Following values are calculated in respect of heights and weights of students of a section of Class

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The following values are calculated in respect of heights and weights of the students of a section of Class $\mathrm{XI}:$

Height Weight

Mean $162.6\,cm$ $52.36\,kg$

Variance $127.69\,c{m^2}$ $23.1361\,k{g^2}$

Can we say that the weights show greater variation than the heights?

Option A

Option B

Option C

Option D

Solution

To compare the variability, we have to calculate their coefficients of variation.

Given $\quad$ Variance of height $=127.69 cm ^{2}$

Therefore Standard deviation of height $=\sqrt{127.69} cm =11.3 cm$

Also $\quad$ Variance of weight $=23.1361 kg ^{2}$

Therefore Standard deviation of weight $=\sqrt{23.1361} kg =4.81 kg$

Now, the coefficient of variations $(C.V.)$ are given by

$(C.V.)$ in heights $=\frac{\text { Standard } \text { Deviation }}{\text { Mean }} \times 100$

$=\frac{11.3}{162.6} \times 100=6.95$

and $\quad$ $(C.V.)$ in weights $=\frac{4.81}{52.36} \times 100=9.18$

Clearly $C.V.$ in weights is greater than the $C.V.$ in heights

Therefore, we can say that weights show more variability than heights

Standard 11

Mathematics

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	Height	Weight
Mean	$162.6\,cm$	$52.36\,kg$
Variance	$127.69\,c{m^2}$	$23.1361\,k{g^2}$

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

The following values are calculated in respect of heights and weights of the students of a section of Class $\mathrm{XI}:$ Height Weight Mean $162.6\,cm$ $52.36\,kg$ Variance $127.69\,c{m^2}$ $23.1361\,k{g^2}$ Can we say that the weights show greater variation than the heights?

Solution

Similar Questions

The mean and standard deviation of the marks of $10$ students were found to be $50$ and $12$ respectively. Later, it was observed that two marks $20$ and $25$ were wrongly read as $45$ and $50$ respectively. Then the correct variance is $…………$.

Statement $1$ : The variance of first $n$ odd natural numbers is $\frac{{{n^2} – 1}}{3}$ Statement $2$ : The sum of first $n$ odd natural number is $n^2$ and the sum of square of first $n$ odd natural numbers is $\frac{{n\left( {4{n^2} + 1} \right)}}{3}$

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The following values are calculated in respect of heights and weights of the students of a section of Class $\mathrm{XI}:$

Height Weight

Mean $162.6\,cm$ $52.36\,kg$

Variance $127.69\,c{m^2}$ $23.1361\,k{g^2}$

Can we say that the weights show greater variation than the heights?

Statement $1$ : The variance of first $n$ odd natural numbers is $\frac{{{n^2} – 1}}{3}$
Statement $2$ : The sum of first $n$ odd natural number is $n^2$ and the sum of square of first $n$ odd natural numbers is $\frac{{n\left( {4{n^2} + 1} \right)}}{3}$

Consider the statistics of two sets of observations as follows :

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 ….. .