Mean and variance of a set of 6 terms is 11 and 24 respectively and mean and variance of another set

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13.Statistics

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Mean and variance of a set of $6$ terms is $11$ and $24$ respectively and the mean and variance of another set of $3$ terms is $14$ and $36$ respectively. Then variance of all $9$ terms is equal to

$40$

$30$

$50$

$35$

Solution

$\sigma^{2}=\frac{n_{1} \sigma_{1}^{2}+n_{2} \sigma_{2}^{2}}{n_{1}+n_{2}}+\frac{n_{1} n_{2}}{\left(n_{1}+n_{2}\right)^{2}}\left(\bar{x}_{1}-\bar{x}_{2}\right)^{2}$

$\sigma^{2}=\frac{6 \times 24+3 \times 36}{6+3}+\frac{6 \times 3}{(6+3)^{2}}(11-14)^{2}$

$\sigma^{2}=\frac{144+108}{9}+\frac{18}{81} \times 9=28+2=30$

Standard 11

Mathematics

If the variance of the frequency distribution is $160$ , then the value of $\mathrm{c} \in \mathrm{N}$ is

$X$ $c$ $2c$ $3c$ $4c$ $5c$ $6c$

$f$ $2$ $1$ $1$ $1$ $1$ $1$

medium

(JEE MAIN-2024)

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medium

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easy

The mean and standard deviation of the marks of $10$ students were found to be $50$ and $12$ respectively. Later, it was observed that two marks $20$ and $25$ were wrongly read as $45$ and $50$ respectively. Then the correct variance is $…………$.

hard

(JEE MAIN-2023)

If the mean of the data : $7, 8, 9, 7, 8, 7, \mathop \lambda \limits^. , 8$ is $8$, then the variance of this data is

hard

(JEE MAIN-2018)

Mean and variance of a set of $6$ terms is $11$ and $24$ respectively and the mean and variance of another set of $3$ terms is $14$ and $36$ respectively. Then variance of all $9$ terms is equal to

Solution

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If the variance of the frequency distribution is $160$ , then the value of $\mathrm{c} \in \mathrm{N}$ is

$X$ $c$ $2c$ $3c$ $4c$ $5c$ $6c$

$f$ $2$ $1$ $1$ $1$ $1$ $1$