Find the find the quadratic average and the geometric average. To do this I have these informations :

The standart deviation, the arithmetic average and the number of values.

The standart deviation, the arithmetic average and the number of values.

Nathalie Fields
2022-07-17
Answered

The standart deviation, the arithmetic average and the number of values.

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Lance Long

Answered 2022-07-18
Author has **8** answers

You can't get the geometric average when $n=3$. Knowing the arithmetic and quadratic average of three numbers $(x,y,z)$ is equivalent to knowing $x+y+z$ and ${x}^{2}+{y}^{2}+{z}^{2}$, while knowing the geometric average means knowing $xyz$. If xyz were a function of the other two, its gradient at any point would be a linear combination of the other two gradients. But

$\begin{array}{rl}\mathrm{\nabla}(x+y+z)& =(1,1,1)\\ \mathrm{\nabla}({x}^{2}+{y}^{2}+{z}^{2})& =2(x,y,z)\\ \mathrm{\nabla}(xyz)& =(yz,xz,xy)\end{array}$

At $(x,y,z)=(1,2,3)$, say, the three gradients are linearly independent.

$\begin{array}{rl}\mathrm{\nabla}(x+y+z)& =(1,1,1)\\ \mathrm{\nabla}({x}^{2}+{y}^{2}+{z}^{2})& =2(x,y,z)\\ \mathrm{\nabla}(xyz)& =(yz,xz,xy)\end{array}$

At $(x,y,z)=(1,2,3)$, say, the three gradients are linearly independent.

prkosnognm

Answered 2022-07-19
Author has **5** answers

It's not possible for the geometric average, there are infinitely many solutions depending on the number of observations.

asked 2022-07-23

A random sample of 5,000 people is selected from local telephone book to participate in financial planning survey to help infer about typical saving and spending habits. three thousand return the questionnaire. Describe the types of bias that may result in at least 4-5 complete sentences.

asked 2022-07-23

From $10$ numbers $a,b,c,...j$ all sets of $4$ numbers are chosen and their averages computed. Will the average of these averages be equal to the average of the $10$ numbers?

asked 2022-07-22

Is a weighted average, assuming the weights sum up to one, always smaller than the unweighted arithmetic average? As if the weights are smaller than one the weighted mean is a convex combination of the individual points. Is there any condition on the individual observation for the statement to hold true?

asked 2022-07-14

Calculating conditional probability given Poisson variable

I encountered a set of problems while studying statistics for research which I have combined to get a broader question. I want to know if this is a solvable problem with enough information specifically under what assumptions or approach.

Given that a minesweeper has encountered exactly 5 landmines in a particular 10 mile stretch, what is the probability that he will encounter exactly 6 landmines on the next 10 mile stretch. (Average number of landmines is 0.6 per mile in the 50 mile stretch)

I have figured that the approach involves finding out the Poisson probabilities of the discrete random variable with the combination of Bayes Conditional probability. But am stuck with proceeding on applying the Bayes rule. i.e $Pr(X=6\mid X=5)$.

I know that $Pr(X=5)={e}^{-6}{5}^{6}/5!.$ Here $\lambda =0.6\cdot 10$ and $X=5$) Similarly for $Pr(X=6)$. Is Bayes rule useful here: $P(Y\mid A)=Pr(A\mid Y)Pr(Y)/(Pr(A\mid Y)Pr(Y)+Pr(A\mid N)Pr(N))$?

Would appreciate any hints on proceeding with these types of formulations for broadening my understanding.

I encountered a set of problems while studying statistics for research which I have combined to get a broader question. I want to know if this is a solvable problem with enough information specifically under what assumptions or approach.

Given that a minesweeper has encountered exactly 5 landmines in a particular 10 mile stretch, what is the probability that he will encounter exactly 6 landmines on the next 10 mile stretch. (Average number of landmines is 0.6 per mile in the 50 mile stretch)

I have figured that the approach involves finding out the Poisson probabilities of the discrete random variable with the combination of Bayes Conditional probability. But am stuck with proceeding on applying the Bayes rule. i.e $Pr(X=6\mid X=5)$.

I know that $Pr(X=5)={e}^{-6}{5}^{6}/5!.$ Here $\lambda =0.6\cdot 10$ and $X=5$) Similarly for $Pr(X=6)$. Is Bayes rule useful here: $P(Y\mid A)=Pr(A\mid Y)Pr(Y)/(Pr(A\mid Y)Pr(Y)+Pr(A\mid N)Pr(N))$?

Would appreciate any hints on proceeding with these types of formulations for broadening my understanding.

asked 2022-07-14

Probability of 1 billion monkeys typing a sentence if they type for 10 billion years

Suppose a billion monkeys type on word processors at a rate of 10 symbols per second. Assume that the word processors produce 27 symbols, namely, 26 letters of the English alphabet and a space. These monkeys type for 10 billion years. What is the probability that they can type the first sentence of Lincoln’s “Gettysburg Address”?

Four score and seven years ago our fathers brought forth on this continent a new nation conceived in liberty and dedicated to the proposition that all men are created equal.

Hint: Look up Boole’s inequality to provide an upper bound for the probability!

This is a homework question. I just want some pointers how to move forward from what I have done so far. Below I will explain my research so far.

First I calculated the probability of the monkey 1 typing the sentence (this question helped me do that); let's say that probability is p:

$P(\text{Monkey 1 types our sentence})=P({M}_{1})=p$

Now let's say that the monkeys are labeled ${M}_{1}$ to ${M}_{{10}^{9}}$, so given the hint in the question I calculated the upper bound for the probabilities of union of all $P({M}_{i})$ (the probability that i-th monkey types the sentence) using Boole's inequality.

Since $P({M}_{i})=P({M}_{1})=p$,

$P\left(\bigcup _{i}{M}_{i}\right)\le \sum _{i=1}^{{10}^{9}}P({M}_{i})=\sum ^{{10}^{9}}p={10}^{9}\phantom{\rule{thinmathspace}{0ex}}p$

Am I correct till this point? If yes, what can I do more in this question? I tried to study Bonferroni inequality for lower bounds but was unsuccessful to obtain a logical step. If not, how to approach the problem?

Suppose a billion monkeys type on word processors at a rate of 10 symbols per second. Assume that the word processors produce 27 symbols, namely, 26 letters of the English alphabet and a space. These monkeys type for 10 billion years. What is the probability that they can type the first sentence of Lincoln’s “Gettysburg Address”?

Four score and seven years ago our fathers brought forth on this continent a new nation conceived in liberty and dedicated to the proposition that all men are created equal.

Hint: Look up Boole’s inequality to provide an upper bound for the probability!

This is a homework question. I just want some pointers how to move forward from what I have done so far. Below I will explain my research so far.

First I calculated the probability of the monkey 1 typing the sentence (this question helped me do that); let's say that probability is p:

$P(\text{Monkey 1 types our sentence})=P({M}_{1})=p$

Now let's say that the monkeys are labeled ${M}_{1}$ to ${M}_{{10}^{9}}$, so given the hint in the question I calculated the upper bound for the probabilities of union of all $P({M}_{i})$ (the probability that i-th monkey types the sentence) using Boole's inequality.

Since $P({M}_{i})=P({M}_{1})=p$,

$P\left(\bigcup _{i}{M}_{i}\right)\le \sum _{i=1}^{{10}^{9}}P({M}_{i})=\sum ^{{10}^{9}}p={10}^{9}\phantom{\rule{thinmathspace}{0ex}}p$

Am I correct till this point? If yes, what can I do more in this question? I tried to study Bonferroni inequality for lower bounds but was unsuccessful to obtain a logical step. If not, how to approach the problem?

asked 2022-07-23

In Traditional Survey Reseach, elaborate what are;

1. Nonresponse Bias

2. Refusal rate

3. Response Bias

1. Nonresponse Bias

2. Refusal rate

3. Response Bias

asked 2022-07-21

There is a box with $12$ dice which all look the same. However there are actually three types of dice:

$6$ normal dice. The probability to get a $6$ is $1/6$ for each dice.

$3$ biased dice. The probability to get a $6$ is $0.85$.$3$ biased dice. The probability to get a $6$ is $0.05$.You take a die from the box at random and roll it.What is the conditional probability that it is of type $b$, given that it gives a $6$?

$6$ normal dice. The probability to get a $6$ is $1/6$ for each dice.

$3$ biased dice. The probability to get a $6$ is $0.85$.$3$ biased dice. The probability to get a $6$ is $0.05$.You take a die from the box at random and roll it.What is the conditional probability that it is of type $b$, given that it gives a $6$?