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(A) Let the assumed mean $A = 65$. Here,the class width $h = 10$.We construct the following table:						Class			Frequency $({f_i})$			Mid-point $({x_i

Vedclass · Accepted Answer

Mean = $62$,Variance = $201$,Standard Deviation = $14.18$

Vedclass · Answer

(A) Let the assumed mean $A = 65$. Here,the class width $h = 10$.We construct the following table:						Class			Frequency $({f_i})$			Mid-point $({x_i})$			${y_i} = \frac{{{x_i} - 65}}{{10}}$			${f_i}{y_i}$			${f_i}{y_i}^2$							$30-40$			$3$			$35$			$-3$			$-9$			$27$							$40-50$			$7$			$45$			$-2$			$-14$			$28$							$50-60$			$12$			$55$			$-1$			$-12$			$12$							$60-70$			$15$			$65$			$0$			$0$			$0$							$70-80$			$8$			$75$			$1$			$8$			$8$							$80-90$			$3$			$85$			$2$			$6$			$12$							$90-100$			$2$			$95$			$3$			$6$			$18$							Total			$N = 50$			-			-			$\sum {f_i}{y_i} = -15$			$\sum {f_i}{y_i}^2 = 105$			Mean $\bar{x} = A + \frac{{\sum {{f_i}{y_i}} }}{N} 	imes h = 65 + \frac{{-15}}{{50}} 	imes 10 = 65 - 3 = 62$.Variance ${\sigma ^2} = \frac{{{h^2}}}{{{N^2}}}\left[ {N\sum {{f_i}{y_i}^2} - {{\left( {\sum {{f_i}{y_i}} } ight)}^2}} ight] = \frac{{100}}{{2500}}\left[ {50(105) - (-15)^2} ight] = \frac{1}{{25}}[5250 - 225] = \frac{{5025}}{{25}} = 201$.Standard Deviation $\sigma = \sqrt{201} \approx 14.18$.

Calculate the mean,variance,and standard deviation for the following distribution.

Classes $30-40$ $40-50$ $50-60$ $60-70$ $70-80$ $80-90$ $90-100$

Frequency $({f_i})$ $3$ $7$ $12$ $15$ $8$ $3$ $2$

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Classes	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$	$80-90$	$90-100$
Frequency $({f_i})$	$3$	$7$	$12$	$15$	$8$	$3$	$2$

Calculate the mean,variance,and standard deviation for the following distribution. Classes $30-40$ $40-50$ $50-60$ $60-70$ $70-80$ $80-90$ $90-100$ Frequency $({f_i})$ $3$ $7$ $12$ $15$ $8$ $3$ $2$