# Tag Info

## Hot answers tagged standard-deviation

2

Your calculations are correct. The extreme $z$ score you get indicates that if the percentage of households was really $33$%, then this sample is highly unrepresentative. The conclusion you would therefore draw is that the actual percentage must in reality be lower than $33$%, assuming that the sample is random and representative of the population as a whole....

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eventually I think I have managed to solve that. The key issue is to calculate the density function $f(x,y)$ of the 2 variables. which because the variables are independent we have: $$f(x,y) = \dfrac{e^{-x^2/2}}{\sqrt{2\pi}} * \dfrac{e^{-y^2/2}}{\sqrt{2\pi}} = \dfrac{e^{-(x^2+y^2)/2}}{2\pi}=g(x^2+y^2)$$

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It is best to use a boxplot to find outliers. The problem with using the sample mean $\bar X$ and the sample SD $S$ is that an outlier seriously affects the values of $\bar X$ and $S$. By contrast, the boxplot uses the median and the interquartile range to detect outliers. These measures of location and dispersion, respectively, are not much affected by ...

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In your problem, there are five independent experiments, each of which is the sum of two die rolls. This is different from ten dice rolls. For example, you would expect a mean of 7 from your experiment, and 3.5 from the single dice rolls. In excel, create two columns of five rows of random die rolls (=INT(RAND()*6)+1 in cells A1..B5), and then add the ...

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