(A) Given: $n = 15$,$\sum x = 170$,and $\sum x^2 = 2830$. Corrected sum of observations: $\sum x = 170 - 20 + 30 = 180$. Corrected sum of squares: $\s

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(A) Given: $n = 15$,$\sum x = 170$,and $\sum x^2 = 2830$. Corrected sum of observations: $\sum x = 170 - 20 + 30 = 180$. Corrected sum of squares: $\sum x^2 = 2830 - (20)^2 + (30)^2 = 2830 - 400 + 900 = 3330$. Corrected variance is given by the formula: $\sigma^2 = \frac{\sum x^2}{n} - \left(\frac{\sum x}{n}\right)^2$. Substituting the values: $\sigma^2 = \frac{3330}{15} - \left(\frac{180}{15}\right)^2$. $\sigma^2 = 222 - (12)^2 = 222 - 144 = 78$.

Class 11 Mathematics Statistics Variance and Standard Deviation

Medium

In an experiment with $15$ observations for $x$,the following results were available: $\sum x^2 = 2830$ and $\sum x = 170$. One observation $20$ was found to be wrong and was replaced by the correct value $30$. Then the corrected variance is:

MHT CET 2024 All MHT CET

A
$78$
B
$210$
C
$225$
D
$88$

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

In an experiment with $15$ observations for $x$,the following results were available: $\sum x^2 = 2830$ and $\sum x = 170$. One observation $20$ was found to be wrong and was replaced by the correct value $30$. Then the corrected variance is:

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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?