The variance of the first 10 multiples of 3 i | vedclass

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Question

(C) The first $10$ multiples of $3$ are $3, 6, 9, \ldots, 30$. Let $x_i = 3i$ for $i = 1, 2, \ldots, 10$. The variance $\sigma^2$ is given by $\frac{\

Vedclass · Accepted Answer

$74.25$

Vedclass · Answer

(C) The first $10$ multiples of $3$ are $3, 6, 9, \ldots, 30$. Let $x_i = 3i$ for $i = 1, 2, \ldots, 10$. The variance $\sigma^2$ is given by $\frac{\sum x_i^2}{n} - (\bar{x})^2$. $\bar{x} = \frac{3(1+2+\ldots+10)}{10} = \frac{3 	imes 10 	imes 11}{10 	imes 2} = 16.5$. $\sum x_i^2 = 3^2(1^2+2^2+\ldots+10^2) = 9 	imes \frac{10(11)(21)}{6} = 9 	imes 385 = 3465$. $\sigma^2 = \frac{3465}{10} - (16.5)^2 = 346.5 - 272.25 = 74.25$.

$x_i$	$5$	$7$	$9$	$11$
$f_i$	$3$	$2$	$1$	$2$

Measure	Height	Weight
Mean	$162.6 \ cm$	$52.36 \ kg$
Variance	$127.69 \ cm^2$	$23.1361 \ kg^2$

The variance of the first $10$ multiples of $3$ is:

Similar Questions

The coefficient of variation for the following data is:
$x_i$ $5$ $7$ $9$ $11$
$f_i$ $3$ $2$ $1$ $2$

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

Measure Height Weight

Mean $162.6 \ cm$ $52.36 \ kg$

Variance $127.69 \ cm^2$ $23.1361 \ kg^2$

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

If the standard deviation of $0, 1, 2, 3, \dots, 9$ is $K$,then the standard deviation of $10, 11, 12, 13, \dots, 19$ is

The variance of the first $20$ natural numbers is

If $1$ is added to each of the first $10$ natural numbers,then the variance of the numbers so obtained is:

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