# Tag Info

1

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 ...

0

Absolutely. It is a known fact that for a sufficiently long list , (denoting mean by $\mu$ and standard deviation by $\sigma$) the range $[\mu-3\sigma,\mu+3\sigma]$ encompasses about (more than) $99.73\%$ of the data points, so if the new value is out of this range then it is $99.7\%$sure to be out of the list You can somewhat use the concept of $p-value$ ...

0

Yes. You can use your Standard Deviation to tell you this. Think about what Standard Deviation is telling you.

5

Maximising the standard deviation or variance for a given mean is equivalent to maximising the sum of squares of the values for a given sum of the values. Meanwhile if $b \ge a$ and $\delta \gt 0$, $$(a-\delta)^2 +(b+\delta)^2$$ $$= a^2-2a\delta+\delta^2+b^2+2b\delta + \delta^2$$ $$= a^2+b^2 +2(b-a)\delta+2\delta^2$$ $$\gt a^2+b^2$$ so the sum of ...

3

First, you might as well assume that the mean is zero, so that $a < 0 < b$; that just simplifies things. (Otherwise: let $a' = a - \mu; b' = b - \mu$ and you can convert your problem for $[a, b], \mu$ to mine for $[a, b], 0$ by subtracting $\mu$ from all the sample values, etc.) The SD is $$s = \sum x_i^2$$ and you want to maximize this subject to ...

1

Obviously if $d$ is the greatest distance from any of the values to the mean, then every $(x_i - \bar{x})^2 \le d^2$; plug into the formula for SD to find that SD $\le d$. Moreover, it can only be $d$ when all distances are equal to $d$, no less; and since the average must still be $d$, this means half the values are at one extreme and half at the other. And ...

3

Let $y_i = x_i - \bar{x}$ then by Cauchy-Schwarz we have the following: $\sum_{i=1}^n |y_i| \frac{1}{n} \leq \|y\|_2 \sqrt{\sum_{i=1}^n \frac{1}{n^2}} = \sqrt{\frac{1}{n}\sum_{i=1}^n |x_i - \bar{x}|^2}$

1

If $X_i$s are independent and you have already their standard deviations, say $\sigma_i$, for $X_i$, which seem to be the same, then the variance of each $X_i$, for $i=1,\cdots, n$ is $\sigma^2_i$, and so the variance of $X$ is equal to $\sum_{i=1}^n\sigma^2_i=n\times \sigma^2_i$. As a result, the standard deviation of $X$ is $\sqrt{n\times \sigma^2_i}=\... 1 You should interpret the statement Experience suggests that the standard deviation is more stable than the mean as the assertion that even when the machine is not working properly, the standard deviation of a package weight remains unchanged. Judging from the context, it's a statement about the population$\sigma$, not the sample standard deviation. So the ... 0 Your histograms look different because your stats package is binning the data. Normalization of a data set by subtracting its mean and dividing by its standard deviation doesn't really change the shape of the data set; it's only scaling the data and shifting the center to achieve a mean of zero and a sd of 1. If you want to confirm this, do a normal QQ plot ... 0 The conditional variance of the exponential moving average is not too hard: $$V(m_t|m_{t-1})=\alpha^2\sigma^2_{x_t}$$ The variance of$m_t$, assuming a stationary process$x_t$would be: $$V(m_t)=\alpha^2\sigma^2_{x_t}+(1-\alpha)^2V(m_{t-1})+2\alpha(1-\alpha)Cov(x_t,m_{t-1})$$ If we assume steady state conditions for our smoothed process (i.e,$m_t$as$...

0

I don't see anything wrong with your method. Here is a computation (in R) with essentially no round-off error. a = 91.29; s = 4.94; n = 27; pm = c(-1,1); t.c = qt(.975,n-1) a + pm*t.c*s/sqrt(n) ## 89.3358 93.2442 t.c ## 2.055529 $SE(\bar X) = s/\sqrt{n} = 0.9507034.$

0

You should ideally be computing bounds statistically using z score like shown below: mean +/- [z residual value * (std dev/sqrt(n))] This will give you the probability range. As far as the z value is within this range, your observation is true or otherwise. Makes sense?

0

A gamma distribution has a strictly positive mean. If $X$ is gamma distributed with shape $a$ and rate $b$, then the mean of $X$ is $$\mu = \operatorname{E}[X] = a/b,$$ and the standard deviation is $$\sigma = \sqrt{\operatorname{Var}[X]} = \sqrt{a}/b.$$ Note that $a$ and $b$ must be positive. It follows from the above that, given a desired mean $\mu$ and ...

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