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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.
Solution
Here, $n_{1}=60, \bar{x}_{1}=650, s_{1}=8$ and $n_{2}=80, \bar{x}_{2}=660, s_{2}=7$
$\therefore \quad \sigma=\sqrt{\frac{n_{1} s_{1}^{2}+n_{2} s_{2}^{2}}{n_{1}+n_{2}}+\frac{n_{1} n_{2}\left(\bar{x}_{1}-\bar{x}_{2}\right)^{2}}{\left(n_{1}+n_{2}\right)^{2}}}$
$=\sqrt{\frac{60 \times(8)^{2}+80 \times(7)^{2}}{60+80}+\frac{60 \times 80(650-660)^{2}}{(60+80)^{2}}}$
$=\sqrt{\frac{6 \times 64+8 \times 49}{14}+\frac{60 \times 80 \times 100}{140 \times 140}}$
$=\sqrt{\frac{192+196}{7}+\frac{1200}{49}=\sqrt{\frac{388}{7}+\frac{1200}{49}}}{\sqrt{\frac{2716+1200}{49}}}$
$=\sqrt{\frac{3915}{49}}=\sqrt{79.9}=8.9$
Similar Questions
The data is obtained in tabular form as follows.
| ${x_i}$ | $60$ | $61$ | $62$ | $63$ | $64$ | $65$ | $66$ | $67$ | $68$ |
| ${f_i}$ | $2$ | $1$ | $12$ | $29$ | $25$ | $12$ | $10$ | $4$ | $5$ |
Let the mean and variance of the frequency distribution
| $\mathrm{x}$ | $\mathrm{x}_{1}=2$ | $\mathrm{x}_{2}=6$ | $\mathrm{x}_{3}=8$ | $\mathrm{x}_{4}=9$ |
| $\mathrm{f}$ | $4$ | $4$ | $\alpha$ | $\beta$ |
be $6$ and $6.8$ respectively. If $x_{3}$ is changed from $8$ to $7 ,$ then the mean for the new data will be:
Find the mean, variance and standard deviation using short-cut method
| Height in cms | $70-75$ | $75-80$ | $80-85$ | $85-90$ | $90-95$ | $95-100$ | $100-105$ | $105-110$ | $110-115$ |
| No. of children | $3$ | $4$ | $7$ | $7$ | $15$ | $9$ | $6$ | $6$ | $3$ |
