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vodomira [7]
4 years ago
7

Consider three events A, B and C with following properties.

Mathematics
1 answer:
lisov135 [29]4 years ago
5 0
By the inclusion/exclusion principle,

\mathbb P(D)=\mathbb P(A\cup B\cup C)
\mathbb P(D)=\mathbb P(A)+\mathbb P(B)+\mathbb P(C)-\bigg(\mathbb P(A\cap B)+\mathbb P(A\cap C)+\mathbb P(B\cap C)\bigg)+\mathbb P(A\cap B\cap C)
\mathbb P(D)=\dfrac14+\dfrac16+\dfrac14-\dfrac39+\dfrac1{13}
\mathbb P(D)=\dfrac{16}{39}\neq1

So the union of A, B, and C does not constitute the entire sample space.
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Step-by-step explanation:

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4 years ago
Darren wants to estimate the mean age in a population of trees. He'll sample nnn trees and build a 90\%90%90, percent confidence
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Using the z-distribution, as we have the standard deviation for the population, it is found that the smallest sample size required to obtain the desired margin of error is of 77.

<h3>What is a z-distribution confidence interval?</h3>

The confidence interval is:

\overline{x} \pm z\frac{\sigma}{\sqrt{n}}

In which:

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  • \sigma is the standard deviation for the population.

The margin of error is given by:

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In this problem, we have that the parameters are given as follows:

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Hence, solving for n, we find the sample size.

M = z\frac{\sigma}{\sqrt{n}}

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\sqrt{n} = \frac{1.645 \times 16}{3}

(\sqrt{n})^2 = \left(\frac{1.645 \times 16}{3}\right)^2

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Rounding up, the smallest sample size required to obtain the desired margin of error is of 77.

More can be learned about the z-distribution at brainly.com/question/25890103

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