Question

The measurements of the diameters (in mm) of the heads of 107 screws are given below

$\begin{array}{cccccc}\text{Diameter (in mm)}& 33-35& 36-38& 39-41& 42-44& 45-47\\ \text{Number of screws}& 17& 19& 23& 21& 27\end{array}$

Calculate the mean and the standard deviation.

Answer 1

We make the table from the given data.

$\begin{array}{ccccccc}\text{Diameter (in mm)}& \text{mid - values}{x}_i& \text{Number of screws}& d_i=\frac{x_i-40}{3}& d_i^2& f_i{d}_i& f_i{d}_i^2\\ 33-35& 34& 17& -2& 4& -34& 68\\ 36-38& 37& 19& -1& 1& -19& 19\\ 39-41& 40& 23& 0& 0& 0& 0\\ 42-44& 43& 21& 1& 1& 21& 21\\ 45-47& 46& 27& 2& 4& 54& 108\\ & & \sum f_i-107& & & \sum f_i{d}_i=22& \sum f_i{d}_i^2=216\end{array}$

Here, A = 40, h = 3, $\sum f_i=107,\sum f_i{d}_i=22$

and $\sum f_i{d}_i^2=216$

$\therefore \text{Mean}=A+\frac{\sum f_i{d}_i}{\sum f_i}\times h=40+\frac{22}{107}\times 3$

= 40 + 1.616 = 40.616

and $Var\left(X\right)=h^2[\frac{\sum f_i{d}_i^2}{\sum f_i}-{\left(\frac{\sum f_i{d}_i}{\sum f_i}\right)}^2]$

$=9[\frac{216}{107}-{\left(\frac{22}{107}\right)}^2]=9[2.0187-0.0420]$

$=9\times 1.976=17.790$

$\therefore SD=\sqrt{Var\left(x\right)}=\sqrt{17.790}=4.217$