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3 years ago

A. 2 B. 3 C. 12 D. 15

Mathematics

1 answer:

satela [25.4K]3 years ago

5 0

count the numbers next to the 5, 6 & 7 = 10. then there are 2 numbers next to 8 less than 2

so 10 +2 = 12 students

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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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sq rt of 29584= 172
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sq rt of 21904 = 148
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We call these perfect squares.

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Answer:

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Step-by-step explanation:

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