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Zinaida [17]
3 years ago
12

.

Mathematics
1 answer:
brilliants [131]3 years ago
7 0
\sf Hello!

\sf Let\: the \:required % \sf be\: x

\sf Then,

\sf x % × \sf 50 = \sf 9

⇒ \dfrac{\sf x}{\sf 100} × \sf 50 = \sf 9

⇒ \dfrac{\sf x}{\sf 2} = \sf 9

⇒ \sf X = \sf 9 × 2

⇒ \boxed{\sf x = 18}

\sf Therefore,
\sf D.\: 18 % \sf of \:50\: is \:9

~ \sf iCarl
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Vedmedyk [2.9K]

Using the z-distribution, we have that:

a) A sample of 601 is needed.

b) A sample of 93 is needed.

c) A.  ​Yes, using the additional survey information from part​ (b) dramatically reduces the sample size.

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of \alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which z is the z-score that has a p-value of \frac{1+\alpha}{2}.

The margin of error is:

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For this problem, we consider that we want it to be within 4%.

Item a:

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  • There is no estimate, hence \pi = 0.5

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.04 = 1.96\sqrt{\frac{0.5(0.5)}{n}}

0.04\sqrt{n} = 1.96\sqrt{0.5(0.5)}

\sqrt{n} = \frac{1.96\sqrt{0.5(0.5)}}{0.04}

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Rounding up:

A sample of 601 is needed.

Item b:

The estimate is \pi = 0.96, hence:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.04 = 1.96\sqrt{\frac{0.96(0.04)}{n}}

0.04\sqrt{n} = 1.96\sqrt{0.96(0.04)}

\sqrt{n} = \frac{1.96\sqrt{0.96(0.04)}}{0.04}

(\sqrt{n})^2 = \left(\frac{1.96\sqrt{0.96(0.04)}}{0.04}\right)^2

n = 92.2

Rounding up:

A sample of 93 is needed.

Item c:

The closer the estimate is to \pi = 0.5, the larger the sample size needed, hence, the correct option is A.

For more on the z-distribution, you can check brainly.com/question/25404151  

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