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: $n_{1}=60, \bar{x}_{1}=650, s_{1}=8$ and $n_{2}=80, \bar{x}_{2}=660, s_{2}=7$.
The formula for the combined standard deviation $\sigma$ is:
$\sigma = \sqrt{\frac{n_{1} s_{1}^{2} + n_{2} s_{2}^{2}}{n_{1} + n_{2}} + \frac{n_{1} n_{2}(\bar{x}_{1} - \bar{x}_{2})^{2}}{(n_{1} + n_{2})^{2}}}$
Substituting the values:
$\sigma = \sqrt{\frac{60(8)^{2} + 80(7)^{2}}{60 + 80} + \frac{60 \times 80(650 - 660)^{2}}{(60 + 80)^{2}}}$
$\sigma = \sqrt{\frac{60(64) + 80(49)}{140} + \frac{4800(-10)^{2}}{(140)^{2}}}$
$\sigma = \sqrt{\frac{3840 + 3920}{140} + \frac{480000}{19600}}$
$\sigma = \sqrt{\frac{7760}{140} + \frac{4800}{196}} = \sqrt{55.428 + 24.489} = \sqrt{79.917} \approx 8.94$ hours.
The formula for the combined standard deviation $\sigma$ is:
$\sigma = \sqrt{\frac{n_{1} s_{1}^{2} + n_{2} s_{2}^{2}}{n_{1} + n_{2}} + \frac{n_{1} n_{2}(\bar{x}_{1} - \bar{x}_{2})^{2}}{(n_{1} + n_{2})^{2}}}$
Substituting the values:
$\sigma = \sqrt{\frac{60(8)^{2} + 80(7)^{2}}{60 + 80} + \frac{60 \times 80(650 - 660)^{2}}{(60 + 80)^{2}}}$
$\sigma = \sqrt{\frac{60(64) + 80(49)}{140} + \frac{4800(-10)^{2}}{(140)^{2}}}$
$\sigma = \sqrt{\frac{3840 + 3920}{140} + \frac{480000}{19600}}$
$\sigma = \sqrt{\frac{7760}{140} + \frac{4800}{196}} = \sqrt{55.428 + 24.489} = \sqrt{79.917} \approx 8.94$ hours.
Explore More
Similar Questions
If the mean deviation about the mean is $m$ and variance is $\sigma^2$ for the following data,then $m + \sigma^2 =$
$x$ $1, 3, 5, 7, 9$ $f$ $4, 24, 28, 16, 8$
| $x$ | $1, 3, 5, 7, 9$ |
| $f$ | $4, 24, 28, 16, 8$ |
If the sum of squares of the deviations from the mean of the data $x_i, (i=1, 2, \ldots, n)$ is $n\bar{x}^2$,where $\bar{x}$ is the mean of $x_i$'s,then the sum of squares of $x_i$'s is
If the mean of two samples of sizes $20$ and $30$ are $25$ and $10$ respectively,and their variances are $9$ and $16$ respectively,then their combined variance is:
Let the mean and variance of $12$ observations be $\frac{9}{2}$ and $4$ respectively. Later on,it was observed that two observations were considered as $9$ and $10$ instead of $7$ and $14$ respectively. If the correct variance is $\frac{m}{n}$,where $m$ and $n$ are co-prime,then $m + n$ is equal to
DifficultJEE MAIN 2023
View SolutionThe mean and $S.D.$ of the marks of $200$ candidates were found to be $40$ and $15$ respectively. Later,it was discovered that a score of $40$ was wrongly read as $50$. The correct mean and $S.D.$ respectively are...
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo