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notka56 [123]
3 years ago
7

Given a mean of 571 and a standard deviation of 236.5, what is the z-score of the value 724 rounded to the nearest tenth? a.0.9

b.–0.9 c.–0.6 d.0.6
Mathematics
2 answers:
tangare [24]3 years ago
6 0
The answer to this question is 0.6.
Nata [24]3 years ago
6 0

Answer:

d. 0.6

Step-by-step explanation:

Since, the formula for calculating the z-score or standard score is,

z=\frac{x-\mu}{\sigma}

Where, x is a score,

\mu is mean score,

\sigma is the standard deviation,

Here,

x = 724,

\mu = 571,

\sigma = 236.5

Hence, the z-score is,

z=\frac{724-571}{236.5}

=\frac{153}{236.5}

=0.646934460888\approx 0.6

Option d is correct.

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Answer:

A sample of n = (\frac{1.96\sqrt{0.3696*0.6304}}{E})^2 is needed, in which E is the desired margin of error, as a proportion. If we find a decimal value, we round up to the next whole number.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

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The margin of error is of:

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In a previous study of 1012 randomly chosen respondents, 374 said that there should be such a law.

This means that n = 1012, \pi = \frac{374}{1012} = 0.3696

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

How large a sample size is needed to be 95% confident with a margin of error of E?

A sample size of n is needed, and n is found when M = E.

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

E = 1.96\sqrt{\frac{0.3696*0.6304}{n}}

E\sqrt{n} = 1.96\sqrt{0.3696*0.6304}

\sqrt{n} = \frac{1.96\sqrt{0.3696*0.6304}}{E}

(\sqrt{n})^2 = (\frac{1.96\sqrt{0.3696*0.6304}}{E})^2

n = (\frac{1.96\sqrt{0.3696*0.6304}}{E})^2

A sample of n = (\frac{1.96\sqrt{0.3696*0.6304}}{E})^2 is needed, in which E is the desired margin of error, as a proportion. If we find a decimal value, we round up to the next whole number.

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