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horsena [70]
3 years ago
10

Whole numbers are written on cards and then placed in a bag. Denzel randomly selects a single card, writes down the number, and

then places it back in the bag. He repeats this 110 times.
Denzel calculates the relative frequency of each number card.

Outcome 1 2 3 4 5
Relative Frequency 0.23 0.41 0.09 0.13 0.15
Which statement about Denzel's experiment is true?

The outcomes appear to be equally likely, so a uniform probability model is a good model to represent probabilities in Denzel's experiment.

The outcomes do not appear to be equally likely, so a uniform probability model is a good model to represent probabilities in Denzel's experiment.

The outcomes appear to be equally likely, so a uniform probability model is not a good model to represent probabilities in Denzel's experiment.

The outcomes do not appear to be equally likely, so a uniform probability model is not a good model to represent probabilities in Denzel's experiment.
Mathematics
2 answers:
mash [69]3 years ago
7 0
<span>Relative Frequency 0.23 0.41 0.09 0.13 0.15
</span>
So t<span>he outcomes do not appear to be equally likely.
</span>
Uniform model is only good for equally likely probabilities so it is not good here.

Correct ans is:

<span>The outcomes do not appear to be equally likely, so a uniform probability model is not a good model to represent probabilities in Denzel's experiment.</span>
goldenfox [79]3 years ago
3 0
Whole numbers are written on cards and then placed in a bag. Denzel randomly selects a single card, writes down the number, and then places it back in the bag. He repeats this 110 times.
Which statement about Denzel's experiment is true?
(D) The outcomes do not appear to be equally likely, so a uniform probability model is not a good model to represent probabilities in Denzel's experiment.
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An automobile manufacturer would like to know what proportion of its customers are not satisfied with the service provided by th
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Answer:

a) A sample size of 5615 is needed.

b) 0.012

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.

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

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

The margin of error is:

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

99.5% confidence level

So \alpha = 0.005, z is the value of Z that has a pvalue of 1 - \frac{0.005}{2} = 0.9975, so Z = 2.81.

(a) Past studies suggest that this proportion will be about 0.2. Find the sample size needed if the margin of the error of the confidence interval is to be about 0.015.

This is n for which M = 0.015.

We have that \pi = 0.2

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

0.015 = 2.81\sqrt{\frac{0.2*0.8}{n}}

0.015\sqrt{n} = 2.81\sqrt{0.2*0.8}

\sqrt{n} = \frac{2.81\sqrt{0.2*0.8}}{0.015}

(\sqrt{n})^{2} = (\frac{2.81\sqrt{0.2*0.8}}{0.015})^{2}

n = 5615

A sample size of 5615 is needed.

(b) Using the sample size above, when the sample is actually contacted, 12% of the sample say they are not satisfied. What is the margin of the error of the confidence interval?

Now \pi = 0.12, n = 5615.

We have to find M.

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

M = 2.81\sqrt{\frac{0.12*0.88}{5615}}

M = 0.012

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