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13.Statistics
medium
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
A
$40$
B
$30$
C
$50$
D
$35$
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
Similar Questions
Calculate the mean, variance and standard deviation for the following distribution:
Class | $30-40$ | $40-50$ | $50-60$ | $60-70$ | $70-80$ | $80-90$ | $90-100$ |
$f_i$ | $3$ | $7$ | $12$ | $15$ | $8$ | $3$ | $2$ |
hard