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. Find the overall standard deviation.
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