Question

Solve problem 3 by step-deviation method.

Answer 1

Take A = 40, h = 3

Diameter (in mm) (Class interval)	Class Marks $\left(x_i\right)$	$$\begin{gathered} d_i=x_i-A \\ {d}_i=x_i-40 \end{gathered}$$	$u_i=\frac{d_i}{h}=\frac{d_i}{3}$	No. of screws $\left(f_i\right)$	$f_i{u}_i$
33-35	34	$-$6	$-$2	10	$-$20
36-38	37	$-$3	$-$1	19	$-$19
39-41	$40\to A$	0	0	23	0
42-44	43	3	1	21	21
45-47	46	6	2	27	54
Total	-	-	-	$\sum _{}^{}{f}_i=100$	$\sum _{}^{}{f}_i{u}_i=36$

Now, mean u_i s is can be calculated thus:

$$\begin{gathered} \overline{u}=\frac{{\displaystyle \sum _{}^{}}{f}_i{u}_i}{{\displaystyle \sum _{}^{}}{f}_i}=\frac{36}{100}=0.36 \\ \mathrm{Mean}=\overline{x}=A+h\overline{u} \\ =40+3\times 0.36 \\ =40+1.08 \\ =41.08 \\ \therefore \mathrm{Mean}\mathrm{diameter}\mathrm{of}\mathrm{the}\mathrm{head}\mathrm{of}\mathrm{the}\mathrm{screw}\mathrm{is}41.08\mathrm{mm}. \end{gathered}$$