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tekilochka [14]
3 years ago
13

According to a poll, 56% of voters support funding a new community center. A surveyor randomly selects 75 voters and asks each v

oter if he or she is in favor of the center. If being in favor of the center is a success, what is the probability of a success for this binomial experiment? 0.34 0.44 0.56 0.75
Mathematics
1 answer:
mestny [16]3 years ago
7 0
The answer is "0.56".

As <span> 56% of voters support funding a new community center, so
56/100 = 0.56.

A binomial experiment refers to a measurable test that has the accompanying properties: The trial comprises of n repeated trials. Every trial can bring about only two conceivable results. We call one of these results a win and the other, a disappointment. The likelihood of accomplishment, meant by p, is the same on each trial.
</span>
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(b) The 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.198, 0.260).

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Let <em>X</em> = number of students who read above the eighth grade level.

(a)

A sample of <em>n</em> = 269 students are selected. Of these 269 students, <em>X</em> = 224 students who can read above the eighth grade level.

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The proportion of students who can read above the eighth grade level is 0.8327.

Compute the proportion of tenth graders reading at or below the eighth grade level as follows:

1-\hat p=1-0.8327

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Thus, the proportion of tenth graders reading at or below the eighth grade level is 0.1673.

(b)

the information provided is:

<em>n</em> = 709

<em>X</em> = 546

Compute the sample proportion of tenth graders reading at or below the eighth grade level as follows:

\hat q=1-\hat p

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The critical value of <em>z</em> for 95% confidence interval is:

z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

Compute the 95% confidence interval for the population proportion as follows:

CI=\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}

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Thus, the 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.198, 0.260).

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