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3

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 ...


2

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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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: ...


1

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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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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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 < ...


1

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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