find the probability that a group of 12 US adult riding the ski gondola would have had a mean weight greater than 167 lbs. so that their total weight would have been greater than the gondola maximum capacity of 2,004 lbs

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Answer 1

The probability of a group of 12 US adults riding the ski gondola having a mean weight greater than 167 lbs, so that their total weight would have been greater than the gondola maximum capacity of 2,004 lbs, is approximately 0.0002 or 0.02%.

To find the probability of a group of 12 US adults riding the ski gondola having a mean weight greater than 167 lbs, we need to use the central limit theorem.
Assuming that the weights of the adults are normally distributed with a mean of μ and a standard deviation of σ, the mean weight of the sample of 12 adults can be approximated by a normal distribution with a mean of μ and a standard deviation of σ/√12.
We know that the maximum capacity of the gondola is 2,004 lbs. Let's assume that the average weight of each adult is 150 lbs, which means that the total weight of the group would be 12 x 150 = 1,800 lbs.
To exceed the maximum capacity, the mean weight of the group would need to be greater than 2,004/12 = 167 lbs.
Using a standard normal distribution table or calculator, we can find the probability of a sample mean greater than 167 lbs with a standard deviation of σ/√12.
P(sample mean > 167) = P(Z > (167-150)/(σ/√12))
Let's assume a standard deviation of σ = 20 lbs.
P(sample mean > 167) = P(Z > (17)/(20/√12))
P(sample mean > 167) = P(Z > 3.6)
Using a standard normal distribution table, we can find that the probability of a Z-score greater than 3.6 is approximately 0.0002.

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Related Questions

Two samples, one of size 28 and the second of size 27, are selected to test the difference between two independent population means. How many degrees of freedom are used to find the critical value

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To test the difference between two independent population means with two samples, you need to calculate the degrees of freedom (df). Here's a step-by-step explanation:

1. Identify the sample sizes: The first sample has a size of 28 (n1 = 28), and the second sample has a size of 27 (n2 = 27).

2. Calculate the degrees of freedom for each sample: For each sample, subtract 1 from the sample size. For sample 1, df1 = n1 - 1 = 28 - 1 = 27. For sample 2, df2 = n2 - 1 = 27 - 1 = 26.

3. Combine the degrees of freedom: Add the degrees of freedom from each sample together to get the total degrees of freedom: df = df1 + df2 = 27 + 26 = 53.

In this case, you will use 53 degrees of freedom to find the critical value for testing the difference between the two independent population means.

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Tammy, John, Allison and Henry paid a total of $45 for movie tickets at the theater. Each movie ticket was the same price. How much did each person pay for a movie ticket

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Each person (Tammy, John, Allison, and Henry) paid $11.25 for a movie ticket.

To determine the cost of each movie ticket, we will divide the total amount paid by the number of people who bought tickets.

Total amount paid: $45
Number of people: Tammy, John, Allison, and Henry (4 people)

Step 1: Divide the total amount paid by the number of people.
$45 ÷ 4 = $11.25

So, Allison and each of her friends paid $11.25 for a movie ticket.

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In a recent poll, 110 people were asked if they liked dogs, and 66% said they did. Find the margin of error of this poll, at the 95% confidence level.

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The margin of error of this poll, at the 95% confidence level, is approximately 0.0993 or 9.93%.

To find the margin of error of a poll, we need to know the sample size and the confidence level. In this case, we have a sample size of 110 and a confidence level of 95%.

First, we need to find the standard error of the proportion:

standard error = [tex]\sqrt{(p\times (1-p)/n)}[/tex]

where p is the proportion who like dogs and n is the sample size.

p = 0.66

n = 110

[tex]standard error = \sqrt{(0.66 \times (1-0.66)/110)}[/tex]

              = 0.0507

Next, we need to find the critical value for a 95% confidence level. Since we have a large sample size (110), we can use the z-score table for a normal distribution. The critical value for a 95% confidence level is 1.96.

Finally, we can find the margin of error:

margin of error = critical value × standard error

               = 1.96 × 0.0507

               = 0.0993

Therefore, the margin of error of this poll, at the 95% confidence level, is approximately 0.0993 or 9.93%. This means that we can be 95% confident that the true proportion of people who like dogs in the population is between 66% +/- 9.93%.

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A(n) ______________________________ is formed by one side of the triangle and the extension of an adjacent side.

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A triangle is a three-sided polygon that consists of three sides and three angles. An extension of a side of a triangle is a line segment that is drawn from one of the endpoints of the side that extends beyond the side.

If we extend one side of the triangle and draw a line that passes through one of the adjacent vertices, the resulting shape is called a triangle's exterior angle. This exterior angle is formed by one side of the triangle and the extension of an adjacent side.

In other words, an exterior angle of a triangle is an angle that is formed by one side of a triangle and the extension of an adjacent side of the triangle.

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If there are seven ​multiple-choice questions on an​ exam, each having four possible​ answers, how many different sequences of answers are​ there?

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Answer: 16384

Work Shown:

4^7 = 16384

There are 16,384 different sequences of answers for an exam with seven multiple-choice questions, each having four possible answers.

To determine the number of different sequences of answers for an exam with seven multiple-choice questions, each having four possible answers, we can use the counting principle.

The counting principle states that if there are 'n' ways to do one thing and 'm' ways to do another, then there are n * m ways to do both.

In this case, there are four possible answers for each of the seven questions. To calculate the total number of sequences, we can multiply the number of choices for each question together:

4 (choices for Q1) * 4 (choices for Q2) * 4 (choices for Q3) * 4 (choices for Q4) * 4 (choices for Q5) * 4 (choices for Q6) * 4 (choices for Q7)

This simplifies to:

4^7 (4 raised to the power of 7)

Calculating this value gives us:

4^7 = 16,384

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1. Hyperbola: y-radius of 1, foci at (0, 3) and (10, 3)

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The equation of the hyperbola is y²/9 - x²/25 = 1

Given data ,

The equation of a hyperbola centered at the origin (0,0), with the foci at (0,3) and (10,3), and a y-radius of 1, can be written in the form:

y²/a² - x²/b² = 1

where "a" is the distance from the center to the vertices along the y-axis, and "b" is the distance from the center to the vertices along the x-axis.

Now , the foci are at (0,3) and (10,3), the distance from the center to the vertices along the y-axis is 3 units (which is the y-radius), so "a" is 3

The distance between the foci along the x-axis is 10 units, so "2b" is 10,which means "b" is 5

Plugging these values into the equation, we get:

y²/3² - x²/5² = 1

Simplifying, we get:

y²/9 - x²/25 = 1

Hence , the equation is y²/9 - x²/25 = 1

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The average score in the NFL is normally distributed. The average score is 43.06 points, with a standard deviation of 15 points. What is the probability that the average of 16 selected games scored larger than 38 points

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To solve this problem, we need to use the formula for the z-score:z = (x - μ) / (σ / sqrt(n)), where x is the sample mean (which is the average score in 16 selected games), μ is the population mean (which is 43.06 points), σ is the population standard deviation (which is 15 points), and n is the sample size (which is 16).

Given the average score in the NFL is 43.06 points with a standard deviation of 15 points. Since we are looking at the average of 16 selected games, we need to calculate the standard error, which is the standard deviation divided by the square root of the sample size (in this case, 16):

Standard error = 15 / √16 = 15 / 4 = 3.75

Now, we will calculate the z-score for 38 points, which represents how many standard errors 38 points is away from the average score:

Z-score = (38 - 43.06) / 3.75 = -1.35

Using a z-score table or calculator, we find the probability of a z-score being less than -1.35 is approximately 0.0885 or 8.85%.

Since we want the probability that the average of 16 selected games scored larger than 38 points, we need to find the complement of this probability:

Probability (average score > 38) = 1 - 0.0885 = 0.9115 or 91.15%

So, the probability that the average of 16 selected games scored larger than 38 points is approximately 91.15%.

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A complete collection of all elements (scores, people, measurements, and so on) to be studied is called the Group of answer choices sample population parameter grade

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The correct option is B. The complete collection of all elements, individuals, measurements, scores, or other entities that are of interest and relevant to a particular study or analysis is called the population.

A collection is a group of items that have been gathered or accumulated together based on a particular theme or purpose. Collections can be made up of physical objects such as books, stamps, coins, or artwork, as well as digital items like photos, music, or videos. Collections can also refer to data structures in computer programming that hold groups of related items.

People often collect items as a hobby or passion, and the act of collecting can bring a sense of fulfillment, enjoyment, and satisfaction. Collections can have both sentimental and monetary value, and they may be displayed in personal or public settings, such as museums or galleries. The process of collecting often involves researching and acquiring new items, as well as organizing and preserving the existing collection.

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When a supervisor for a survey telephones a subset of the respondents to verify certain information, it is an example of _________.

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When a supervisor for a survey telephones a subset of the respondents to verify certain information, it is an example of quality control or data validation. This process helps ensure the accuracy and reliability of the survey responses.

When a supervisor for a survey telephones a subset of the respondents to verify certain information, it is an example of follow-up or validation. This is a common practice in survey research to ensure the accuracy and validity of the data collected.

During the follow-up process, a subset of the respondents are contacted again to confirm or validate their responses. This may involve asking additional questions or asking the respondent to clarify or elaborate on their previous answers. The purpose of the follow-up is to ensure that the data collected is reliable and accurate, and to identify and correct any errors or discrepancies.

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quizleet Harrison wants to compare the creativity scores of people who prefer Star Wars to those who prefer Star Trek. He explains that he is going to use a Pearson's r statistical test for analyze his data. What would you tell Harrison

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It is important for Harrison to ensure that his sample is representative of the population he is interested in and that his measures of creativity are reliable and valid. He should also consider potential confounding variables, such as age or gender, that may affect his results.

I would advise Harrison that Pearson's r is a measure of correlation, which examines the linear relationship between two continuous variables. However, it may not be the best statistical test for comparing the creativity scores of people who prefer Star Wars to those who prefer Star Trek, as creativity scores may not necessarily have a linear relationship with preference for either franchise.

Instead, a more appropriate statistical test may be an independent samples t-test or ANOVA, which can compare the means of two or more groups on a continuous variable. These tests are more appropriate when the variable of interest is continuous and the groups being compared are categorical.

Additionally, it is important for Harrison to ensure that his sample is representative of the population he is interested in and that his measures of creativity are reliable and valid. He should also consider potential confounding variables, such as age or gender, that may affect his results.

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Greg has 4 shirts: a white one, a black one, a red one, and a blue one. He also has two pairs of pants, one blue and one tan. What is the probability, if Greg gets dressed in the dark, that he winds up wearing the white shirt and tan pants

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The probability of Greg winding up wearing the white shirt and tan pants when getting dressed in the dark is 1/8 or 12.5%.

The probability of Greg wearing the white shirt and tan pants when getting dressed in the dark can be calculated by finding the probability of each event occurring independently and then multiplying those probabilities together.

First, let's find the probability of Greg choosing the white shirt. He has 4 shirts to choose from, so the probability of picking the white shirt is 1/4 (one white shirt out of four total shirts).

Next, let's find the probability of Greg choosing the tan pants. He has 2 pairs of pants to choose from, so the probability of picking the tan pants is 1/2 (one tan pair out of two total pairs).

Now, we can multiply the two probabilities together to find the overall probability of Greg wearing the white shirt and tan pants: (1/4) * (1/2) = 1/8.

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What is the probability that the number of times Luke will hit the inner ring of the target out of the 5 attempts is less than the mean of X ?

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The probability that Luke will hit the inner ring fewer than 3 times is  0.069.

How to calculate the probability the number of times Luke will hit the inner ring of the target out?

To calculate the probability that the number of times Luke will hit the inner ring of the target out of the 5 attempts is less than the mean of X, we need to know the distribution of X.

Assuming that each shot is independent and has the same probability p of hitting the inner ring, X follows a binomial distribution with parameters n=5 and p.

The mean of a binomial distribution is given by μ = np, so in this case, the mean of X is 5p.

To find the probability that X is less than 5p, we can use the cumulative distribution function (CDF) of the binomial distribution. Let F(k) denote the CDF of the binomial distribution with parameters n=5 and p, evaluated at k.

Then the probability that X is less than 5p is:

P(X < 5p) = F(4p)

Note that we use 4p instead of 5p in the argument of F, since we want the probability that X is strictly less than 5p, not less than or equal to 5p.

Using a binomial table or calculator, we can look up or compute the value of F(4p) for a given value of p.

For example, if p=0.6 (which corresponds to Luke hitting the inner ring 60% of the time), we get:

P(X < 5p) = F(2.4) ≈ 0.069

So the probability that Luke will hit the inner ring fewer than 3 times (which is less than 5p=3) out of the 5 attempts is about 0.069, assuming he hits the inner ring with a probability of 0.6 on each shot.

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For a given population, the mean of all the sample means Picture of sample size n, and the mean of all (N) population observations (X) are _______.

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the answer is D equal to the population mean…

The mean of all the sample means (which is equal to the population mean) and the mean of all (N) population observations (X) are both equal to the population mean μ.

The mean of all the sample means of a given population is equal to the population mean, which is denoted by the symbol μ. This is a consequence of the central limit theorem, which states that the distribution of the sample means becomes approximately normal as the sample size n becomes larger, with mean equal to the population mean μ.

The mean of all (N) population observations (X) is simply the population mean, which is also denoted by the symbol μ. It represents the average value of the variable of interest across all individuals in the population.

Therefore, the mean of all the sample means (which is equal to the population mean) and the mean of all (N) population observations (X) are both equal to the population mean μ.

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Determine whether the following interaction plot suggests that significant interaction exists among the factors.

Does significant interaction exist among the factors?

a).No, because the lines cross more than once.

b). No, because the lines are relatively parallel.

c). Yes, because there are significant differences in the slopes of the lines.

d). Yes, because the lines are almost a mirror image of each other.

Answers

The correct answer is c) Yes, because there are significant differences in the slopes of the lines.

we cannot ignore the interaction between the factors when interpreting the results of the experiment.



An interaction plot is a graphical representation of the interaction between two factors in an experiment. It shows how the response variable changes across different levels of the two factors. If there is no interaction between the factors, the lines on the plot will be relatively parallel. If there is a significant interaction, the lines will cross or have different slopes.

In this case, the fact that there are significant differences in the slopes of the lines suggests that there is a significant interaction between the factors. This means that the effect of one factor on the response variable depends on the level of the other factor. Therefore, we cannot ignore the interaction between the factors when interpreting the results of the experiment.

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ARCHERY The height, in feet, of an arrow can be modeled by the expression 89- 161², where is the time in
seconds. Factor the expression.

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The factored expression is 89 - 161² = - ( 161 + √89 ) ( 161 - √89 ).

How to determine factored expression?

The expression can be factored as:

89 - 161² = -161² + 89

Use the difference of squares formula, which states that:

a² - b² = ( a + b )( a - b )

In this case,:

a = 161 and b = √89

So, write:

-161² + 89 = - ( 161 + √89 ) ( 161 - √89 )

Therefore, the factored expression is:

89 - 161² = - ( 161 + √89 )( 161 - √89 )

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A machine produces bolts which are 6% defective. A random sample of 100 bolts produced by this machine are collected. a) Find the exact probability that there are at most 3 defectives in the sample. Write your answer in decimal form. b) Find the probability that there are at most 3 defectives by normal approximation. Write your answer in decimal form. c) Find the probability that between 4 and 7, inclusive, are defective by normal approximation. Write your answer in decimal form.

Answers

a)The exact probability that there are at most 3 defectives in the sample is 0.4234. b) The exact probability that there are at most 3 defectives in the sample is 0.4234. c) the probability that between 4 and 7, inclusive, are defective by normal approximation is 0.2

a) To find the exact probability that there are at most 3 defectives in the sample, we can use the binomial distribution formula. The probability of getting at most 3 defectives is the sum of the probabilities of getting 0, 1, 2, or 3 defectives.

P(X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

Where X is the number of defective bolts in the sample.

Using the binomial distribution formula, we get:

P(X = k) = (n choose k) * p^k * (1-p)^(n-k)

Where n is the sample size (100), p is the probability of a bolt being defective (0.06), and k is the number of defective bolts.

P(X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
         = 0.4234

Therefore, the exact probability that there are at most 3 defectives in the sample is 0.4234.

b) To find the probability that there are at most 3 defectives by normal approximation, we need to calculate the mean and standard deviation of the binomial distribution.

Mean = np = 100 * 0.06 = 6
Standard deviation = sqrt(np(1-p)) = sqrt(100 * 0.06 * 0.94) = 2.424

We can then use the normal distribution to approximate the binomial distribution:

P(X ≤ 3) ≈ P(Z ≤ (3.5 - 6)/2.424)

Where Z is a standard normal random variable.

Using a standard normal table or calculator, we get:

P(Z ≤ -1.23) = 0.1093

Therefore, the probability that there are at most 3 defectives by normal approximation is 0.1093.

c) To find the probability that between 4 and 7, inclusive, are defective by normal approximation, we can use the same approach as in part b.

Mean = np = 100 * 0.06 = 6
Standard deviation = sqrt(np(1-p)) = sqrt(100 * 0.06 * 0.94) = 2.424

We can then use the normal distribution to approximate the binomial distribution:

P(4 ≤ X ≤ 7) ≈ P(3.5 ≤ X ≤ 7.5) ≈ P((3.5 - 6)/2.424 ≤ Z ≤ (7.5 - 6)/2.424)

Where Z is a standard normal random variable.

Using a standard normal table or calculator, we get:

P(-1.23 ≤ Z ≤ 0.62) = 0.2816

Therefore,  the probability that between 4 and 7, inclusive, are defective by normal approximation is 0.2

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Which item does the author use to describe the appearance of the two Things?

A
a horse's head

B
a fisherman's basket

C
a thin hail

D
a gap in the lightning

Answers

I'm sorry, but I would need more context about what "two Things" the author is referring to in order to answer your question accurately. Can you please provide more information or context?

Suppose that there are no restrictions on how many pages a printer can print. How many ways are there for the 100 pages to be assigned to the four printers

Answers

There are 176,851 ways to assign the 100 pages to the four printers.

Since there are no restrictions on how many pages a printer can print, we can think of this problem as distributing 100 identical pages among 4 distinct printers. This is an example of a "balls and urns" problem, which can be solved using the stars and bars formula.

The stars and bars formula states that the number of ways to distribute k identical objects among n distinct containers is:

C(k+n-1, n-1)

where C represents the combination function. In this case, we have k = 100 identical pages and n = 4 distinct printers. Therefore, the number of ways to assign the pages to the printers is:

C(100+4-1, 4-1) = C(103, 3) = 176,851

So there are 176,851 ways to assign the 100 pages to the four printers.

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what is the probability that someone has two aces if you know they have an ace versus if you know they have the ace of spades

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The probability that someone has two aces given that they have one ace is 3/51, or about 5.88%. The probability that someone has two aces given that they have the ace of spades is 2/50, or about 4%.

To see why these probabilities are different, consider that knowing someone has one ace doesn't give you any information about whether they have a second ace.

The probability of drawing an ace from a standard deck of 52 cards is 4/52, or 1/13. Therefore, the probability of drawing two aces is (1/13) * (3/51), which is about 0.0588.

On the other hand, if you know that someone has the ace of spades, then you know that one of the two aces has already been accounted for. The probability of drawing the other ace is now 2/50, or 1/25.

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: A satellite system consists of 4 components and can operate adequately if at least 2 of the 4 components are functional. If each component is, independently, functional with probability 0.6, what is the probability that the system operates adequately

Answers

The probability that the satellite system operates adequately is 0.7056.

The probability that a component is not functional is 0.4. Therefore, the probability that a component is functional is 1-0.4=0.6.
Using the rule of combinations, there are 6 possible combinations of functional and non-functional components:
1. All 4 components are functional: (0.6)^4=0.1296
2. 3 components are functional: (0.6)^3(0.4)=0.3456
3. 2 components are functional: (0.6)^2(0.4)^2=0.2304
4. 1 component is functional: (0.6)(0.4)^3=0.0256
5. No components are functional: (0.4)^4=0.0256
6. At least 2 components are functional: P(2 or 3 or 4) = 0.1296 + 0.3456 + 0.2304 = 0.7056
Therefore, the probability that the satellite system operates adequately is 0.7056.

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If you have a sample size of 16 and a population standard deviation of 40, what is your standard error of the mean

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The standard error of the mean is 10.

The standard error of the mean (SEM) is calculated by dividing the

population standard deviation (σ) by the square root of the sample size (n):

SEM = σ / √n

In this case, the population standard deviation is 40 and the sample size is 16.

Substituting these values into the formula, we get:

SEM = 40 / √16

SEM = 40 / 4

SEM = 10

Therefore, the standard error of the mean is 10.

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there are 24 chairs in 4 rows how many chairs are in one row

Answers

Answer:

6

Step-by-step explanation:

total number=24

number of row = 4

now,

number in one row=24/4

=6

Penny flips three fair coins into a box with two compartments. Each compartment is equally likely to receive each of the coins. What is the probability that either of the compartments has at least two coins that landed heads

Answers

The probability that either of the compartments has at least two coins that landed heads is 161/192.

To solve this problem, we can use the principle of inclusion-exclusion.

Let A be the event that the first compartment has at least two coins that landed heads, and let B be the event that the second compartment has at least two coins that landed heads. We want to find the probability of the union of these events: P(A ∪ B).

To compute P(A ∪ B), we need to compute the probabilities of A, B, and A ∩ B.

The probability of A is the probability that at least two of the three coins landed heads in the first compartment, and the remaining coin landed tails in either compartment. There are three ways this can happen:

HHT (with probability 1/8)

HTH (with probability 3/8)

THH (with probability 3/8)

So, the probability of A is (1/8) + (3/8) + (3/8) = 7/8.

Similarly, the probability of B is also 7/8.

To compute the probability of A ∩ B, we can use the multiplication rule:

P(A ∩ B) = P(A) × P(B | A)

where P(B | A) is the probability that the second compartment has at least two coins that landed heads, given that the first compartment has at least two coins that landed heads.

To compute P(B | A), we can condition on the number of heads in the first compartment:

If the first compartment has exactly two heads, then there is only one way to distribute the remaining coin, which is to put it in the second compartment. So, the probability of B in this case is 1/2.

If the first compartment has three heads, then there are three ways to distribute the remaining coin, two of which result in the second compartment having at least two heads. So, the probability of B in this case is 2/3.

Therefore,

P(B | A) = (1/2) × P(first compartment has exactly two heads) + (2/3) × P(first compartment has three heads)

= (1/2) × (3/8) + (2/3) × (1/8)

= 7/24.

Hence,

P(A ∩ B) = (7/8) × (7/24) = 49/192.

Now, we can apply the inclusion-exclusion principle:

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

= (7/8) + (7/8) - (49/192)

= 161/192.

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A random sample of 100 stores from a large chain of 1,000 garden supply stores was selected to determine the average number of lawnmowers sold at an end-of-season clearance sale. The sample results indicated an average of 6 and a standard deviation of 2 lawnmowers sold. A 95% confidence interval (5.623 to 6.377) was established based on these results. True or False: Of all possible samples of 100 stores drawn from the population of 1,000 stores, 95% of the sample means will fall between 5.623 and 6.377 lawnmowers

Answers

True, The 95% confidence interval (5.623 to 6.377) means that if we were to repeat this sampling process multiple times,

About 95% of the intervals we construct will contain the true population average number of lawnmowers sold. Since the sample size is 100, the Central Limit Theorem applies, and the distribution of sample means will be approximately normal, regardless of the population distribution.

Therefore, 95% of all possible samples of 100 stores drawn from the population of 1,000 stores will have a sample mean between 5.623 and 6.377 lawnmowers sold.


True. Based on the 95% confidence interval (5.623 to 6.377) calculated from the sample of 100 stores, it can be concluded that 95% of all possible sample means drawn from the population of 1,000 stores will fall within this range, with an average of 6 lawnmowers sold.

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Part a.) If you apply the distributive property first to solve the equation, what operation will you need to do last? Part b.) If instead you divide first to solve the equation, what operation would you need to use last?

Answers

The equation is solved and the distributive property is used

Given data ,

a)

If you use the distributive property to solve the equation first, either addition or subtraction will need to be done last, depending on the equation. This is so that you may isolate the variable on one side of the equation or combine like terms after applying the distributive principle, which usually requires addition or subtraction as the last step.

b)

Instead, if you divide first to answer the problem, you would need to apply multiplication as the last operation. This is because, in order to "undo" the division operation and find the variable, you would need to multiply by the reciprocal of the value by which you had divided to isolate the variable. As the inverse operation of division, multiplication would be utilized as the last step in the equation's solution.

Hence , the equations are solved

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t was also reported that 32% of those with an allergy are allergic to multiple foods. If a child younger than 18 is randomly selected, what is the probability that he or she is allergic to multiple foods

Answers

The probability of a child under 18 being allergic to multiple foods is approximately 32%. This means that out of every 100 children with an allergy, we would expect 32 of them to be allergic to more than one food.

To solve this problem, we need to use the information provided in the question. We know that 32% of those with an allergy are allergic to multiple foods. This means that out of every 100 people with an allergy, 32 of them are allergic to more than one food.

If we assume that allergies are equally common in all age groups, then we can use this percentage to estimate the probability of a child under 18 being allergic to multiple foods. However, it is important to note that this assumption may not be entirely accurate, as some allergies are more common in children than in adults.

Assuming that allergies are equally common in all age groups, the probability of a child under 18 being allergic to multiple foods is approximately 32%. This means that out of every 100 children with an allergy, we would expect 32 of them to be allergic to more than one food.

It is important to note that this is just an estimate based on the information provided in the question. In reality, the probability of a child being allergic to multiple foods may be higher or lower depending on a variety of factors such as genetics, environment, and diet.

In order to get a more accurate estimate of the probability of a child being allergic to multiple foods, we would need to collect more data and analyze it using statistical methods. However, based on the information provided in the question, we can conclude that there is a significant likelihood that a child under 18 with an allergy is also allergic to multiple foods.

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A fair coin is tossed 10 times, given that there were 4 heads in the 10 tosses, what is the probability that the first toss was head

Answers

The probability of getting a head on the first toss is 0.5, regardless of the results of the other tosses.

Given that there were 4 heads in the 10 tosses, we can use Bayes' Theorem to find the probability that the first toss was a head:

P(First toss was head | 4 heads in 10 tosses) = P(4 heads in 10 tosses | First toss was head) * P(First toss was head) / P(4 heads in 10 tosses)

We know that P(First toss was head) = 0.5, and we can calculate P(4 heads in 10 tosses | First toss was head) using the binomial distribution:

P(4 heads in 10 tosses | First toss was head) = (10 choose 4) * (0.5)^4 * (0.5)^6 = 210 * 0.0625 * 0.015625 = 0.206

To calculate P(4 heads in 10 tosses), we can use the binomial distribution again:

P(4 heads in 10 tosses) = (10 choose 4) * (0.5)^4 * (0.5)^6 = 210 * 0.0625 * 0.015625 + 210 * 0.9375 * 0.984375 = 0.2051

Therefore, we can calculate the probability that the first toss was a head given that there were 4 heads in 10 tosses:

P(First toss was head | 4 heads in 10 tosses) = 0.206 * 0.5 / 0.2051 = 0.503

So the probability that the first toss was a head given that there were 4 heads in 10 tosses is approximately 0.503.

the volume of a box is 96 cubic inches. the length is 8 inches more than the height. the width is 2 inches less than the height. find the dimensions of the box

Answers

The dimensions of the box are 12 inches by 2 inches by 4 inches.

Let's use variables to represent the dimensions of the box:

Let h be the height of the box (in inches).

Then, the length of the box is 8 inches more than the height, so it is h + 8.

The width of the box is 2 inches less than the height, so it is h - 2.

The volume of the box is given as 96 cubic inches, so we can set up an equation:

Volume = Length × Width × Height

96 = (h + 8) × (h - 2) × h

h = 4

Height: h = 4 inches

Length: h + 8 = 12 inches

Width: h - 2 = 2 inches

Therefore, the dimensions of the box are 12 inches by 2 inches by 4 inches.

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The mean points obtained in an aptitude examination is 103103 points with a variance of 169169. What is the probability that the mean of the sample would differ from the population mean by less than 2.82.8 points if 6363 exams are sampled

Answers

The probability that the mean of the sample would differ from the population mean by less than 2.8 points if 63 exams are sampled is 0.9592 or approximately 95.92%.

It is necessary to determine the region under the normal distribution curve between -2.8/0.5916 and 2.8/0.5916 standard deviations from the mean in order to determine the likelihood that the sample mean deviates from the population mean by less than 2.8 points. We can use a standard normal distribution table or a calculator to find this probability.

Mean of the distribution of sample means = 103

Variance of the distribution of sample means = 169/63 = 2.68

Using the z-score method, we can determine the likelihood that the sample mean deviates from the population mean by no more than 2.8 points.:

z = (x - μ) / (σ / √(n))

where n is the sample size, x is the sample mean, is the population mean, is the population standard deviation (square root of the variance), and is the population mean.

Plugging in the values, we get:

z = (x - 103) / (√(169/63))

z = (x - 103) / 0.5916

We calculate the chance to be roughly 0.9592 using a conventional normal distribution table.

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______ analysis is the statistical process of estimating the relationship between dependent and ______ variables. Multiple Choice Time series, outcome Regression, outcome Time series, latent Regression, independent

Answers

The statistical method of determining the link between dependent and outcome variables is called regression analysis. Here option B is the correct answer.

Regression analysis is a statistical method used to estimate the relationship between a dependent variable and one or more independent variables. The dependent variable is the outcome variable of interest, while the independent variable is the variable used to explain the variation in the dependent variable. Regression analysis is widely used in many fields such as economics, finance, psychology, engineering, and social sciences.

In a typical regression analysis, the relationship between the dependent variable and the independent variable(s) is estimated using a regression equation. The regression equation provides a mathematical formula to predict the value of the dependent variable based on the values of the independent variable(s). The regression equation can be used to identify the strength and direction of the relationship between the dependent and independent variable(s).

Regression analysis is useful for many purposes, such as understanding the factors that influence an outcome variable, predicting the value of the outcome variable based on the values of the independent variable(s), and evaluating the effectiveness of an intervention or treatment.

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Complete question:

______ analysis is the statistical process of estimating the relationship between dependent and ______ variables.

a) Time series, outcome

b) Regression, outcome

c) Time series, latent

d) Regression, independent

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