The mean and standard deviation of 20 observa | vedclass

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(D) Given: $n = 20$,$\bar{x} = 10$,$\sigma = 2.5$. Incorrect sum of observations: $\Sigma x_i = n 	imes \bar{x} = 20 	imes 10 = 200$. Correct sum of

Vedclass · Accepted Answer

$(10.5, 26)$

Vedclass · Answer

(D) Given: $n = 20$,$\bar{x} = 10$,$\sigma = 2.5$. Incorrect sum of observations: $\Sigma x_i = n 	imes \bar{x} = 20 	imes 10 = 200$. Correct sum of observations: $\Sigma x_i = 200 - 25 + 35 = 210$. Correct mean $\alpha = \frac{210}{20} = 10.5$. Incorrect sum of squares: $\sigma^2 = \frac{\Sigma x_i^2}{n} - (\bar{x})^2 \implies (2.5)^2 = \frac{\Sigma x_i^2}{20} - 10^2$. $6.25 = \frac{\Sigma x_i^2}{20} - 100 \implies \Sigma x_i^2 = 20 	imes 106.25 = 2125$. Correct sum of squares: $\Sigma x_i^2 = 2125 - 25^2 + 35^2 = 2125 - 625 + 1225 = 2725$. Correct variance $\beta = \frac{\Sigma x_i^2}{n} - (\alpha)^2 = \frac{2725}{20} - (10.5)^2 = 136.25 - 110.25 = 26$. Thus,$(\alpha, \beta) = (10.5, 26)$.

Class	$4-8$	$8-12$	$12-16$	$16-20$
Frequency	$3$	$\lambda$	$4$	$7$

Marks:	$2$	$3$	$5$	$7$
Frequency:	$(x+1)^2$	$2x-5$	$x^2-3x$	$x$

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