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The mean of two samples of size $200$ and $300$ were found to be $25, 10$ respectively their $S.D.$ is $3$ and $4$ respectively then variance of combined sample of size $500$ is :-
$64$
$65.2$
$67.2$
$64.2$
Solution
$\mathrm{x}_{1}=200 \quad \mathrm{x}_{2}=300$
$\overline{\mathrm{x}}_{1}=25 \quad \overline{\mathrm{x}}_{2}=10$
$\sigma_{1}=3 \quad \sigma_{2}=4$
combined mean $=\frac{25 \times 200+10 \times 300}{500}=16$
$\sigma_{1}^{2}=9=\frac{1}{200}\left(\sum x_{i}^{2}\right)-625$
$126800=\sum x_{i}^{2}$
$\sigma_{2}^{2}=16=\frac{1}{300} \sum y_{1}^{2}-100$
$34800=\sum y_{1}^{2}$
$\sigma^{2}=\frac{1}{500}(126800+34800)-(16)^{2}$
$=323.2-256=67.2$
Similar Questions
The following values are calculated in respect of heights and weights of the students of a section of Class $\mathrm{XI}:$
Height | Weight | |
Mean | $162.6\,cm$ | $52.36\,kg$ |
Variance | $127.69\,c{m^2}$ | $23.1361\,k{g^2}$ |
Can we say that the weights show greater variation than the heights?
Find the mean and variance for the following frequency distribution.
Classes | $0-10$ | $10-20$ | $20-30$ | $30-40$ | $40-50$ |
Frequencies | $5$ | $8$ | $15$ | $16$ | $6$ |