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Ierofanga [76]
2 years ago
11

Select the correct answer. which data set is the farthest from a normal distribution? a. 2, 3, 3, 4, 4, 4, 5, 5, 6 b. 3, 4, 5, 6

, 6, 7, 7, 7, 8, 8, 9, 10 c. 0.9, 1.0, 1.0, 1.1, 1.1, 1.1, 1.2, 1.2, 1.3 d. 4, 4, 4, 5, 5, 6, 7, 7, 8, 8, 8 e. 2, 4, 5, 5, 6, 6, 7, 7, 7, 8, 8, 9, 9, 10
Mathematics
1 answer:
tigry1 [53]2 years ago
4 0

The answer choice which is the farthest from a normal distribution is; Choice E; 2, 4, 5, 5, 6, 6, 7, 7, 7, 8, 8, 9, 9, 10.

<h3>Which data set is farthest from a normal distribution?</h3>

A normal distribution, is a data set which when graphed must follow a bell-shaped symmetrical curve centered around the mean. Additionally, such distribution must adhere to the empirical rule that indicates the percentage of the data set that falls within (plus or minus) 1, 2 and 3 standard deviations of the mean.

On this note, upon evaluation of the data sets, it follows that answer choice E represents the data set that's most farthest from a normal distribution.

Read more on normal distribution;

brainly.com/question/26678388

#SPJ4

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The difference is 6. Each number increases by 6 in the sequence.
2 + 6 = 8
8 + 6 = 14
14 + 6 = 20
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Etc.
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When grading an exam, 90% of a professor's 50 students passed. If the professor randomly selected 10 exams, what is the probabil
Damm [24]

Using the binomial distribution, it is found that there is a:

a) 0.9298 = 92.98% probability that at least 8 of them passed.

b) 0.0001 = 0.01% probability that fewer than 5 passed.

For each student, there are only two possible outcomes, either they passed, or they did not pass. The probability of a student passing is independent of any other student, hence, the binomial distribution is used to solve this question.

<h3>What is the binomial probability distribution formula?</h3>

The formula is:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

In this problem:

  • 90% of the students passed, hence p = 0.9.
  • The professor randomly selected 10 exams, hence n = 10.

Item a:

The probability is:

P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10)

In which:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 8) = C_{10,8}.(0.9)^{8}.(0.1)^{2} = 0.1937

P(X = 9) = C_{10,9}.(0.9)^{9}.(0.1)^{1} = 0.3874

P(X = 10) = C_{10,10}.(0.9)^{10}.(0.1)^{0} = 0.3487

Then:

P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) = 0.1937 + 0.3874 + 0.3487 = 0.9298

0.9298 = 92.98% probability that at least 8 of them passed.

Item b:

The probability is:

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

Using the binomial formula, as in item a, to find each probability, then adding them, it is found that:

P(X < 5) = 0.0001

Hence:

0.0001 = 0.01% probability that fewer than 5 passed.

You can learn more about the the binomial distribution at brainly.com/question/24863377

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Answer:

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Answer:

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Answer:

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Step-by-step explanation:

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We need to find the length x of the hypotenuse.

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