Root mean square deviation Puns

When you take the square of all of the deviations from a mean, average them, and take the square root of that mean, you're giving a higher weight to the outliers. Why is that the norm instead of using the absolute values of the negative deviations? Why is that weight given to outliers a good thing?

Hello, I am a high school student taking AP Statistics (intro college-level stats in a high school class) and was looking over the idea of standard deviation. My textbook explains that we need to square/square root the data to compensate for the positive/negative distances from the mean, which I completely understand. The book also goes on to say that statisticians instead could've decided to just take the absolute values of each deviation and summed those up instead, but refuses to elaborate further (saying that the decision was "for mathematical reasons beyond the scope of this book").

So, I had a couple of questions:

1. Why do we find the square root of variance instead of using absolute values for standard deviation?

2. If we summed up the absolute value of all of the deviations and divided by n-1, I assume this would give average deviation. However, because of how standard deviation is calculated, we end up with different results than the absolute value method. Why is this value considered the correct value for SD instead?

Firstly, apologies if this is not the right forum. This is not a homework question.

I am trying to improve my statistics, and time and disuse have really had their effect. I am currently reading Statistics Without Tears, by Derek Rowntree. I suppose this book would be too basic (or childish even) for most of you, but I figured it would be best to start from scratch, as I was never too strong with mathematics in the first place.

In Chapter 6, while explaining a different concept, the author mentions that to "combine" 2 different standard deviations, you need to follow the steps in the title. He doesn't really go into the logic of why these steps need to be followed.

I am puzzled here by 2 different things:

1. What exactly does it mean to "combine" the standard deviations of 2 different samples?
2. What is the underlying logic of the steps in the title? To put it another way, why not do something simpler, like simply averaging them?

Thank you for your help.

I have heard the explanation that the Standard Deviation is the sum of the average distances of each data in a set to the mean of the data set. I get that you square the numbers and square root the numbers later to avoid using absolute values (which you would use to be measuring distances in all positive numbers). However, in the equation, the square root extends to averaging out the samples, the n on bottom and I am having trouble understanding why. Or at least the explanation I have heard is leaving something out. What am I missing?

EDIT: So I realized that I was under the wrong impression from a youtube video describing the nature of Standard Deviation in terms of the SD being the average distance to the mean, but I realized that's not actually true. It's really NOT the literal average distance to the mean. To his credit, he keeps saying things like "kind of" or "sort of", but really it's the square root of the average variance (definition found everywhere), but what I needed to understand is that is not the average distance to the mean. I got thrown off, because I thought the math was supposed to represent average distance to the mean, but he was just approximating, which in my opinion is a bit misleading. At least he says "kind of" or "sort of". All the actual math he does is correct still.

https://www.youtube.com/watch?v=dq_D30kyR1A&t=919s

If you were trying to find the average squared distance of the mean, you really would only need to take absolute values (or square root of the indivual's square), add them, and divide by the population number. That's the simple average of the distance to the mean, but that's not the SD.

So the SD is useful because it's not really an average but becomes a unit for an individual data set when comparing the spread of data. So if an individual is 1, 2, or 3 SD's away, it falls within a predicted percentage of a bell curve.

I’m practicing for my act, and I keep seeing the square root symbol in so many problems, and in most of them, some of the answers are with the square root symbol too. I don’t understand, does it mean more than square root? Example:

√11 6 3 √11 — * — = ——— x √11 11

“What is the value of x”

A. 6 B. 11 C. 121 D. √11 E. 2 √11

Hello, tell me, please - Am I finding the root-mean-square error correctly?

import pandas as pd
import numpy as np
from sklearn.model_selection import train_test_split
from sklearn import metrics 
from sklearn.linear_model import LinearRegression
data = pd.read_csv('data.csv')


X_train, X_test, y_train, y_test = train_test_split(
    data.drop(columns="target"), 
    data["target"],
    test_size=0.33,
    random_state=42)


model = LinearRegression().fit(x_train, y_train)

y_pred = model.predict(x_test)

print('Root Mean Squared Error:', np.sqrt(metrics.mean_squared_error(y_test, y_pred)))

Result: Root Mean Squared Error: 65.48034654566514
Dataser: https://disk.yandex.ru/i/0fT4twJ2P-ln9w

The coordinates of the vertices of a triangle in the plane are independent random variables with standard normal distribution. What is the root mean square of area of the triangle?

Hard variant: (n+1) points in Euclidean n-space have i.i.d. standard normal distributed coordinates. What's the root mean square of measure of their convex hull?

note: this problem is easier variant of another question, which asked for average measure, hopefully it still put up a fun challenge.

Demand variability isn’t an issue but supply lead time is. Could actual vs expected supply lead times be used in place of actual demand vs forecast in the RMSE formula to generate a safety stock need??

Root Mean Square Error In R, The root mean square error (RMSE) allows us to measure how far predicted values are from observed values in…

https://finnstats.com/index.php/2021/07/23/how-to-calculate-root-mean-square-error-rmse-in-r/

I am doing an simulation of cylinder flow of around Re 500,000

I noted from most of the publications that people would compare values of strouhal number, root mean square lift coefficient and as well mean drag coefficient in their simulations

I am a bit confused, like why would not root mean square drag coefficient and mean lift coefficient be used instead? Or in other words, why are mean square lift coefficient and mean drag coefficient better? Wouldn’t it more fair if people compare mean drag coefficient & mean lift coefficient (or the pair of root mean square coefficient) ? Any reason to pick these parameters (strouhal number, root mean square lift coefficient and as well mean drag coefficient ) to compare?

I am new to CFD so I am really confused.

Rms=sqrt(x_1_^2 + x_2_^2 + x_3_^2 . . .+x_n_^2 /n)
What is x?