Hot answers tagged


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....


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: $$f(x,y) = \dfrac{e^{-x^2/2}}{\sqrt{2\pi}} * \dfrac{e^{-y^2/2}}{\sqrt{2\pi}} = \dfrac{e^{-(x^2+y^2)/2}}{2\pi}=g(x^2+y^2)$$


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 ...


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 ...

Only top voted, non community-wiki answers of a minimum length are eligible