Find the mean,variance,and standard deviation | vedclass

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Question

(A) The calculations are performed using the step-deviation method where $h = 5$ and assumed mean $A = 92.5$.      Class Interval    Frequency $f_i$

Vedclass · Accepted Answer

Mean $= 93$,Variance $= 105.58$,Standard Deviation $= 10.27$

Vedclass · Answer

(A) The calculations are performed using the step-deviation method where $h = 5$ and assumed mean $A = 92.5$.      Class Interval    Frequency $f_i$    Mid-point $x_i$    $y_i = \frac{x_i - 92.5}{5}$    $f_i y_i$    $f_i y_i^2$        $70-75$    $3$    $72.5$    $-4$    $-12$    $48$        $75-80$    $4$    $77.5$    $-3$    $-12$    $36$        $80-85$    $7$    $82.5$    $-2$    $-14$    $28$        $85-90$    $7$    $87.5$    $-1$    $-7$    $7$        $90-95$    $15$    $92.5$    $0$    $0$    $0$        $95-100$    $9$    $97.5$    $1$    $9$    $9$        $100-105$    $6$    $102.5$    $2$    $12$    $24$        $105-110$    $6$    $107.5$    $3$    $18$    $54$        $110-115$    $3$    $112.5$    $4$    $12$    $48$        Total    $N = 60$    -    -    $\sum f_i y_i = 6$    $\sum f_i y_i^2 = 254$  Mean $\bar{x} = A + \left( \frac{\sum f_i y_i}{N} ight) 	imes h = 92.5 + \left( \frac{6}{60} ight) 	imes 5 = 92.5 + 0.5 = 93$.Variance $\sigma^2 = \frac{h^2}{N^2} [N \sum f_i y_i^2 - (\sum f_i y_i)^2] = \frac{25}{3600} [60 	imes 254 - (6)^2] = \frac{25}{3600} [15240 - 36] = \frac{25}{3600} 	imes 15204 = 105.58$.Standard Deviation $\sigma = \sqrt{105.58} \approx 10.27$.

	Section $A$	Section $B$
Number of students	$50$	$60$
Average marks in the test	$45$	$45$
Variance of distribution of marks	$64$	$81$

Find the mean,variance,and standard deviation using the short-cut method for the following data:

Height in cms $70-75$ $75-80$ $80-85$ $85-90$ $90-95$ $95-100$ $100-105$ $105-110$ $110-115$

No. of children $3$ $4$ $7$ $7$ $15$ $9$ $6$ $6$ $3$

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Students of two sections $A$ and $B$ of a class show the following results in a test conducted for $100$ marks. Then

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Number of students $50$ $60$

Average marks in the test $45$ $45$

Variance of distribution of marks $64$ $81$

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Height in cms	$70-75$	$75-80$	$80-85$	$85-90$	$90-95$	$95-100$	$100-105$	$105-110$	$110-115$
No. of children	$3$	$4$	$7$	$7$	$15$	$9$	$6$	$6$	$3$

Find the mean,variance,and standard deviation using the short-cut method for the following data: Height in cms $70-75$ $75-80$ $80-85$ $85-90$ $90-95$ $95-100$ $100-105$ $105-110$ $110-115$ No. of children $3$ $4$ $7$ $7$ $15$ $9$ $6$ $6$ $3$