Find the mean of the following data by all the three methods:

Class	$50-70$	$70-90$	$90-110$	$110-130$	$130-150$	$150-170$
Frequency	$10$	$18$	$7$	$6$	$5$	$4$

(A) To find the mean,we first calculate the midpoint $(x_i)$ for each class interval.

Class	Frequency $(f_i)$	Midpoint $(x_i)$	$d_i = x_i - A$	$u_i = \frac{x_i - A}{c}$	$f_i x_i$	$f_i d_i$	$f_i u_i$
$50-70$	$10$	$60$	$-20$	$-1$	$600$	$-200$	$-10$
$70-90$	$18$	$80 = A$	$0$	$0$	$1440$	$0$	$0$
$90-110$	$7$	$100$	$20$	$1$	$700$	$140$	$7$
$110-130$	$6$	$120$	$40$	$2$	$720$	$240$	$12$
$130-150$	$5$	$140$	$60$	$3$	$700$	$300$	$15$
$150-170$	$4$	$160$	$80$	$4$	$640$	$320$	$16$
Total	$\Sigma f_i = 50$	-	-	-	$\Sigma f_i x_i = 4800$	$\Sigma f_i d_i = 800$	$\Sigma f_i u_i = 40$

Here,$A = 80$ and $c = 20$.
$1$. Direct Method: $\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{4800}{50} = 96$.
$2$. Assumed Mean Method: $\bar{x} = A + \frac{\Sigma f_i d_i}{\Sigma f_i} = 80 + \frac{800}{50} = 80 + 16 = 96$.
$3$. Step Deviation Method: $\bar{x} = A + \left( \frac{\Sigma f_i u_i}{\Sigma f_i} \right) \times c = 80 + \left( \frac{40}{50} \right) \times 20 = 80 + 16 = 96$.
Thus,the mean of the data is $96$.