Question

To check the quality of two brands of light bulbs, their life in burning hours was estimated as under for 100 brand.

$\begin{array}{ccc}\text{Life (in hrs)}& \text{Number of bulbs (Brand A)}& \text{Number of bulbs (Brand B)}\\ 0-50& 15& 2\\ 50-100& 20& 8\\ 100-150& 18& 60\\ 150-200& 25& 25\\ 200-250& 22& 5\\ \text{Total}& 100& 100\end{array}$

(i) Which brand gives higher life ?

(ii) Which brand is more dependable ?

Answer 1

For Brand A

$\begin{array}{cccccccc}\text{Life}& \text{Number of Bulbs}& \text{MX}& A=125& d=\frac{X-A}{t}& d^2& fd& f{d}^2\\ \text{in hrs}& \text{(f)}& \text{X}& \text{X - A = d}& & & & \\ 0-5& 15& 25& -100& -2& 4& -30& 60\\ 50-100& 20& 75& -50& -1& 1& -20& 20\\ 100-150& 18& 125& 0& 0& 0& 0& 0\\ 150-200& 25& 175& 50& 1& 1& 25& 25\\ 200-250& 22& 225& 100& 2& 4& 44& 88\\ & \sum f=100& & & & & \sum fd=19& \sum f{d}^2=193\end{array}$

$\overline{X}=A+\frac{\sum fd}{\sum f}\times i=125+\frac{19}{100}\times 50=134.5\sigma =\sqrt{\frac{\sum f{d}^2}{\sum f}-{\left(\frac{\sum fd}{\sum f}\right)}^2}\times i=\sqrt{\frac{193}{100}-{\left(\frac{19}{100}\right)}^2}\times 50=\sqrt{\frac{193}{100}-\frac{361}{10000}}\times 50=\sqrt{\frac{18939}{10000}}\times 50=68.5CV=\frac{\sigma }{\overline{X}}\times 100=\frac{68.5}{13.5}\times 100=50.93$

For Brand B

$\begin{array}{cccccccc}\text{Life}& \text{Number of Bulbs(f)}& \text{Mid Value m}& A=125& d=\frac{X-A}{t}& d^2& fd& f{d}^2\\ 0-5& 2& 25& -100& -2& 4& -4& 8\\ 50-100& 8& 75& -50& -1& 1& -8& 8\\ 100-150& 60& 125& 0& 0& 0& 0& 0\\ 150-200& 25& 175& 50& 1& 1& 25& 25\\ 200-250& 5& 225& 100& 2& 4& 10& 20\\ & \sum f=100& & & & \sum fd=23& \sum f{d}^2=61& \end{array}$

$\overline{X}=A+\frac{\sum ifd}{\sum f}\times i=125+\frac{23}{100}\times 50=136.5\sigma =\sqrt{\frac{\sum f{d}^2}{\sum f}-{\left(\frac{\sum fd}{\sum f}\right)}^2}\times i=\sqrt{\frac{61}{100}-{\left(\frac{23}{100}\right)}^2}\times 50=\sqrt{\frac{61}{100}-\frac{529}{10000}}\times 50=\sqrt{\frac{5571}{10000}}\times 50=37.5=\frac{\sigma }{\overline{X}}\times 100=\frac{37.5}{136.5}\times 100=27.47$

(i) The average life of bulb of Brand B is compartively higher than that of brand A.

(ii) The bulbs of Brand B are more depandable as CV of Brand B is lesser than CV of Brand A.