Find the mean,median and mode of the followin | vedclass

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(N/A) $1$. Mean: $\sum f_i = 100$. Midpoints $(x_i)$: $5, 15, 25, 35, 45, 55, 65$. $\sum f_i x_i = (5 	imes 5) + (12 	imes 15) + (18 	imes 25) + (4

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(N/A) $1$. Mean: $\sum f_i = 100$. Midpoints $(x_i)$: $5, 15, 25, 35, 45, 55, 65$. $\sum f_i x_i = (5 \times 5) + (12 \times 15) + (18 \times 25) + (40 \times 35) + (15 \times 45) + (7 \times 55) + (3 \times 65) = 25 + 180 + 450 + 1400 + 675 + 385 + 195 = 3310$. Mean $= \frac{\sum f_i x_i}{\sum f_i} = \frac{3310}{100} = 33.1$.
$2$. Median: $N=100, N/2 = 50$. Cumulative frequencies: $5, 17, 35, 75, 90, 97, 100$. Median class is $30-40$. $l=30, cf=35, f=40, h=10$. Median $= l + \left( \frac{N/2 - cf}{f} \right) \times h = 30 + \left( \frac{50 - 35}{40} \right) \times 10 = 30 + \left( \frac{15}{4} \right) = 30 + 3.75 = 33.75$.
$3$. Mode: Modal class is $30-40$ (highest frequency $40$). $l=30, f_1=40, f_0=18, f_2=15, h=10$. Mode $= l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h = 30 + \left( \frac{40 - 18}{2(40) - 18 - 15} \right) \times 10 = 30 + \left( \frac{22}{80 - 33} \right) \times 10 = 30 + \left( \frac{220}{47} \right) \approx 30 + 4.68 = 34.68$.

Class	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	$60-70$
Frequency	$5$	$12$	$18$	$40$	$15$	$7$	$3$

Class	$0-20$	$20-40$	$40-60$	$60-80$	$80-100$
Frequency	$5$	$x$	$20$	$y$	$2$

Marks	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$
Number of students	$2$	$4$	$7$	$6$	$1$

Find the mean,median and mode of the following frequency distribution:

Class $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$

Frequency $5$ $12$ $18$ $40$ $15$ $7$ $3$

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Find the mean,median and mode of the following frequency distribution: Class $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$ Frequency $5$ $12$ $18$ $40$ $15$ $7$ $3$