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Eupraxis was smart to say that a normal distribution was assumed. In the real world, we aren't so lucky. For example, go to a financial time series. Take the daily percent change of a security. You can calculate the mean and standard deviation, but you will find some data points beyond 6 sigma (which, according to current dogma, is supposed to be ...

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You may want to consider measures of dispersion other than standard deviation. For example the difference between symmetric quantiles is also a measure of "spread". Any quantile will do, if the spread between all of the others is equal. This immediately shows your Set 1 to have more "spread" than Set 2 merely by inspection; no calculation needed. I'm ...

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If the standard deviation of 3x+7 is 4, then the standard deviation of x is 4/3. So the variance must be 16/9. When you multiply data by a constant, you multiply the standard deviation by that same constant. You can prove this to yourself by using the formula defining standard deviation. Left as an exercise for the student, of course.

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Back in the 1970's we realized that when the standard deviation is small compared to the mean, there is great loss of significant digits. Then-existing handheld calculators (personal computers had not yet arrived) used the naïve computer formula that accumulated the sum and the sum of the squares. So if this is the case with your data, you must use a ...

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Squaring the Deviations The variance of a sample measures the spread of the values in a sample or distribution. We could do this with any function of $|x_k-\bar{x}|$. The reason that we use $(x_k-\bar{x})^2$ is because the variance computed this way has very nice properties. Here are a couple: $1$. The variance of the sum of independent variables is the ...

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Hint: You can measure the spread by Minimum Absolute Deviation |(x-xbar)| or the squared differences. Take an example of the sequence, -3,0,3. Xbar for this is 0. Assume that you don't measure the Absolute deviation, then the spread is 0 if you just took x-xbar. The mean squared will avoid this situation and give you an objective measure of spread (to ...

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Squaring $x_i-\bar x$: If we didn't square it, we would just be adding up $x_i - \bar x$, and that will always give us zero. What we want instead is to total "how far" each $x_i$ is from $\bar x$. So, we need to make sure we're taking the average of some positive quantity representing how far $x_i$ is from $\bar x$; one good choice is $(x_i - \bar ... 4 The square is used to remove the effect of the sign of$x_i - \overline{x}$. Suppose your mean was 0, and you had measurements at -2 and +2. These would cancel, but squaring gets rid of that issue. Now, you might ask, "why not use absolute value?" Great question! The reason is that if we use absolute value, variances are no longer additive. In other words, ... 1 It is necessary to square the deviations from the mean as you want to measure both positive as well as negative deviations (note that$\sum (x_i - \bar{x})$is just zero). Another possibility is to take absolute values, but the above formula turns out to have nicer properties (such as additivity of variances, as pointed out by Arkamis). Regarding the$N-1$... 0 Since you have 15 people, it is not too bad to use a Gaussian approximation to the average score assigned by people for a particular category. Calculate the average of the scores for a category over the 15 customers, and then compute the standard deviation of the 15 scores for the category. Then your statistic for determining whether the score is unusually ... 0 First, don't be down on yourself!! I agree with you that standard deviation is an excellent measure of what you want to quantify. It is a measure of the variation, or how "spread out" the data is. Alternatively (but equivalently), you can use the variance. 0 Try analysing the coefficient of variation $$\rho[X]=\frac{E[X]}{\sigma[X]}.$$ In your case $$\rho_1=2\rho_2,$$ which says that the sample$1$is twice as dispersed as the second one. 1 The above result means that the standard normal distribution cannot fit adequately enough to your data. That is becaues the percentiles of your data do not match to the percentiles that you would have if the distribution was standard normal. In order to determine the appropriate distribution for your data note that the mean is less than the median. This ... 1 Let$z=y/\sin(2\pi x)$and compute the values of$z$. Then draw$z$as a function of$x$on logarithmic scales, i.e.$\log(z)$as a function of$\log(x)$. What do you think of a linear regression then ? 0 That is a chi-square test for independence. See the section "Calculating the test statistic" to find what you want. And here is an example. 1 Actually, I don't think$n=12$;$12$is the number of categories, and I think you're interested in the number of respondents.$ \alpha =5$% is your significance level. I assume your initial hypothesis is that there is no connection between self-image and choice of model. This means that the proportion of trait -to-model is$25$%, or$ 4$-to-$1$. Now, you ... 0 The fundamental reason for the presence of log, the Kullback-Liebler divergence. That is why MLE is consistent, among other things. Pointing out trivial consequences like "log turns product into sum..." is very misleading. 0 The M/M/1 model assumes that job sizes are exponentially distributed with rate parameter$\mu$, in your setup the job sizes are uniformly distributed so you need to consider an M/G/1 queueing model. What is the difference between 'Mean time it takes to send a packet through the line' and 'Average latency'? They sound like a description of the same phenomena ... 1 There are at least two senses in which you can use the standard deviation in quality measurement: Manufacturing precision: How closely are we able to meet a manufacturing specificiation (e.g., bearing diameter). Fraction falling within a given tolerance interval: Continuing the example, if bearings must be within$\pm 1$mm, how many standard deviations ... 0 Sometimes we don't use the log (natural or otherwise). Mostly it's a matter of convenience - taking logs in many cases makes it simpler to find the argmax. However, results like Wilks' theorem may make it more convenient to work with logs in more situations than might otherwise be apparent. Taking logs is not so much help with the triangular distribution, ... 1 Since natural log is a strictly increasing function, the max of the density in question will be the same as the max of the natural log transformation, given that it exists. The natural log simplifies densities that involve exponentials. Also since densities usually involve products, the transformation will simplify all that potentially messy calculation. 3 Likelihood function is generally ratio of two densities. Since working with ratios is not convenient when taking max, one normally takes the logs. This is most effective when densities involve exponential functions and is extremely convenient in the case of normal (Gaussian) densities. Taking logs does not always help. In fact if the underlying densities ... 1 An intuitive way of thinking about it is that standard deviation is a measure of spread. So you need some way of saying for every value in the data set, on average how far away is that value from the mean. So you take the differences$d_i = x_i - \bar{x}$. What do you do with the differences? If you simply add them up, you'll get zero. You want to give ... 1 The sum of cubes of differences with the mean (or integral, in the case of a continuous distribution) gives the third central moment, known as a measure of skewness, rather than variance in the second moment case. More technically, the skewness$\gamma_1$is defined by: $$\gamma_1 = E[(X-\mu)^3]/\sigma$$ where$\mu$is mean of the distribution of random ... 1 Technically the smallest power that will accomplish smoothness at zero is 1+ϵ, as the derivative will be left with x^ϵ which then vanishes, basically minimally blunting the sharp 'tip' of the absolute value function. I mention this not to be pedantic, but because it can be useful if you really do want something that largely behaves like the absolute value ... 0 Yes if you keep the sum of the squares.$\sum(x_i-\bar{x})^2=\sum x_i^2 -n \bar{x}^2$0 Do you know the difference between pooled and unpooled? Hint: pooled tests assumes that two populations have equal variances. Use this to calculate standard error. 0 With these few values, you can just run the calculation by hand. Alpha gives$\frac {2\sqrt 5}3 \approx 1.4907$2 The average of your data is $$\bar x = \frac{1+2+3+4+5+1}{6}=\frac83.$$ Thus $$s_n^2=\frac1n\sum(x_i-\bar x)^2= \frac16\left(\frac{25}{9}+\frac49+\frac19+\frac{16}9+\frac{49}{9}+\frac{25}{9}\right)=\frac{20}{9}$$ and $$s_{n-1}^2=\frac1{n-1}\sum(x_i-\bar x)^2=\frac 83.$$ This makes $$s_n\approx 1.4907\qquad s_{n-1}\approx 1.63299$$ It seems you are ... 0 Hint: Have you tried using the 68/95/99.7 rule? 9 inches greater than the mean is 2 standard deviations. The question is asking for the area to the right of$2 \sigma$. Since the interval within$2\sigma$from the mean is 95%, the left and right combined is 5%, so the area only to the right is 2.5%. 0 The given formula is a rewriting of the following expression for the variance (see comments above): $$\sigma^2 = E(y^2)-\mu^2.$$ 0 Before you started the survey, the expected percentage saying YES was$48\%$and the standard deviation from a sample of$50$would have been$\sqrt{\dfrac{p(1-p)}{n}} \approx 7\%$. After the survey, you "know" the percentage of the sample saying YES was$45\%$. This suggests that$22.5$people said YES and$27.5\$ people said NO.

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