Question

Life of bulbs producedby two factories A and B are given below :
$\begin{array}{cccccc}\text{Length of life}& 550-650& 650-750& 750-850& 850-950& 950-1050\\ \text{(in hours)}:& & & & & \\ \text{Factory A}:& & & & & \\ \text{(Number of bulbs)}& 10& 22& 52& 20& 16\\ \text{Factory B}:& & & & & \\ \text{(Number of bulbs)}& 8& 60& 24& 16& 12\end{array}$

The bulbs of which factory are more consistent from the point of view of length of

Answer 1

For Factory A

Let the assumed mean A = 800 and h = 100

$\begin{array}{ccccccc}\text{Length of life}& Mid-values& u_i=\frac{x_i-800}{100}& u_i^2& Number& f_i{u}_i& f_i{u}_i^2\\ \text{(in hours):}& \left(x_i\right)& & & \text{of bulbs}|f_i|& & \\ 550-650& 600& -2& 4& 10& -20& 40\\ 650-750& 700& -1& 1& 22& -22& 22\\ 750-850& 800& 0& 0& 52& 0& 0\\ 850-950& 900& 1& 1& 20& 20& 20\\ 950-1050& 1000& 2& 4& 16& 32& 64\\ & & & & \sum f_i=120& \sum f_i{u}_i=10& \sum f_i{u}_i^2=146\end{array}$

Mean, ${\overline{X}}_A=A+h\left(\frac{\sum f_i{u}_i}{\sum f_i}\right)=800+100\times \frac{10}{120}=808.33$

Standard Deviation,

${\sigma }_A=h\sqrt{\frac{\sum f_i{u}_i^2}{\sum f_i}-{\left(\frac{\sum f_i{u}_i}{\sum f_i}\right)}^2}=100\sqrt{\frac{146}{120}-{\left(\frac{10}{126}\right)}^2}=100\times 1.0998=109.98$

$\therefore$ Coefficient of variation $\frac{{\sigma }_A}{\overline{A}}\times 100=\frac{109.98}{80833}\times =13.61$

For Factory B

Let the assumed mean A = 800 and

h = 100

$\begin{array}{ccccccc}\text{Length of life}& Mid-values& u_i=\frac{x_i-800}{100}& u_i^2& Number& f_i{u}_i& f_i{u}_i^2\\ \text{(in hours):}& \left(x_i\right)& & & \text{of bulbs}|f_i|& & \\ 550-650& 600& -2& 4& 8& -16& 32\\ 650-750& 700& -1& 1& 60& -60& 60\\ 750-850& 800& 0& 0& 24& 0& 0\\ 850-950& 900& 1& 1& 16& 20& 16\\ 950-1050& 1000& 2& 4& 12& 24& 48\\ & & & & \sum f_i=120& \sum f_i{u}_i=-36& \sum f_i{u}_i^2=156\end{array}$

Mean , ${\overline{X}}_B$ $=A+h\left(\frac{\sum f_i{u}_i}{\sum f_i}\right)=800+100\times \left(\frac{-36}{120}\right)=800-30=770$

Standard Deviation ,

$h\sqrt{\frac{\sum f_i{u}_i^2}{\sum f_i}-{\left(\frac{\sum f_i{u}_i}{\sum f_i}\right)}^2}=100\sqrt{\frac{156}{120}-{\left(\frac{-36}{120}\right)}^2}=100\times 1.1=110$

$\therefore$ Coefficient of variation $\frac{{\sigma }_B}{{\overline{X}}_B}\times =\frac{110}{770}\times 100=14.29$

Hence Coeff. of varience of factory B is greater than the coefficient of varience of Factory A.

$\therefore$ Factory B has more variability then A. This means bulbs of factory A are more consistant from the point of view of length of life .