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I would disagree with Michael's statement that "Pointing out trivial consequences like "log turns product into sum..." is very misleading." In fact, those consequences are precisely WHY the logs are used when finding the MLE for a parameter. The MLE has the value that it has in order to minimize the Kullback-Liebler divergence. However, I think this ...


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A late answer, just for completeness with a different view on the thing. You might look at your data as measured in a multidimensional space, where each subject is a dimension and each item is a vector in that space from the origin towards the items' measurement over the full subject's space. Additional remark: this view of things has an additional ...


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Standardize so that you can use the standard normal distribution to find the numerical value for the answer. So, we have $$\begin{align} \mathbb{P}(X \geq 203) &= \mathbb{P}\left(\frac{X - 200}{10} \geq \frac{203 -200}{10}\right) \\ &= \mathbb{P}\left(Z \geq \frac{3}{10}\right) \\ &= 1 - \mathbb{P}\left(Z < \frac{3}{10}\right) \\ &= ...


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If $X$ is a random variable, then $E(X + k) = E(X) + k,$ $E(cX) = c E(X),$ $Var(X + k) = Var(X),$ and $Var(cX) = c^2 Var(X).$ (These are facts for any random variable, including normal variables.) If you already know these facts, you can use them to determine the mean and variance of $Y - \mu,$ then use them again to find your desired result.


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EDIT: if there is some expensive decision riding on this, consider trying to get funding for an hour of consulting from a graduate student or professor in statistics there. These issues are always about interpretation, and need the hand of a master. The last time I saw something like this, it was a medical doctor doing a study on lung cancer or the like, but ...


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Let $x$ be the maximum height of an adult that admits a clearance of at least $17$ cm for the doorway. Let $h$ be the height of such a doorway, so $h = x + 17$. Now we wish to find the value of $x$ such that $99\%$ of adults have height less than or equal to $x$; i.e., this is $\Pr[X \le x] = 0.99$ = invnorm(0.99,187.5,9.5) = 209.600. Therefore, $x + 17$ ...


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You have to use the fact that the weights are normally distributed. You're given the standard deviation, and a value for the left 4% of the distribution. You can calculate (or look up) how many standard deviations away from the mean you are at 4%. Given this, you multiply the number of standard deviations from the mean by the standard deviation (in lbs) ...


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Hint: The equation is $P(X \leq 20)=\Phi\left( \frac{20-\mu}{4.2} \right)=0.04$, where $\Phi(.)$ is the standard normal distribution. Solve the equation for $\mu$. The value for $\Phi^{-1}(0.04)$ can be looked up in the table for the standard normal distribution. Additional: It might be helpful, that $1-\Phi(z)=\Phi(-z)$. Insert the value for ...


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Let the weight of the chickens be $X$. Then $X \sim N(\mu,\sigma^2)$ where $\sigma = 4.2$ is the standard deviation of the weights and $\mu$ is the mean of the weights. Then, the statement you're given is $P(X < 20) = 0.04$. Now, rewrite $P(X<20) = P(X-\mu < 20-\mu) = P( \frac{X-\mu}{\sigma} < \frac{20-\mu}{\sigma}) = P(Z < ...


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Note that the article you reference does not guarantee the unbiasedness of the estimator in your post. A good approach in this situation is to jacknife or bootstrap correct the bias. Unlike the sample variance estimate, the sample standard deviation estimator bias is very sensitive to the actual distribution (or family of distributions).


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The $x$ variable will be $\chi$-distributed. http://en.wikipedia.org/wiki/Chi_distribution


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It's not clear to me why you say the mean is 41 and the standard deviation is 2. Because the box-and-whisker plot is not skewed it follows that your dataset is quite symmetric. This means the mean is close to the average of 39 and 63, which is 51. As for the standard deviation, you can find for example here: ...


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You made a minor simplification mistake. With $X\sim N(1,\sigma^2)$ $$ 0.9=Pr(0.98<X\leq 1.02)=Pr\left(\frac{0.98-1}{\sigma}<\frac{X-1}{\sigma}\leq\frac{1.02-1}{\sigma}\right)=Pr\left(\frac{-0.02}\sigma<Z\leq\frac{0.02}{\sigma}\right) $$ where $Z=(X-1)/\sigma$ has a standard normal distribution, the CDF of which we denote by $\Phi$. Then, $$ ...


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You can perform a transformation on each standard deviation $SD_i$ so that all transformed values ($tSD_i$) are distributed in the range $0-1$. Simply call $x_{min}$ and $x_{max}$ the smallest and the largest among all $SD_i$, respectively, and then use $$ \displaystyle tSD_i= \frac{SD_i-x_{min}}{x_{max}-x_{min}}$$


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Your final observation about the $25$th, $50$th, and $75$th percentiles just says the distribution is not symmetric. Do you have a reason to believe that it should be? Using the standard deviation assumes the distribution is Gaussian, which the evidence contradicts. All you really know is that $350.2$ is somewhere between the $75$th and $90$th percentile. ...


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Since you know the upper ($H$) and lower ($L$) bounds of your index range, the "ideal" index distribution would be discrete-uniform on $[L,H]$. Unfortunately, the standard deviation, skew, or kurtosis will only partially characterize what you are looking for. I have actually had to deal with the same issue as you (that of finding a maximally "even" ...



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