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serious [3.7K]
4 years ago
9

A researcher is interested in determining if the more than two thirds of students would support making the Tuesday before Thanks

giving a holiday. The researcher asks 1,000 random selected students if they would support making the Tuesday before Thanksgiving a holiday. Seven hundred students said that they would support the extra holiday.
Define the parameter.

a. p = the population proportion of UF students who would support making the Tuesday before Thanksgiving break a holiday.

b. phat = the sample proportion of 700 UF students who would support making the Tuesday before Thanksgiving break a holiday.

c. phat = the population proportion of UF students who would support making the Tuesday before Thanksgiving break a holiday.

d. p = the population proportion of 700 UF students who would support making the Tuesday before Thanksgiving break a holiday.
Mathematics
1 answer:
klio [65]4 years ago
8 0

Answer:

a. p = the population proportion of UF students who would support making the Tuesday before Thanksgiving break a holiday.

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they are in favor of making the Tuesday before Thanksgiving a holiday, or they are against. This means that we can solve this problem using concepts of the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which C_{n,x} is the number of different combinatios of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

So, the binomial probability distribution has two parameters, n and p.

In this problem, we have that n = 1000 and p = \frac{700}{1000} = 0.7. So the parameter is

a. p = the population proportion of UF students who would support making the Tuesday before Thanksgiving break a holiday.

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Answer:

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Step-by-step explanation:

Given

Base\ Amount = 12

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We roll two fair 6-sided dice. Each of the 36 possible outcomes is assumed to be equally likely. (a) Given that the roll results
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Answer:

a) \frac{1}{7}

b) \frac{2}{15}

Step-by-step explanation:

Given : We roll two fair 6-sided dice. Each of the 36 possible outcomes is assumed to be equally likely.

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(1, 1) (1, 2)  (1, 3)  (1, 4)  (1, 5)  (1, 6)

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(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)

(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)

(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)

(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)

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Let P(A) be the probability of event that the roll results in a sum of 3 or more.

Except (1,1) rest sum is 3 or greater than 3.

So, P(A)=\frac{35}{36}

Now, P(A and B) is that the roll results in a sum of 3 or more and that doubles (the first and the second rolls result in the same number) are rolled.

i.e. (2,2), (3,3), (4,4), (5,5), (6,6) - 5

P(A\cap B)=\frac{5}{36}

The conditional probability is given by,

P(B|A)=\frac{P(A\cap B)}{P(A)}

P(B|A)=\frac{\frac{5}{36}}{\frac{35}{36}}

P(B|A)=\frac{5}{35}

P(B|A)=\frac{1}{7}

(b) Given that the two dice land on different numbers, find the conditional probability that the sum is 5.

Let P(A) be the probability of event that two dice land on different numbers.

Except (1,1),(2,2), (3,3), (4,4), (5,5), (6,6) rest two dice land on different numbers.

So, P(A)=\frac{30}{36}

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P(A\cap B)=\frac{4}{36}

The conditional probability is given by,

P(B|A)=\frac{P(A\cap B)}{P(A)}

P(B|A)=\frac{\frac{4}{36}}{\frac{30}{36}}

P(B|A)=\frac{4}{30}

P(B|A)=\frac{2}{15}

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