8. Solve problem 3 by 'Step Deviation Method'.
The measurements (in mm) of the diameters of the head of screws are given below:
| Diameter (in mm) | 33 - 35 | 36 - 38 | 39 - 41 | 42 - 44 | 45 - 47 |
|---|---|---|---|---|---|
| No. of screws | 10 | 19 | 23 | 21 | 27 |
Sol.
Class width ($h$) = 3, Assumed mean ($A$) = 40
Diameter
(in mm)Class Mark
($x_i$)$u_i = \frac{d_i}{h}$ No. of screws
($f_i$)$f_i u_i$ 33 - 35 34 $- 2$ 10 $- 20$ 36 - 38 37 $- 1$ 19 $- 19$ 39 - 41 40 $\rightarrow A$ 0 23 0 42 - 44 43 1 21 21 45 - 47 46 2 27 54 Total 100 36
$\bar{u}$ $=$ $\frac{\Sigma f_i u_i}{\Sigma f_i}$ $\therefore \bar{u}$ $=$ $\frac{36}{100}$ $\therefore \bar{u}$ $=$ $0.36$
Mean ($\bar{x}$) $=$ $A + h\bar{u}$ $=$ $40 + 3(0.36)$ $=$ $40 + 1.08$ $=$ $41.08$