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Phoenix [80]
3 years ago
9

About 1414​% of the population of a large country is nervous around strangersnervous around strangers. If two people are randoml

y​ selected, what is the probability both are nervous around strangersnervous around strangers​? What is the probability at least one is nervous around strangersnervous around strangers​?
Mathematics
1 answer:
seraphim [82]3 years ago
6 0

One person being nervous around strangers is assumed  independent of other people behaviour. Then, the multiplication rule can be applied as

both nervous = 0.14*0.14 = 0.0196 = 1.96%

one is nervous, one is not nervous = 0.14*(1-0.14) = 0.1204 = 12.04%

At least one nervous = 1.96%  + 12.04% = 14%

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Answer:

A sample of 18 is required.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1 - 0.92}{2} = 0.04

Now, we have to find z in the Z-table as such z has a p-value of 1 - \alpha.

That is z with a pvalue of 1 - 0.04 = 0.96, so Z = 1.88.

Now, find the margin of error M as such

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

In which \sigma is the standard deviation of the population and n is the size of the sample.

A previous study indicated that the standard deviation was 2.2 days.

This means that \sigma = 2.2

How large a sample must be selected if the company wants to be 92% confident that the true mean differs from the sample mean by no more than 1 day?

This is n for which M = 1. So

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

1 = 1.88\frac{2.2}{\sqrt{n}}

\sqrt{n} = 1.88*2.2

(\sqrt{n})^2 = (1.88*2.2)^2

n = 17.1

Rounding up:

A sample of 18 is required.

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<em>Answer</em>


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<em>Explanation</em>

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