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aev [14]
3 years ago
7

A researcher studying public opinion of proposed Social Security changes obtains a simple random sample of 35 adult Americans an

d asks them whether or not they support the proposed changes. To say that the distribution of the sample proportion of adults who respond​ yes, is approximately​ normal, how many more adult Americans does the researcher need to sample in the following​ cases?
(a) 20% of all adult Americans support the changes(b) 25% of all adults Americans support the changes
Mathematics
1 answer:
Liula [17]3 years ago
6 0

Answer:

Adult required in the case of “a” 28 and in the case of “b” the adult requirement is 19.

Step-by-step explanation:

(a) The percentage of adult that support the change is 20 percent.

Now calculate the number of adult required.

Given p = 0.20

Use the below condition:

np(1 – p) \geq 10 \\n \times 0.20 (1 – 0.20) = 10 \\n = 63 round off

Since 35 adults are already there so required adults are 63 -35 = 28

(b) The percentage of adult that support the change is 25 percent.

Now calculate the number of adult required.

Given p = 0.25

Use the below condition:

np(1 – p) \geq 10 \\n \times 0.25 (1 – 0.25) = 10 \\n = 54 (round off)

Since 35 adults are already there so required adults are 54 -35 = 19 .

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scoundrel [369]

Answer:

answer D

Step-by-step explanation:

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2 years ago
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Radda [10]
36 pints of green paint.
3 0
2 years ago
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A multiple-choice examination has 15 questions, each with five answers, only one of which is correct. Suppose that one of the st
Alex

Answer:

0.0111% probability that he answers at least 10 questions correctly

Step-by-step explanation:

For each question, there are only two outcomes. Either it is answered correctly, or it is not. The probability of a question being answered correctly is independent from other questions. So we use the binomial probability distribution 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.

A multiple-choice examination has 15 questions, each with five answers, only one of which is correct.

This means that n = 15, p = \frac{1}{5} = 0.2

What is the probability that he answers at least 10 questions correctly?

P(X \geq 10) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)

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

P(X = 10) = C_{15,10}.(0.2)^{10}.(0.8)^{5} = 0.0001

P(X = 11) = C_{15,11}.(0.2)^{11}.(0.8)^{4} = 0.000011

P(X = 12) = C_{15,12}.(0.2)^{12}.(0.8)^{3} \cong 0

P(X = 13) = C_{15,13}.(0.2)^{13}.(0.8)^{2} \cong 0

P(X = 14) = C_{15,14}.(0.2)^{14}.(0.8)^{1} \cong 0

P(X = 15) = C_{15,15}.(0.2)^{15}.(0.8)^{0} \cong 0

P(X \geq 10) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) = 0.0001 + 0.000011 = 0.000111

0.0111% probability that he answers at least 10 questions correctly

3 0
2 years ago
Can someone help me out with my math homework my grades are ba
kupik [55]
<h3>Answer:  18</h3>

==========================================================

Explanation:

2 cm on paper (or on screen) represents 3 meters in real life.

The backyard is 2 cm horizontally across on the screen, so it's 3 meters across in real life. This is the shorter dimension.

The longer dimension is 6 meters. We can think of it like this

2 cm : 3 meters

2*2 cm : 2*3 meters .... multiply both sides by 2

4 cm : 6 meters

The real backyard is 6 meters by 3 meters to give an area of 6*3 = 18 square meters.

7 0
2 years ago
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