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(D) Given: $n_{1}=10, n_{2}=10$Means: $m_{1}=60, m_{2}=40$Standard deviations: $\sigma_{1}=4, \sigma_{2}=6$The formula for the combined standard devia

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$11.2$

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(D) Given: $n_{1}=10, n_{2}=10$Means: $m_{1}=60, m_{2}=40$Standard deviations: $\sigma_{1}=4, \sigma_{2}=6$The formula for the combined standard deviation $\sigma$ is:$\sigma=\sqrt{\frac{n_{1} \sigma_{1}^{2}+n_{2} \sigma_{2}^{2}}{n_{1}+n_{2}}+\frac{n_{1} n_{2}(m_{1}-m_{2})^{2}}{(n_{1}+n_{2})^{2}}}$Substituting the values:$\sigma=\sqrt{\frac{10 	imes 4^{2}+10 	imes 6^{2}}{10+10}+\frac{10 	imes 10(60-40)^{2}}{(10+10)^{2}}}$$\sigma=\sqrt{\frac{160+360}{20}+\frac{100(20)^{2}}{400}}$$\sigma=\sqrt{\frac{520}{20}+\frac{100 	imes 400}{400}}$$\sigma=\sqrt{26+100}=\sqrt{126} \approx 11.22$

Classes	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$
Frequency	$11$	$29$	$18$	$4$	$5$	$3$

Marks of $A$	$30, 20, 40$
Marks of $B$	$70, 0, 5$

Variate $(x)$	$x_{1}$	$x_{2}$	$x_{3} \ldots x_{15}$
Frequency $(f)$	$f_{1}$	$f_{2}$	$f_{3} \ldots f_{15}$

The average marks of $10$ students in a class was $60$ with a standard deviation of $4$,while the average marks of another $10$ students was $40$ with a standard deviation of $6$. If all the $20$ students are taken together,their combined standard deviation will be:

Similar Questions

If the standard deviation of the numbers $2, 3, a$,and $11$ is $3.5$,then which of the following is true?

The variance of the following frequency distribution is:
Classes $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$
Frequency $11$ $29$ $18$ $4$ $5$ $3$

In an experiment with $15$ observations,the results were available as $\sum X^2 = 2830$ and $\sum X = 170$. One observation that was $20$ was found to be wrong and was replaced by the correct value $30$. The corrected variance is:

The marks obtained by students $A$ and $B$ in $3$ examinations are given below:

Marks of $A$ $30, 20, 40$

Marks of $B$ $70, 0, 5$

The ratio of the coefficient of variation of marks of $A$ and the coefficient of variation of marks of $B$ is:

For the frequency distribution:

Variate $(x)$ $x_{1}$ $x_{2}$ $x_{3} \ldots x_{15}$

Frequency $(f)$ $f_{1}$ $f_{2}$ $f_{3} \ldots f_{15}$

where $0 < x_{1} < x_{2} < x_{3} < \ldots < x_{15} = 10$ and $\sum_{i=1}^{15} f_{i} > 0$,the standard deviation cannot be:

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