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Fittoniya [83]
3 years ago
9

What is mila’s average swimming speed?

Mathematics
1 answer:
skad [1K]3 years ago
5 0

Answer:

she swam 1500 meters in 1/4 of the time and he swam 750 in half of his time. Hope this helps

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Two sisters, sister A and sister B, play SCRABBLE with each other every evening. Sister A is a statistician, and she draws a ran
snow_tiger [21]

Answer:

First question: LCL = 522, UCL = 1000.5

Second question: A sample size no smaller than 418 is needed.

Step-by-step explanation:

First question:

Lower bound:

0.36 of 1450. So

0.36*1450 = 522

Upper bound:

0.69 of 1450. So

0.69*1450 = 1000.5

LCL = 522, UCL = 1000.5

Second question:

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}}

The project manager believes that p will turn out to be approximately 0.11.

This means that \pi = 0.11

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.

The project manager wants to estimate the proportion to within 0.03

This means that the sample size needed is given by n, and n is found when M = 0.03. So

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

0.03 = 1.96\sqrt{\frac{0.11*0.89}{n}}

0.03\sqrt{n} = 1.96\sqrt{0.11*0.89}

\sqrt{n} = \frac{1.96\sqrt{0.11*0.89}}{0.03}

(\sqrt{n})^2 = (\frac{1.96\sqrt{0.11*0.89}}{0.03})^2

n = 417.9

Rounding up

A sample size no smaller than 418 is needed.

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Step-by-step explanation:

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