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snow_lady [41]
3 years ago
12

Using the table for many repetitions of the experiment of tossing a coin 10 times, what are the mean number of heads in 50 repet

itions of this experiment and 500 repetitions and which of these sample means is closer to the population mean? Number of Heads 0 1 2 3 4 5 6 7 8 9 10
In 50 Repetitions 0 0 1 5 9 16 10 6 2 1 0
In 200 Repetitions 0 2 8 22 38 55 41 24 9 1 0
In 500 Repetitions 2 5 24 57 111 111 110 56 21 3 0
In 1000 Repetitions 1 8 43 117 207 248 203 121 45 6 1

4.8, 5.03; 50 repetitions should be closer to the population mean
4.9, 4.99; 50 repetitions should be closer to the population mean
5.2, 4.94; 500 repetitions should be closer to the population mean
5.1, 4.9; 500 repetitions should be closer to the population mean
Mathematics
1 answer:
Artyom0805 [142]3 years ago
6 0

Answer:

5.2, 4.94; 500 repetitions should be closer to the population mean  

Step-by-step explanation:

A. Mean of 50 repetitions

\begin{array}{rrr}\\\mathbf{f} &\mathbf{x} &\mathbf{f\cdot x}\\0 & 0 & 0\\1 & 0 & 0\\2 & 1 & 2\\3 & 5 & 15\\4 & 9 & 36\\5 & 16 & 80\\6 & 10 & 60\\7 & 6 & 42\\8 & 2 & 16\\9 & 1 &9\\10 & 0 & 0\\\text{Sum =}&\mathbf{50} &\mathbf{260}\\\end{array}\\\\\text{Mean} = \dfrac{\sum{f \cdot x}}{n} = \dfrac{260}{50} = \mathbf{5.20}

B. Mean of 500 repetitions

\begin{array}{rrr}\\\mathbf{f} &\mathbf{x} &\mathbf{f\cdot x}\\0 & 2 & 0\\1 & 5 & 5\\2 & 24 & 48\\3 & 57 & 171\\4 & 111 & 444\\5 & 111 & 555\\6 & 110 & 660\\7 & 56 & 392\\8 & 21 & 168\\9 & 3 &27\\10 & 0 & 0\\\text{Sum =}&\mathbf{500} &\mathbf{2470}\\\end{array}\\\\\text{Mean} = \dfrac{\sum{f \cdot x}}{n} = \dfrac{2470}{500} = \mathbf{4.94}

500 repetitions should be closer to the population mean because. Per the Central Limit Theorem, the mean of the sample should approach the mean of the population as the sample size increases.

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1003.4375

Step-by-step explanation:

16055/16 is 1003.4375

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A group of friends wants to go to the amusement
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Answer:

The maximum amount of people that can go to the amusement park is 15.

Step-by-step explanation:

First you create an equation to represent the question

8.75+25.75x≤420

You put the less than or equal to sign because 420 is the maximum amount of money.

Now you just solve the equation.

25.75x≤420-8.75

25.75x≤411.25

x≤411.25/25.75

x≤15.9708738

Then you have to round down because you cant have .9 of a person.

So x≤15

Then, Check your solution

8.75+(25.75 x 15)≤420

8.75+386.25≤420

395≤420

That is true. But to make sure that is the maximum add another 25.75

395+25.75≤420

420.75≤420

This is equation is false so, our answer is correct.

The maximum amount of people that can go to the amusement park is 15.

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When the sample size and the sample proportion remain the same, a 90 percent confidence interval for a population proportion p w
deff fn [24]

Answer:

A 90% confidence interval for <em>p</em> will be <u>narrower </u>than the 99% confidence interval.

Step-by-step explanation:

The formula to compute the (1 - <em>α</em>) % confidence interval for a population proportion is:

CI=\hat p\pm z_{\alpha /2}\sqrt{\frac{\hat p(1-\hat p)}{n}}

Here \hat p is the sample proportion.

The margin of error of the confidence interval is:

MOE= z_{\alpha /2}\sqrt{\frac{\hat p(1-\hat p)}{n}}

The MOE is dependent on:

  1. Confidence level
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  3. Sample size

The MOE is directly related to the confidence level and standard deviation.

So if any of the two increases then the MOE also increases, thus widening the confidence interval.

And the MOE is inversely related to the sample size.

So if the sample increases the MOE decreases and vice versa.

It is provided that the sample size and the sample proportion are not altered.

The critical value of <em>z</em> for 90% confidence level is:

z_{\alpha/2}= z_{0.10/2}=z_{0.05}=1.645

And the critical value of <em>z</em> for 99% confidence level is:

z_{\alpha/2}= z_{0.05/2}=z_{0.05}=1.96

So as the confidence level increases the critical value increases.

Thus, a 90% confidence interval for <em>p</em> will be narrower than the 99% confidence interval.

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