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Pepsi [2]
2 years ago
9

Jasmine created this spinner to predict whether a randomly selected student in her school will be traveling during the upcoming

holiday break.
Jasmine’s spinner is designed to reflect data she gathered by surveying 36 students in her gym class.

How many of the students Jasmine surveyed indicated they would be traveling during the holiday break?
2
9
12
27

Mathematics
2 answers:
riadik2000 [5.3K]2 years ago
4 0

Answer: 9

Step-by-step explanation:

There are 8 equal sections which are meant to reflect the data Jasmine collected from surveying the students.

This means that each section represents;

= 36/8

= 4.5 students.

There are 2 sections that represent the number of students who said they would be traveling during the holiday break.

= 4.5 * 2

= 9 students

Keith_Richards [23]2 years ago
3 0

Answer:

The number of students Jasmine surveyed indicated they would be traveling during the holiday break is 9.

Step-by-step explanation:

The spinner created by Jasmine, to predict whether a randomly selected student in her school will be traveling during the upcoming holiday break, has 8 equal sections labeled:

{yes, yes, no, no, no, no, no, no}

So, there are two sections labeled yes and six labelled as no.

Compute the probability of a yes as follows:

Jasmine selected n = 36 students in her gym class.

If a student in her school will be traveling during the upcoming holiday break is independent of the other students.

Let the random variable X be defined as the number f students that will be traveling during the upcoming holiday break.

The random variable X follows a Binomial distribution with parameters n = 36 and p = 0.25.

Compute the expected value of X as follows:

Thus, the number of students Jasmine surveyed indicated they would be traveling during the holiday break is 9.

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​41% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the
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Answer:

a) 0.2087 = 20.82% probability that the number of U.S. adults who have very little confidence in newspapers is exactly​ five.

b) 0.1834 = 18.34% probability that the number of U.S. adults who have very little confidence in newspapers is at least​ six.

c) 0.3575 = 35.75% probability that the number of U.S. adults who have very little confidence in newspapers is less than four.

Step-by-step explanation:

For each adult, there are only two possible outcomes. Either they have very little confidence in newspapers, or they do not. The answers of each adult are independent, which means that the binomial probability distribution is used to solve this question.

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 combinations 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.

​41% of U.S. adults have very little confidence in newspapers.

This means that p = 0.41

You randomly select 10 U.S. adults.

This means that n = 10

(a) exactly​ five

This is P(X = 5). So

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

P(X = 5) = C_{10,5}.(0.41)^{5}.(0.59)^{5} = 0.2087

0.2087 = 20.82% probability that the number of U.S. adults who have very little confidence in newspapers is exactly​ five.

(b) at least​ six

This is:

P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)

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

P(X = 6) = C_{10,6}.(0.41)^{6}.(0.59)^{4} = 0.1209

P(X = 7) = C_{10,7}.(0.41)^{7}.(0.59)^{3} = 0.0480

P(X = 8) = C_{10,8}.(0.41)^{8}.(0.59)^{2} = 0.0125

P(X = 9) = C_{10,9}.(0.41)^{9}.(0.59)^{1} = 0.0019

P(X = 10) = C_{10,10}.(0.41)^{10}.(0.59)^{0} = 0.0001

Then

P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.1209 + 0.0480 + 0.0125 + 0.0019 + 0.0001 = 0.1834

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(c) less than four.

This is:

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

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

P(X = 0) = C_{10,0}.(0.41)^{0}.(0.59)^{10} = 0.0051

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P(X = 2) = C_{10,2}.(0.41)^{2}.(0.59)^{8} = 0.1111

P(X = 3) = C_{10,3}.(0.41)^{3}.(0.59)^{7} = 0.2058

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P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0051 + 0.0355 + 0.1111 + 0.2058 = 0.3575

0.3575 = 35.75% probability that the number of U.S. adults who have very little confidence in newspapers is less than four.

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