final probability would be 2 * 0.0023 = 0.0046 or rounded to four decimal places, 0.0046.

Answer:

55%

Step-by-step explanation:

You could first find

10% of 1500=150

50%=750

25%=525

5%=75

750+75=825

The answer is 55% are female

The percentage of the audience that are female at the concert is 55%. This is calculated by dividing the number of females (825) by the total audience number (1500), and then multiplying that result by 100.

To calculate the percentage of women in the audience, we need to divide the number of women by the total number of people in the audience, then multiply by 100 to convert the decimal into a percentage.

Thus, the calculation would be as follows:

(825 / 1500) * 100 = 55%

So, the percentage of the audience that are female is 55%.

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