Solution:

Given that the mean life of a sample of 60 bulbs was 650 hours and the standard deviation was 8 hours. A second sample of 80 bulbs has a mean life of 660 hours and standard deviation 7 hours

We now need to find the overall standard deviation

As per the criteria given, in the first set of samples,

No. of sample bulbs, $n_{1}=60$

Standard deviation, $s_{1}=8 h r s$

Mean life, $\overline{\mathrm{x}}_{1}=650$

And in the second set of samples,

No. of sample bulbs, $\mathrm{n}_{2}=80$

Standard deviation, $\mathrm{s}_{2}=7 \mathrm{hrs}$

Mean life, $\overline{\mathrm{x}}_{2}=660$

The standard deviation for combined two series is

$SD(\sigma)=\sqrt{\frac{n_{1} s_{1}^{2}+n_{2} s_{2}^{2}}{n_{1}+n_{2}}+\frac{n_{1} n_{2}\left(\bar{x}_{1}-\bar{x}_{2}\right)^{2}}{\left(n_{1}+n_{2}\right)^{2}}}$

Substitute the corresponding values, we obtain

$SD(\sigma)=\sqrt{\frac{(60)(8)^{2}+(80)(7)^{2}}{60+80}+\frac{(60 \times 80)(650-660)^{2}}{(60+80)^{2}}}$

Adding the denominator

$SD(\sigma)=\sqrt{\frac{(60) 64+(80) 49}{140}+\frac{(4800)(10)^{2}}{(140)^{2}}}$

$SD(\sigma)=\sqrt{\frac{(60) 64+(80) 49}{140}+\frac{(4800) 100}{19600}}$

Simplify

$\mathrm{SD}(\sigma)=\sqrt{\frac{(6) 64+(8) 49}{14}}+\frac{(4800) 1}{196}$

$SD(\sigma)=\sqrt{\frac{384+392}{14}+\frac{4800}{196}}$

$SD(\sigma)=\sqrt{\frac{388}{7}}+\frac{1200}{49}$

$SD(\sigma)=\sqrt{\frac{388 \times 7+1200}{49}}$

$SD(\sigma)=\sqrt{\frac{2716+1200}{49}}$

$SD(\sigma)=\sqrt{\frac{3916}{49}}$

Or, $\sigma=8.9$

As a result, $8.9$ is the standard deviation of the set obtained by combining the given two sets.