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Let a random variable $X$ be normally distributed with a mean of $500$ and a standard deviation of $60.$ That is, $X \sim N(500, 60^2)$, where $N(\mu, \sigma ^2)$ is a normal distribution with mean $\mu$ and variance $\sigma ^2$. a) We need to find $P(X > 530)$. We can do this by normalizing and using a standard normal table. \begin{aligned} P(X>530) ...

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If we sample from $X\sim N(150, 40^2)$ 100 times, then the sample mean is $$\bar X = \frac{X_1+\dotsb+X_{100}}{100}.$$ Then \begin{align*} \text{Var}(\bar X) &= \text{Var}\left(\frac{X_1+\dotsb+X_{100}}{100}\right) \\ &=\frac{1}{100^2}\text{Var}(X_1+\dotsb+X_{100})\\ &= \frac{1}{100^2}\cdot100\cdot\text{Var}(X_1) \\ &= \frac{40^2}{100} ...

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First, you will want to calculate the mean, $E(X)$, by adding together the values and dividing by 18. The next step is to find the variance. For this you will want to calculate $E(X^2)$, which can be found by summing the squares of each value and dividing by 18. In other words: $E(X^2)=(25^2+25^2+32^2+45^2+...)/18$ $Var=E(X^2)-E(X)^2$ and standard ...

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First you have a sequence of i.i.d. Bernoulli trials $X_i \sim \text{Bernoulli}(p)$, then the sum of them (the total number of "success") follows a Binomial distribution: $$\sum_{i=1}^n X_i \sim \text{Binomial}(n, p)$$ and thus the variance is given by $np(1 - p)$. In particular when $p = 0.5$, you have your stated result. When you generalized to more than ...

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Notice that $15$inches$-10$inches is two standard deviations since $\sigma=2.5$ inches. Using the 68-95-99 rule tells us that only $5\%$ of all the trees will lie outside of two standard deviations from the mean. Since we are only looking at being bigger than the mean instead of bigger or smaller, that means we cut that $5\%$ number in half to get $2.5\%$. ...

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For general distributions, Chebyshev's inequality is applicable https://en.wikipedia.org/wiki/Chebyshev%27s_inequality. It says that $1-\frac{1}{k^2}$ of the data falls within $k$ standard deviations of the mean. (E.g. $\frac34$ of the data falls within $2$ standards deviations of the mean.)

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The comment and @bubba 's answer offer useful technical information. For your use case Let's say for example you are software developer and you need to measure latency of the system over time and present this to the management I'd recommend some care in collecting measurements. The latency is probably highly dependent on system load, perhaps ...

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For any distribution that's roughly normal, you can get a good feel from the 68, 95, 99.7 rule mentioned in the comment. More generally, for arbitrary distributions, the standard deviation measures the variability of the data. More specifically, it measures the average distance of values from the mean. A small standard deviation means that most values are ...

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The studentized residuals are $$t_i=\frac{\epsilon_i}{\hat\sigma\sqrt{1-h_{ii}}}$$ Where $\epsilon_i$ is the residual, $h_{ii}$ the leverage and $\hat\sigma$ is the estimate of the standard deviation of residuals, that is $$\hat\sigma^2=\frac1{n-m}\sum_{i=1}^n\epsilon_i^2$$ Where $n$ is the number of observations (here $4$) and $m$ the number of ...

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We know $\bar x\pm \frac{z^*\sigma}{\sqrt n}$ is the confidence interval. Call the margin of error $m$. The value $z^*$ is the corresponding z-score to a 95% CI (value that has area 0.025 to the right of it). $$m=\frac{z^*\sigma}{\sqrt n}\implies n=(\frac{z^*\sigma}{m})^2$$

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The population $coefficient\; of\; variation,$ $\sigma/\mu$ is defined only for a distributions that has a $positive$ mean $\mu.$ And the sample coefficient of variation $S/\bar X$ is defined only when the sample mean $\bar X$ is positive. Indeed, the most common uses of the coefficient of variation (population or sample) are for populations with $only\; ... 1 For the mean: Yes, this approach is correct, but actually it should be $$500\cdot12+200\cdot5=7000$$ since in the second sample you have$5$balls and not$10$. For the standard deviation. One approach is to use the pooled standard deviation$s_p$which is the root of the pooled variance$s_p^2\$ with ...

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