(A) The coefficient of variation $(CV)$ is defined as $CV = \frac{\sigma}{\bar{x}} \times 100$. For group $A$ to be more consistent than group $B$,the

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(A) The coefficient of variation $(CV)$ is defined as $CV = \frac{\sigma}{\bar{x}} \times 100$. For group $A$ to be more consistent than group $B$,the coefficient of variation of $A$ must be less than the coefficient of variation of $B$. Thus,$CV_A Substituting the values,we get $\frac{\sigma_A}{\bar{x}} Given $\sigma_A = 2$ and $\sigma_B = 3$,we have $\frac{2}{\bar{x}} Rearranging the inequality to solve for $\frac{\bar{y}}{\bar{x}}$,we get $\frac{\bar{y}}{\bar{x}} < \frac{3}{2}$.

Class 11 Mathematics Statistics Word problem -Statistics

Easy

The means of two groups of observations $A$ and $B$ are $\bar{x}$ and $\bar{y}$ respectively,and their standard deviations are $2$ and $3$ respectively. In order for group $A$ to be more consistent than group $B$,$\frac{\bar{y}}{\bar{x}} < $

TS EAMCET 2020 All TS EAMCET

A
$\frac{3}{2}$
B
$\frac{5}{1}$
C
$\frac{2}{3}$
D
$\frac{6}{5}$

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Consider the statistics of two sets of observations as follows:

Set Size Mean Variance

Observation $I$ $10$ $2$ $2$

Observation $II$ $n$ $3$ $1$

If the variance of the combined set of these two observations is $\frac{17}{9}$,then the value of $n$ is equal to:

DifficultJEE MAIN 2021

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The $A.M.$ of $n$ observations is $M$. If the sum of $n - 4$ observations is $a$,then the mean of the remaining $4$ observations is:

Easy

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In a Mathematics test,the average marks of boys is $x \%$ and the average marks of girls is $y \%$ with $x \neq y$. If the average marks of all students is $z \%$,the ratio of the number of girls to the total number of students is

DifficultKVPY 2017

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If the mean of the distribution is $2.6$,then the value of $y$ is:
Variate $x$ $1$ $2$ $3$ $4$ $5$
Freq $f$ of $x$ $4$ $5$ $y$ $1$ $2$

Medium

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The mean and standard deviation of $20$ observations are found to be $10$ and $2$,respectively. It was later discovered that one observation was mistakenly recorded as $8$ instead of $12$. The correct standard deviation is:

MediumJEE MAIN 2024

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Set	Size	Mean	Variance
Observation $I$	$10$	$2$	$2$
Observation $II$	$n$	$3$	$1$

The means of two groups of observations $A$ and $B$ are $\bar{x}$ and $\bar{y}$ respectively,and their standard deviations are $2$ and $3$ respectively. In order for group $A$ to be more consistent than group $B$,$\frac{\bar{y}}{\bar{x}} < $

Similar Questions

Consider the statistics of two sets of observations as follows: Set Size Mean Variance Observation $I$ $10$ $2$ $2$ Observation $II$ $n$ $3$ $1$ If the variance of the combined set of these two observations is $\frac{17}{9}$,then the value of $n$ is equal to:

The $A.M.$ of $n$ observations is $M$. If the sum of $n - 4$ observations is $a$,then the mean of the remaining $4$ observations is:

In a Mathematics test,the average marks of boys is $x \%$ and the average marks of girls is $y \%$ with $x \neq y$. If the average marks of all students is $z \%$,the ratio of the number of girls to the total number of students is

If the mean of the distribution is $2.6$,then the value of $y$ is:Variate $x$$1$$2$$3$$4$$5$Freq $f$ of $x$$4$$5$$y$$1$$2$

The mean and standard deviation of $20$ observations are found to be $10$ and $2$,respectively. It was later discovered that one observation was mistakenly recorded as $8$ instead of $12$. The correct standard deviation is:

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Consider the statistics of two sets of observations as follows:

Set Size Mean Variance

Observation $I$ $10$ $2$ $2$

Observation $II$ $n$ $3$ $1$

If the variance of the combined set of these two observations is $\frac{17}{9}$,then the value of $n$ is equal to:

If the mean of the distribution is $2.6$,then the value of $y$ is:
Variate $x$ $1$ $2$ $3$ $4$ $5$
Freq $f$ of $x$ $4$ $5$ $y$ $1$ $2$