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Likurg_2 [28]
3 years ago
14

Listed below are the highest amounts of net worth (in millions of dollars) of celebrities. The celebrities are Tom Cruise, Will

Smith, Robert De Niro, Drew Carey, George Clooney, John Travolta, Samuel L. Jackson, Larry King, Demi Moore, and Bruce Willis. Find the range, variance and standard deviation. Are the measures of variation typical for all celebrities?
{ 250, 200, 185, 165, 160, 160, 150, 150, 150, 150 }
Mathematics
1 answer:
emmainna [20.7K]3 years ago
8 0

<u><em>The range's difference is the highest and lowest value of the data</em></u>

<u><em>RANGE= 250-150= 100</em></u>

<u><em>The mean would be the sum of all values divided by how many values </em></u>

<u><em>250+200+185+165+160+150+150+150+150 / 10 = 1720/10 = 172</em></u>

<u><em>The variance is the sum of squared deviations </em></u>

<u><em>*n=10 </em></u>

<u><em>s^2 = (250-172)^2+(150-172)^2 / 10-1</em></u>

<u><em>which equals an esitmate of 1034.4444 </em></u>

<u><em>take the square root and you get your final answer of 32.1628</em></u>

<u><em>The measure of the variation isn't typical for all celebs, since the numbers are only based on ten highest networths </em></u>

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Answer:

The population growth is 6.26 per square mile per year

Step-by-step explanation:

We are given

he US Census found at the population per square mile in the state rose from 128.7 to 191.3 during the last 10 years

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Initial population =128.7

Final Population =191.3

Total time =10 years

now, we can find population grow per year

=\frac{final-initial}{T}

now, we can plug values

=\frac{191.3-128.7}{10}

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The population growth is 6.26 per square mile per year

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4 years ago
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Answer:

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Alex787 [66]

Answer:

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Step-by-step explanation:

The estimated proportion of interest is \hat p=0.45

We need to find a critical value for the confidence interval using the normla standard distributon. For this case we have 95% of confidence, then the significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. And the critical value is:

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The confidence interval for the true population proportion is interest is given by this formula:  

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

Replacing the values provided we got:

0.45 - 1.96\sqrt{\frac{0.45(1-0.45)}{36}}=0.287

0.45 + 1.96\sqrt{\frac{0.45(1-0.45)}{36}}=0.613

And we can conclude at 95% of confidence that the true proportion of interest for this case is between 0.287 and 0.613

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