The mean life of a sample of 60 bulbs was 650

Solution:

Given 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

Now we have to find the overall standard deviation

As per given criteria, in first set of samples,

Number of sample bulbs, n1=60

Standard deviation, s1=8hrs

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

And in second set of samples,

Number of sample bulbs, $\mathrm{n}_{2}=80$

Standard deviation, $s_{2}=7 \mathrm{hr}$ s

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

We know the standard deviation for combined two series is

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

Substituting the corresponding values, we get

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

By adding the denominator

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

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

On simplifying we get

S. D $(\sigma)=\sqrt{\frac{(6) 64+(8) 49}{14}+\frac{(4800) 1}{196}}$

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

S. D $(\sigma)=\sqrt{\frac{388}{7}+\frac{1200}{49}}$

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

S. D $(\sigma)=\sqrt{\frac{2716+1200}{49}}$

S. D $(\sigma)=\sqrt{\frac{3916}{49}}$

Or, σ=8.9

Hence the standard deviation of the set obtained by combining the given two sets is 8.9