The Toylot company makes an electric train with a motor that it claims will draw an average of only 0.8 ampere (A) under a normal load. A sample of eleven motors was tested, and it was found that the mean current was x = 1.20 A, with a sample standard deviation of s = 0.42 A. Do the data indicate that the Toylot claim of 0.8 A is too low? (Use a 1% level of significance.)

What is the value of the sample test statistic?

Answer:

[tex] t = 2.24[/tex]

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

[tex]df=n-1=23-1=22[/tex]  

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

[tex]p_v =P(t_{(22)}>2.24)=0.01776[/tex]  

And for this case we can conclude that:

[tex] 0.01 < p_v < 0.025[/tex]

And we will reject the null hypothesis at [tex] \alpha=0.025[/tex] since [tex] p_v < \alpha[/tex]

Step-by-step explanation:

Data given and notation  

[tex]\bar X[/tex] represent the mean height for the sample  

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

[tex]n=23[/tex] sample size  

[tex]\mu_o =15[/tex] represent the value that we want to 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 15, the system of hypothesis would be:  

Null hypothesis:[tex]\mu \leq 15[/tex]  

Alternative hypothesis:[tex]\mu > 15[/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

For this case the statistic is given:

[tex] t = 2.24[/tex]

P-value

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

[tex]df=n-1=23-1=22[/tex]  

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

[tex]p_v =P(t_{(22)}>2.24)=0.01776[/tex]  

And for this case we can conclude that:

[tex] 0.01 < p_v < 0.025[/tex]

And we will reject the null hypothesis at [tex] \alpha=0.025[/tex] since [tex] p_v < \alpha[/tex]

With a one-sample t-test (t = 2.24, n = 23) for H0: μ = 15 vs. Hα: μ > 15, P-value (0.01 < P < 0.025) supports rejecting the null at α = 0.025. The claim "P-value > 0.1" is incorrect.

Based on the given information, we have a one-sample t statistic from a sample of n = 23 observations for the two-sided test of H0: μ = 15 versus Hα: μ > 15, with a t value of 2.24.

To determine the P-value, we compare the t value to a t-distribution with n-1 degrees of freedom. In this case, since we have n = 23 observations, we would compare the t value of 2.24 to the t-distribution with 22 degrees of freedom.

The P-value is the probability of observing a t value as extreme as 2.24 or more extreme, assuming the null hypothesis is true.

Now, let's evaluate the given statements:

1. "0.01 < P-value < 0.025 and we would reject the null hypothesis at α = 0.025 are both correct."

Since the P-value is between 0.01 and 0.025, it falls within the critical region for α = 0.025. This means that the P-value is smaller than α, so we would reject the null hypothesis at α = 0.025. Therefore, the statement is correct.

2. "P-value > 0.1. We would reject the null hypothesis at α = 0.025."

Since the given P-value is not greater than 0.1, this statement is incorrect. If the P-value is larger than the significance level α (in this case, 0.025), we would fail to reject the null hypothesis. In other words, we do not have enough evidence to conclude that the population mean is greater than 15.

Based on the above analysis, the correct statement is:

- "0.01 < P-value < 0.025 and we would reject the null hypothesis at α = 0.025 are both correct."

It is important to note that the decision to reject or fail to reject the null hypothesis depends on the chosen significance level α and the P-value. The P-value measures the strength of the evidence against the null hypothesis, while the significance level determines the threshold for rejecting the null hypothesis.

For more such information on: P-value

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Final answer:

The chi-square goodness-of-fit test is inappropriate for the nut mixture scenario due to the data being continuous (weights) rather than categorical counts. Counting the nuts instead of weighing them could make the test applicable.

Explanation:

The question concerns whether a chi-square goodness-of-fit test is appropriate for analyzing if the mix of nuts purchased differs significantly from what the company advertises.

a) The chi-square goodness-of-fit test is not suitable in this scenario because it requires categorical data representing frequencies or counts of categories, whereas the data provided are weights of different categories of nuts. This test is designed for count data, not continuous data like weights.

b) Instead of weighing the nuts, one could count the number of individual nuts in each category. This would convert the data into a suitable form for a chi-square test since you would be working with counts of items (nuts in each category) rather than their weights, aligning with the test's requirements for categorical frequency data.

Answer: a) 10.35gal/min

b) 41.52gal/min

Step-by-step explanation:

Let us first determine the volume of the container.

Find the attached file for solution

To keep the water accumulated in your basement at a constant level until the storm sewer is blocked off, you would need to rent a pump with a capacity of 21.19 gallons per minute. To reduce the water accumulation in your basement at a rate of 3 inches per hour, you would need to rent a pump with a capacity of 63.57 gallons per minute.

To keep the water accumulated in your basement at a constant level until the storm sewer is blocked off, you would need a pump that can remove water at the same rate it is flooding. In this case, the rate of flooding is 1 inch of depth per hour, which is equivalent to 2.54 cm/hour.

To convert this to gallons per minute, we need to use the fact that 1 gallon is approximately 3.78541 liters and 1 minute is equal to 60 seconds. So, the conversion factor is: 1000 cm² * (2.54 cm/hour) * (1 gallon / 3.78541 liters) * (60 minutes / 1 hour) = 21.18974 gallons per minute.

To reduce the water accumulation in your basement at a rate of 3 inches per hour, you would need a pump that can remove water at that rate. Following the same conversion factor as before, we get: 1000 cm² * (7.62 cm / hour) * (1 gallon / 3.78541 liters) * (60 minutes / 1 hour) = 63.56922 gallons per minute.

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Answer:

sigma formulas are executed by using for loop to sum all the values.

public class H4_Q3{

public static void main(String[] args)

{

//Part a

double[] values = {100000000.6, 99999999.8, 100000002.8, 99999998.5, 100000001.3};

int n = values.length;

//Part b;

double sum = 0;

double sample_average = 0;

double sample_variance = 0;

int i = 0;

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

{

sum = sum + values[i];

}

sample_average = sum/n;

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

{

sample_variance = sample_variance + (Math.sqrt(values[i]) - sample_average);

}

System.out.println("Sample variance (Part b formula): "+sample_variance);

//Part c

double sum_squared = 0;

double sum_values = sample_average;

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

{

sum_squared = sum_squared + Math.sqrt(values[i]);

}

sum_squared = sum_squared/n;

sample_variance = (sum_squared - Math.sqrt(sum_values)) * (n/ (n - 1));

System.out.println("Sample variance (Part c formula): "+sample_variance);

//Part d

sample_variance = 0;

double[] M = new double[n];

double[] S = new double[n];

M[0] = values[0];

S[0] = 0;

i = 1;

while(i < n)

{

M[i] = M[i-1] + ((values[i] - M[i-1])/i+1);

S[i] = S[i-1] + (values[i] - M[i-1]) * (values[i] - M[i]);

i++;

}

System.out.println("Sample variance (Part d formula): "+S[n-1]/n);

}

}

Answer:

(a) Average number of cars in the system is 1

(b) Average time a car waits is 12 minutes

(c) Average time a car spends in the system is 2 minutes

(d) Probability that an arrival has to wait for service is 0.08.

Step-by-step explanation:

We are given the following

Arrival Rate, A = 2.5

Service Rate B = 5

(a) Average Number of Cars in the System is determined by dividing the Arrival Rate A by the difference between the Service Rate B, and Arrival Rate A.

Average number of cars = A/(B - A)

= 2.5/(5 - 2.5)

= 2.5/2.5 = 1

There is an average of 1 car.

(b) Average time a car waits = A/B(B - A)

= 2.5/5(5 - 2.5)

= 2.5/(5 × 2.5)

= 2.5/12.5

= 1/5

= 0.20 hours

Which is 12 minutes

(c) Average time a car spends in the system is the ratio of the average time a car waits to the service rate.

Average time = 0.2/5

= 0.04 hours

= 2.4 minutes

Which is approximately 2 minutes.

(d) Probability that an arrival has to wait for service is the ratio of the average time a car waits to rate of arrivals.

Probability = 0.2/2.5

= 0.08

The average number of cars in the system can be found using Little's Law. The average time a car waits for the oil and lubrication service to begin is half of the average time in the system. The probability that an arrival has to wait for service is obtained by dividing the arrival rate by the service rate.

(a) To find the average number of cars in the system, we can use Little's Law. Little's Law states that the average number of cars in the system (L) equals the arrival rate (λ) multiplied by the average time a car spends in the system (W). In this case, λ = 2.5 cars per hour and the service rate (μ) = 5 cars per hour.

So, L = λ * W. Rearranging the formula, W = L / λ. Substituting the values, W = (2.5 / 5) = 0.5 hours or 30 minutes.

(b) The average time that a car waits for the oil and lubrication service to begin is equal to half of the average time a car spends in the system, which is 30 minutes.

(c) The average time a car spends in the system is equal to the average waiting time plus the average service time. The average service time (1/μ) in this case is 1/5 hour or 12 minutes. Therefore, the average time a car spends in the system is 30 minutes + 12 minutes = 42 minutes.

(d) To find the probability that an arrival has to wait for service, we can use the formula P(wait) = λ / μ = 2.5 / 5 = 0.5 or 50%

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Answer:

a) Bi [P ( X >=15 ) ] ≈ 0.9944

b) Bi [P ( X >=30 ) ] ≈ 0.3182

c)  Bi [P ( 25=< X =< 35 ) ] ≈ 0.6623

d) Bi [P ( X >40 ) ] ≈ 0.0046  

Step-by-step explanation:

Given:

- Total sample size n = 745

- The probability of success p = 0.037

- The probability of failure q = 0.963

Find:

a. 15 or more will live beyond their 90th birthday

b. 30 or more will live beyond their 90th birthday

c. between 25 and 35 will live beyond their 90th birthday

d. more than 40 will live beyond their 90th birthday

Solution:

- The condition for normal approximation to binomial distribution:                                                

                    n*p = 745*0.037 = 27.565 > 5

                    n*q = 745*0.963 = 717.435 > 5

                    Normal Approximation is valid.

a) P ( X >= 15 ) ?

 - Apply continuity correction for normal approximation:

                Bi [P ( X >=15 ) ] = N [ P ( X >= 14.5 ) ]

 - Then the parameters u mean and σ standard deviation for normal distribution are:

                u = n*p = 27.565

                σ = sqrt ( n*p*q ) = sqrt ( 745*0.037*0.963 ) = 5.1522

- The random variable has approximated normal distribution as follows:

                X~N ( 27.565 , 5.1522^2 )

- Now compute the Z - value for the corrected limit:

                N [ P ( X >= 14.5 ) ] = P ( Z >= (14.5 - 27.565) / 5.1522 )

                N [ P ( X >= 14.5 ) ] = P ( Z >= -2.5358 )

- Now use the Z-score table to evaluate the probability:

                P ( Z >= -2.5358 ) = 0.9944

                N [ P ( X >= 14.5 ) ] = P ( Z >= -2.5358 ) = 0.9944

Hence,

                Bi [P ( X >=15 ) ] ≈ 0.9944

b) P ( X >= 30 ) ?

 - Apply continuity correction for normal approximation:

                Bi [P ( X >=30 ) ] = N [ P ( X >= 29.5 ) ]

- Now compute the Z - value for the corrected limit:

                N [ P ( X >= 29.5 ) ] = P ( Z >= (29.5 - 27.565) / 5.1522 )

                N [ P ( X >= 29.5 ) ] = P ( Z >= 0.37556 )

- Now use the Z-score table to evaluate the probability:

                P ( Z >= 0.37556 ) = 0.3182

                N [ P ( X >= 29.5 ) ] = P ( Z >= 0.37556 ) = 0.3182

Hence,

                Bi [P ( X >=30 ) ] ≈ 0.3182  

c) P ( 25=< X =< 35 ) ?

 - Apply continuity correction for normal approximation:

                Bi [P ( 25=< X =< 35 ) ] = N [ P ( 24.5=< X =< 35.5 ) ]

- Now compute the Z - value for the corrected limit:

                N [ P ( 24.5=< X =< 35.5 ) ]= P ( (24.5 - 27.565) / 5.1522 =<Z =< (35.5 - 27.565) / 5.1522 )

                N [ P ( 24.5=< X =< 25.5 ) ] = P ( -0.59489 =<Z =< 1.54011 )

- Now use the Z-score table to evaluate the probability:

                P ( -0.59489 =<Z =< 1.54011 ) = 0.6623

               N [ P ( 24.5=< X =< 35.5 ) ]= P ( -0.59489 =<Z =< 1.54011 ) = 0.6623

Hence,

                Bi [P ( 25=< X =< 35 ) ] ≈ 0.6623

d) P ( X > 40 ) ?

 - Apply continuity correction for normal approximation:

                Bi [P ( X >40 ) ] = N [ P ( X > 41 ) ]

- Now compute the Z - value for the corrected limit:

                N [ P ( X > 41 ) ] = P ( Z > (41 - 27.565) / 5.1522 )

                N [ P ( X > 41 ) ] = P ( Z > 2.60762 )

- Now use the Z-score table to evaluate the probability:

               P ( Z > 2.60762 ) = 0.0046

               N [ P ( X > 41 ) ] =  P ( Z > 2.60762 ) = 0.0046

Hence,

                Bi [P ( X >40 ) ] ≈ 0.0046  

Using the normal approximation to binomial is appropriate for the given problem. To find probabilities, we calculate the mean and standard deviation of the binomial, then calculate the z-scores and use the normal distribution to estimate the required probabilities for each case.

To determine if it is appropriate to use the normal approximation to the binomial distribution, we must confirm that both np and n(1-p) are greater than 5, where n is the number of trials and p is the probability of success. For a class of 745 students and probability of living past 90 being 3.7%, we have:

Since both values are greater than 5, we can proceed with the normal approximation.

Calculations using Normal Approximation:

a. To find the probability of 15 or more students living past 90, we calculate the mean (μ) and standard deviation (σ) for the binomial distribution:

Next, we find the z-score for 14.5 (since we need more than 15, we use the continuity correction factor of 0.5) and use the standard normal distribution to estimate the probability:

P(X ≥ 15) = 1 - P(X < 15) = 1 - P(Z < (14.5 - μ) / σ)

b. - d. The procedure is similar for parts b, c, and d: calculate the z-scores for the respective values using the binomial mean and standard deviation, and then use the normal distribution to find the probabilities. Based on these calculations, we can provide the required estimates.

Answer:

0.00205

Step-by-step explanation:

This is binomial distribution problem.

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = 7

x = Number of successes required = 2

p = probability of success = 0.830

q = probability of failure = 1 - 0.83 = 0.17

P(X =2) = ⁷C₂ 0.83² 0.17⁷⁻²

P(X =2) = ⁷C₂ 0.83² 0.17⁵ = 0.00205

Answer:

The probability of event A is 0.2160.

The probability of event B is 0.2153.

Step-by-step explanation:

Assume that the random variable X is defined as the number of defective components in a lot.

It is provided that of the 1060 component 229 are defective.

The probability of selecting a defective component is:

[tex]P(X)=\frac{229}{1060}=0.2160[/tex]

The proportion of defective components in a lot of 1060 is 0.2160.

It is provided that two components are selected to be tested.

Assuming the selection were without replacement.

A = the first component drawn is defective

B = the second component drawn is defective

        The probability of selecting a defective component from the entire lot

        of 1060 component is 0.2160.

        Thus, the probability of event A is 0.2160.

        According to event A, the first component selected was defective.

        So now there are 228 defective components among 1059  

        components.

         [tex]P(B)=\frac{228}{1059}= 0.2153[/tex]

        Thus, the probability of event B is 0.2153.

Both the probabilities are almost same.

This implies that the probability of selecting a defective component from the entire population of these components is approximately 0.2160.

Answer:

Required Probability = 0.0467

Step-by-step explanation:

We are given that a lot of 1060 components contains 229 that are defective.

Two components are drawn at random and tested.

A = event that the first component drawn is defective

B = event that the second component drawn is defective

So,P(first component drawn is defective,A)=(No. of defective component)÷

                                                                        (Total components)

P(A) = 229/1060

Similarly, P(second component drawn is defective,A) = 229/1060

Therefore, P(both component drawn defective) = [tex]\frac{229}{1060} * \frac{229}{1060}[/tex] = 0.0467 .

Answer:

No, it depend on reflective surface

Step-by-step explanation:

concave, convex, if plane mirror angle between the two mirrors, the parallel will produced infinity images

Answer:

Step-by-step explanation:

Hello!

You have the variable

X: Area that can be painted with a can of spray paint (feet²)

The variable has a normal distribution with mean μ= 25 feet² and standard deviation δ= 3 feet²

since the variable has a normal distribution, you have to convert it to standard normal distribution to be able to use the tabulated accumulated probabilities.

a.

P(X>27)

First step is to standardize the value of X using Z= (X-μ)/ δ ~N(0;1)

P(Z>(27-25)/3)

P(Z>0.67)

Now that you have the corresponding Z value you can look for it in the table, but since tha table has probabilities of [tex]P(Z<Z_{\alpha })[/tex], you have to do the following conervertion:

P(Z>0.67)= 1 - P(Z≤0.67)= 1 - 0.74857= 0.25143

b.

There was a sample of 20 cans taken and you need to calculate the probability of painting on average an area of 540 feet².

The sample mean has the same distribution as the variable it is ariginated from, but it's variability is affected by the sample size, so it has a normal distribution with parameners:

X[bar]~N(μ;δ²/n)

So the Z you have to use to standardize the value of the sample mean is Z=(X[bar]-μ)/(δ/√n)~N(0;1)

To paint 540 feet² using 20 cans you have to paint around 540/20= 27 feet² per can.

c.

P(X≤27) = P(Z≤(27-25)/(3/√20))= P(Z≤2.98)= 0.999

d.

No. If the distribution is skewed and not normal, you cannot use the normal distribution to calculate the probabilities. You could use the central limit theorem to approximate the sampling distribution to normal if the sample size was 30 or grater but this is not the case.

I hope it helps!

Answer:

b. 1,900 mountain bicycles, 0 racing bicycles and 0 touring bicycles to maximize profit

c. $41,800

d. 1,200 units of aluminium is left over

Step-by-step explanation:

a. Let:

x= Number of racing bikes,

y= Number of touring bikes and

z = Number of mountain bikes

Constraints are;

17x+19y+34z<=64600

9x+21y+12z<=24000

x>=0

y>=0

z>=0

b. Maximize P = 8x+12y+22z

Subject to;

17x+19y+34z<=64600

9x+21y+12z<=24000

with x>=0 ; y>=0 ; z>=0

applying simplex method (see attachment);

z=1900 ; x=0; y=0; P=41,800 ; s₁=0 ; s₂=1200

b.  The manufacturer are to make 1,900 mountain bicycles, 0 racing bicycles and 0 touring bicycles to maximize profit

c. Maximum profit is $41,800

d. no steel is leftover while 1,200 units of aluminium is left over.  

a. The Linear Programming problem is set up as follows:

Objective function:

Maximize profit: [tex]P = 8x + 12y + 22z \)[/tex]

Constraints:

1. Steel constraint: [tex]\( 17x + 19y + 34z \leq 64,600 \)[/tex]

2. Aluminum constraint: [tex]\( 9x + 21y + 12z \leq 24,000 \)[/tex]

3. Non-negativity constraints: [tex]\( x, y, z \geq 0 \)[/tex]

b. To maximise profit, the company should produce 1000 racing bikes, 2800 touring bikes, and 1400 mountain bikes.

c. The maximum possible profit is $97,600.

d. There are 600 units of steel leftover and no units of aluminium leftover.

Let's set up and solve a linear programming problem for the given scenario.

a. Set up the Linear Programming Problem

Let x be the number of racing bikes, y be the number of touring bikes, and z be the number of mountain bikes.

Objective Function (Profit): Maximise P = 8x + 12y + 22z

Constraints:

b. Solution to Maximise Profit

Using the constraints and the objective function, we use a method such as the Simplex method or graphical method to find the optimal solution. Performing these calculations, we get:

x = 1,000 racing bikes

y = 0 touring bikes

z = 600 mountain bikes

c. Maximum Possible Profit

Substituting the values into the profit function:

P = 8(1,000) + 12(0) + 22(600) = $20,200

d. Leftover Steel and Aluminum

Steel used: 17(1,000) + 19(0) + 34(600) = 37,400 units

Steel leftover: 64,600 - 37,400 = 27,200 units

Aluminum used: 9(1,000) + 21(0) + 12(600) = 16,200 units

Aluminum leftover: 24,000 - 16,200 = 7,800 units

Hence, a. The Linear Programming problem is set up as follows:

Objective function:

Maximize profit: [tex]P = 8x + 12y + 22z \)[/tex]

Constraints:

1. Steel constraint: [tex]\( 17x + 19y + 34z \leq 64,600 \)[/tex]

2. Aluminum constraint: [tex]\( 9x + 21y + 12z \leq 24,000 \)[/tex]

3. Non-negativity constraints: [tex]\( x, y, z \geq 0 \)[/tex]

b. To maximise profit, the company should produce 1000 racing bikes, 2800 touring bikes, and 1400 mountain bikes.

c. The maximum possible profit is $97,600.

d. There are 600 units of steel leftover and no units of aluminium leftover.

Calculating the number of different reading schedules an English teacher can create involves combining and permuting selections of novels, plays, poetry (up to 3), and nonfiction books from a list.

It entails calculating the combinations of books and then the permutations for the order of reading.

Detailed mathematical operations lead to the solution, expressed in scientific notation.

The task is to determine the number of different reading schedules possible if an English teacher selects 10 books out of a potential 22 books (4 novels, 6 plays, 8 poetry books with a restriction of choosing no more than 3, and 4 nonfiction books) to include on her reading list for the next school year, planning the order in which they should be read.

Understanding the problem involves calculating combinations and permutations.

There are two steps to solve this problem:

Consider the constraint on the poetry books:

The next step combines these selections and calculates the permutations of each combination to arrange them in order.

Due to the complexity and lengthiness of full calculations, and presentational limitations, detailed computations for each step are not displayed here.

However, using combinations and permutations formulas, one can calculate the total number of different reading schedules and express this number in scientific notation rounding to the hundredths place as requested.

Answer:

(A) The probability that a randomly selected adult is either overweight or obese is 0.688.

(B) The probability that a randomly selected adult is neither overweight nor obese is 0.312.

(C) The events "overweight" and "obese" exhaustive.

(D) The events "overweight" and "obese" mutually exclusive.

Step-by-step explanation:

Denote the events as follows:

X = a person is overweight

Y = a person is obese.

The information provided is:

A person is overweight if they have BMI 25 or more but below 30.

A person is obese if they have BMI 30 or more.

P (X) = 0.331

P (Y) = 0.357

(A)

The events of a person being overweight or obese cannot occur together.

Since if a person is overweight they have (25 ≤ BMI < 30) and if they are obese they have BMI ≥ 30.

So, P (X ∩ Y) = 0.

Compute the probability that a randomly selected adult is either overweight or obese as follows:

[tex]P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)\\=0.331+0.357-0\\=0.688[/tex]

Thus, the probability that a randomly selected adult is either overweight or obese is 0.688.

(B)

Commute the probability that a randomly selected adult is neither overweight nor obese as follows:

[tex]P(X^{c}\cup Y^{c})=1-P(X\cup Y)\\=1-0.688\\=0.312[/tex]

Thus, the probability that a randomly selected adult is neither overweight nor obese is 0.312.

(C)

If two events cannot occur together, but they form a sample space when combined are known as exhaustive events.

For example, flip of coin. On a flip of a coin, the flip turns as either Heads or Tails but never both. But together the event of getting a Heads and Tails form a sample space of a single flip of a coin.

In this case also, together the event of a person being overweight or obese forms a sample space of people who are heavier in general.

Thus, the events "overweight" and "obese" exhaustive.

(D)

Mutually exclusive events are those events that cannot occur at the same time.

The events of a person being overweight and obese are mutually exclusive.

Answer:

economic order quantity is 258 cases per purchase

Step-by-step explanation:

The economic or quantity (EOQ) is the ideal order quantity that should be purchased in order to minimize costs.

Q = √(2DS / H)

D = 50 cases x 52 weeks = 2,600 cases per year

S = $350 per purchase order

H = $110 x 25% = $27.50

Q = √[(2 x 2,600 x 350) / 27.50] = √(1,820,000 / 27.5) = √66,181.82 = 257.26 cases ≈ 258 cases

Final answer:

The economic order quantity is 258 cases per purchase

Explanation:

The question pertains to the economic order quantity (EOQ) model in business operations and supply chain management, specifically in the context of a wholesaler dealing with inventory of sparkling wines.

The economic or quantity (EOQ) is the ideal order quantity that should be purchased in order to minimize costs.

[tex]Q = \sqrt{(2DS / H)[/tex]

Where,

D = annual demand in units

S = order cost per purchase order

H = holding cost per unit, per year

D = 50 cases x 52 weeks = 2,600 cases per year

S = $350 per purchase order

H = $110 x 25% = $27.50

Q = [tex]\sqrt{[(2 x 2,600 x 350) / 27.50][/tex]

[tex]= \sqrt{(1,820,000 / 27.5)[/tex]

= [tex]\sqrt{66,181.82[/tex]

= 257.26 cases

= 258 cases

Answer:

The typical error between a predicted college GPA using this regression model and an actual college GPA for a given student will be about 0.374 grade points in size (absolute value).

Step-by-step explanation:

The linear regression line for College GPA based on High school GPA is:

College GPA = 1.07 + 0.62 High-school GPA

It is provided that the standard error of the regression line is,

[tex]s_{e}=0.374[/tex]

The standard error of a regression line is the average distance between the predicted value and the regression equation.

It is the square root of the average squared deviations.

It is also known as the standard error of estimate.

The standard error of 0.374 implies that:

The typical error between a predicted college GPA using this regression model and an actual college GPA for a given student will be about 0.374 grade points in size (absolute value).

The standard error of the regression tells us the average amount that the actual college GPA deviates from the predicted college GPA. A smaller standard error indicates a better fit of the model to the data. In this case, the small standard error suggests that the linear regression model provides a good prediction of college GPA based on high school GPA for the students at the local college.

The standard error of the regression tells us the average amount that the actual college GPA deviates from the predicted college GPA based on the high school GPA. In this case, the standard error of the regression is 0.374. This means that, on average, the actual college GPA for a student deviates from the predicted college GPA by approximately 0.374.

This value gives us an idea of the accuracy of the linear regression model in predicting college GPA based on high school GPA. A smaller standard error indicates a better fit of the model to the data, implying that the predicted college GPA is closer to the actual college GPA. Conversely, a larger standard error suggests that the model's predictions are less accurate.

In this case, the standard error of the regression is relatively small (0.374), which indicates that the linear regression model provides a good prediction of college GPA based on high school GPA for the students at the local college.

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Step-by-step explanation:

Below is an attachment containing the solution.

Answer:

Probability of receiving the signal incorrectly is ".05792"

Step-by-step explanation:

Since it is mentioned here that the majority decoding is used so it implies that if the message is decoded as zero if there have been minimum of 3 Zeros in the messgae.

As we can see that there have been 5 bits, so the incorrection will occur only if there have been atleast 3 incorrect bits. Here we are asked to find out the probability of receiveing the signal incorrectly so i will take incorrectly transmiited bit as "Success".

The binomial distribution gives the probability of exactly m successes in n trials where the probability of  each individual trial succeeding is p.

With this method, we get the following pattern,

Binomial at (m=3, n=5, p=0.2) for probability of incorrection due to exactly 3 fails,

Binomial at (m=4, n=5, p=0.2) for probability of incorrection due to exactly 4 fails,

Binomial at (m=5 ,n=5, p=0.2) for probability of incorrection due to exactly 5 fails.

hence the probability "P" of receiving the signal incorrectly will be,

           P=(⁵₃) p³ (1 − p)⁵⁻³   +     (⁵₄) p⁴(1 − p)⁴⁻³      +      (⁵₅) p³(1 − p)³⁻³

whereas,

(⁵₃)= 5! / (5-3)! = 10

(⁵₄)= 5! / (5-4)! = 5

(⁵₅)= 5! / (5-5)! = 1

Putting these values in above equation we get,

P= 10 (0.2)³(1-0.2)²  + 5(0.2)⁴(1-0.2)¹  + (0.2)⁵

P=.05792

So the probablility "P" of receiving the signal incorrectly becomes ".05792"

Answer:

0% probability that at least 1,500 will agree to respond

Step-by-step explanation:

I am going to use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

[tex]E(X) = np[/tex]

The standard deviation of the binomial distribution is:

[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]

Normal probability distribution

Problems of normally distributed samples can be 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.

When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].

In this problem, we have that:

[tex]n = 15000, p = 0.09[/tex]

So

[tex]\mu = E(X) = np = 15000*0.09 = 1350[/tex]

[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{15000*0.09*0.91} = 35.05[/tex]

What is the probability that at least 1,500 will agree to respond

This is 1 subtracted by the pvalue of Z when X = 1500-1 = 1499. So

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

[tex]Z = \frac{1499 - 1350}{35.05}[/tex]

[tex]Z = 4.25[/tex]

[tex]Z = 4.25[/tex] has a pvalue of 1.

1 - 1 = 0

0% probability that at least 1,500 will agree to respond

The probability that at least 1,500 will agree to respond is 0.000009.

From the information given,

n = 15000

p = 0.09

Therefore, np = 15000 × 0.09

= 1350

nq = 15000 × (0.91)

= 13650

The probability that at least 1,500 will agree to respond will be:

= 1 - P (Z < (1500 - 1350)/35.05]

= 1 - P(Z < 4.2796)

= 0.000009

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Answer:

The first player should add 1/8 more dollar to the pot

Step-by-step explanation:

The probability that the first player wins on the first turn is equivalent to the probability of getting 5 in the dices. Out of the 36 possible outcomes for a dice, only 4 are favourable in the event of obtaining 5 in the sum:

- first dice 1, second dice 4

- first dice 2, second dice 3

- first dice 3, second dice 2

- first dice 4, second dice 1

Thus, the probability that the first player wins on the first turn is 4/36 = 1/9.

Note that if the first player doesnt win in the first turn, then the second player will be at the same position the first player was at the start of the game. If we call P the probability that the first player wins, then, the second player will have a probability of P of winning after 'surviving' on the first turn.

As a result, the probability that the second player wins is P multiplied by 8/9 (the probability that the first player doenst win in the first turn). In short

Probability that the first player wins = P

Probability that the second player wins = 8/9 * P

Since the sum should be 1 (because both events are complementary from each other), then P + P*8/9 = 1, thus

17/9 P = 1

P = 9/17

Lets call E the extra amount of money player A should include into the pot (in dollard). We have that both players should have expected gain equal to 0, and we have that

- First player gains 1 if he wins (with probability 9/17)

- First player gains -1-E if the loses (with probability 8/17)

Therefore

9/17*1+ 8/17 (-1-E) = 0

9/17 = 8/17 + E 8/17

E = (1/17) / (8/17) = 1/8

The first player should add 1/8 more dollar to the pot.

The test statistic for ANOVA is called an F test statistic (or F-ratio), which is calculated by dividing the variance between the samples by the variance within the samples.

The test statistic for ANOVA is called an F test statistic (also called an F-ratio). It is calculated by dividing the variance between the samples (variation due to treatment) by the variance within the samples (variation due to error or chance).

The F statistic follows an F distribution with (number of groups - 1) as the numerator degrees of freedom and (number of observations - number of groups) as the denominator degrees of freedom.

For ANOVA (Analysis of Variance), the test statistic is called an F test statistic (also called an F-ratio), which is the variance between samples (a.k.a., variation due to treatment) divided by the variance within samples (a.k.a., variation due to error or chance).

Fill in blank (2): between, Fill in blank (3): within

Analysis of Variance (ANOVA)

ANOVA is a statistical method used to compare the means of three or more samples to see if at least one of them is significantly different from the others. It does this by analyzing the variances within the data.

F-Test Statistic

The test statistic used in ANOVA is called the F-test statistic or the F-ratio. This F-ratio helps to determine whether the variances between the sample means are significantly larger than the variances within the samples. The F-ratio is calculated as follows:

[tex]\[ F = \frac{\text{variance between samples}}{\text{variance within samples}} \][/tex]

Variance Between Samples

The variance between samples (also known as between-group variance or treatment variance) measures the variability among the sample means. This variability reflects how much the group means differ from the overall mean. If the group means are very different from each other, the between-group variance will be large. This part of the variance is often attributed to the effect of the different treatments or conditions being compared.

- Blank (2): The term used here is between.

Variance Within Samples

The variance within samples (also known as within-group variance or error variance) measures the variability within each of the groups. This variability reflects how much the individual data points within each group differ from their respective group mean. This part of the variance is usually attributed to random error or chance.

- Blank (3): The term used here is within.

In the context of ANOVA:

- The F-test statistic (F-ratio) is used to compare the variance between samples to the variance within samples.

- The variance between samples represents the variation due to treatment.

- The variance within samples represents the variation due to error or chance.

The complete question is

For ANOVA, the test statistic is called an_ test statistic (also called an_-ratio), which is the variance (2) samples (a.k.a., variation due to treatment) divided by the variance _(3)_samples (a.k.a., variation due to error or chance) The first two blanks are completed with the letter this author uses for the ANOVA test statistic. What is this letter? Fill in blank (2): Fill in blank (3)

It appears that the question is incomplete but the answer to the part given is as below.

Answer:

(a) 0.0313

Step-by-step explanation:

In a normal distribution curve, the mean divides the curve into two equal parts. Hence, the probability of being lesser or higher than the mean is 1/2. This is the probability for a single day.

From the question, the probability of a day is independent of that of another day. For 5 days, the probability is

[tex](\frac{1}{2})^5 = 0.5^5 = 0.03125 = 0.0313[/tex] to 3 significant digits.

Answer:

The calculated 99% confidence interval is wider than the 95% confidence interval.      

Step-by-step explanation:

We are given the following in the question:

95% confidence interval for the population proportion

(0.65, 0.69)

Let [tex]\hat{p}[/tex] be the sample proportion

Confidence interval:

[tex]p \pm z_{stat}(\text{Standard error})[/tex]

[tex]z_{critical}\text{ at}~\alpha_{0.05} = 1.96[/tex]

Let x be the standard error, then, we can write

[tex]\hat{p} - 1.96x = 0.65\\\hat{p}+1.96x = 0.69[/tex]

Solving the two equations, we get,

[tex]2\hat{p} = 0.65 + 0.69\\\\\hat{p} = \dfrac{1.34}{2} = 0.67\\\\x = \dfrac{0.69 - 0.67}{1.96} \approx 0.01[/tex]

99% Confidence interval:

[tex]p \pm z_{stat}(\text{Standard error})[/tex]

[tex]z_{critical}\text{ at}~\alpha_{0.01} = 2.58[/tex]

Putting values, we get,

[tex]0.67 \pm 2.58(0.01)\\=0.67 \pm 0.0258\\=(0.6442,0.6958)[/tex]

Thus, the calculated 99% confidence interval is wider than the 95% confidence interval .

Answer:

From the calculation the interest income in 2018 = 10% of $2,612,000= $261,200

Step-by-step explanation:

The complete question states:

What amount of interest income should Benedict record in 2018 (the second year of the lease period) as a result of the lease?

To answer the question, we look at the given information

The present value given = $3,520,000

The Annual Instalmental payments = $600,000

The Period is 8 years and teh given interest rate at 10%

Based on the information we prepare the following schedule

Year 0 = Instalment + interest = Principal

$600,000 + 0 (interest) = $600,000

Balance = Present value- Principal

Balance = $3520,000 - $600,000 = $2,920,000

Year 1 (2017) = Instalment + interest = Principal

$600,000 + $292,000 (10% of $2,920,000) = $308,000

Balance = Present value- Principal

Balance = $2,920,000- $308,000= $2,612,000

Year 2 (2018) = Instalment + interest = Principal

$600,000 + $261,200(10% of $2,612,000) = $338,000

Balance = Present value- Principal

Balance = $2,612,000- $338,000= $2,350,800

From the calculation the interest income in 2018 = 10% of $2,612,000= $261,200

Answer:

Her speed on the mountain route was 52 miles per hour.

She travelled 12 miles per hour faster on the highway route.

Step-by-step explanation:

Speed is given by distance ÷ time taken

On the mountain,

distance = 312 miles

time = 6 hours

speed = 312 ÷ 6 = 52 miles per hour

On the highway,

distance = 320 miles

time = hours

speed = 320 ÷ 5 = 64 miles per hour

This is greater than the mountain route speed by 64 - 52 = 12 miles per hour.

Answer:

Speed at mountain route = 52 miles/hour

Speed at highway route = 64 miles/hour

Difference in speed = 12 miles/hour

Difference in speed = 18.75 % greater than mountain route

Step-by-step explanation:

As we know the speed is given by

Speed = Distance/time

Albuquerque to New Mexico using mountain route:

Speed = 312/6

Speed = 52 miles/hour

Albuquerque to New Mexico using highway route:

Speed = 320/5

Speed = 64 miles/hour

Difference in speed:

difference = 64 - 52 = 12 miles/hours

difference = 12/64*100 = 18.75 %

Therefore, Nella's speed at the highway route was 12 miles/hour greater than her speed at the mountain route.

Nella's speed at the highway route was 18.75 % greater than the speed at mountain route.

Options: A. Chi squared Statistics

B. ANOVA

C. Independent samples t-test.

D. z-test of a population proportion.

Answer:A. Chi squared Statistics

Step-by-step explanation: A Chi Squared Statistics is a Statistical technique or test measure that determines how an expectation compares to actual observation. Chi squared Statistics data is expected to have the following features

Such as the data must be RAW, RANDOM, MUTUALLY EXCLUSIVE, OBTAINED FROM INDEPENDENT VARIABLES, AND LARGE SAMPLES WHICH WILL BE ENOUGH.

the survey of one hundred adults from each ethnic/racial groups means the data possess all the characters of a Chi Squared Statistics or test measure.

To find the probability that a part is longer than 90.3 millimeters or shorter than 89.7 millimeters, calculate the cumulative probability for each scenario and subtract them from 1. The resulting probability is approximately 0.8413.

To find the probability that a part is longer than 90.3 millimeters or shorter than 89.7 millimeters, we need to calculate the cumulative probability for each scenario and then subtract them from 1.

Step 1: Calculate the z-scores for both values using the formula:

z = (x - mean) / standard deviation

For 90.3 millimeters:

z = (90.3 - 90.2) / 0.1 = 1

For 89.7 millimeters:

z = (89.7 - 90.2) / 0.1 = -5

Step 2: Use a standard normal distribution table or a calculator to find the cumulative probability for each z-score.

For a z-score of 1, the cumulative probability is approximately 0.8413.

For a z-score of -5, the cumulative probability is approximately 0.0000003.

Step 3: Subtract the cumulative probability for the shorter length from 1 and add the cumulative probability for the longer length.

Probability = (1 - 0.0000003) + 0.8413 = 0.8413

Therefore, the probability that a part is longer than 90.3 millimeters or shorter than 89.7 millimeters is approximately 0.8413.

Answer:

5.0 ft-lbf

Step-by-step explanation:

The force is

[tex]F = \dfrac{9}{6^x}[/tex]

This force is not a constant force. For a non-constant force, the work done, W, is

[tex]W = \int\limits^{x_2}_{x_1} {F(x)} \, dx[/tex]

with [tex]x_1[/tex] and [tex]x_2[/tex] the initial and final displacements respectively.

From the question, [tex]x_1 =0[/tex] and [tex]x_2 = 12[/tex].

Then

[tex]W = \int\limits^{12}_0 {\dfrac{9}{6^x}} \, dx[/tex]

Evaluating the indefinite integral,

[tex]\int\limits \dfrac{9}{6^x} \, dx =9 \int\limits\!\left(\frac{1}{6}\right)^x \, dx[/tex]

From the rules of integration,

[tex]\int\limits a^x\, dx = \dfrac{a^x}{\ln a}[/tex]

[tex]9 \int\limits \left(\frac{1}{6}\right)^x \, dx = 9\times\dfrac{(1/6)^x}{\ln(1/6)} = -5.0229\left(\dfrac{1}{6}\right)^x[/tex]

Returning the limits,

[tex]\left.-5.0229\left(\dfrac{1}{6}\right)^x\right|^{12}_0 = -5.0229(0.1667^{12} - 0.1667^0) = 5.0229 \approx 5.0 \text{ ft-lbf}[/tex]

Answer:

21 minutes

Step-by-step explanation:

Her expected commute time is given by the probability of having to wait for the train (0.2) multiplied by the commute time in this scenario (25 min), added to the probability of not having to wait for the train (1 - 0.2) multiplied by the commute time in this scenario (20 min). The expected commute time is:

[tex]E(X) = 0.2*25+(1-0.2)*20\\E(X) = 21\ minutes[/tex]

Her expected commute time is 21 minutes.

Answer:

a) 9.52% probability​ that, in a​ year, there will be 4 hurricanes.

b) 4.284 years are expected to have 4 ​hurricanes.

c) The value of 4 is very close to the expected value of 4.284, so the Poisson distribution works well here.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

In which

x is the number of sucesses

e = 2.71828 is the Euler number

[tex]\mu[/tex] is the mean in the given time interval.

6.9 per year.

This means that [tex]\mu = 6.9[/tex]

a. Find the probability​ that, in a​ year, there will be 4 hurricanes.

This is P(X = 4).

So

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

[tex]P(X = 4) = \frac{e^{-6.9}*(6.9)^{4}}{(4)!}[/tex]

[tex]P(X = 4) = 0.0952[/tex]

9.52% probability​ that, in a​ year, there will be 4 hurricanes.

b. In a 45​-year ​period, how many years are expected to have 4 ​hurricanes?

For each year, the probability is 0.0952.

Multiplying by 45

45*0.0952 = 4.284.

4.284 years are expected to have 4 ​hurricanes.

c. How does the result from part​ (b) compare to a recent period of 45 years in which 4 years had 4 ​hurricanes? Does the Poisson distribution work well​ here?

The value of 4 is very close to the expected value of 4.284, so the Poisson distribution works well here.

Final answer:

BJ must deposit approximately $41,191.39 today to reach their goal of $50,000 at the end of Year 5, considering a 4.5% interest rate.

Explanation:

To find out how much BJ must deposit today to reach their goal of $50,000 at the end of Year 5, we can use the concept of compound interest. The formula to calculate the future value of a lump sum deposit is:

Future Value = Principal Amount x (1 + Interest Rate)^Number of Periods

In this case, BJ wants to find out the principal amount, which is the deposit they need to make today. We have the future value ($50,000), the interest rate (4.5%), and the number of periods (5).

Let's substitute the values into the formula and solve for the principal amount:

$50,000 = Principal Amount x (1 + 0.045)^5

Simplifying the equation:

$50,000 = Principal Amount x 1.21550625

Dividing both sides by 1.21550625:

Principal Amount = $41,191.39

Therefore, BJ must deposit approximately $41,191.39 today to reach their goal.