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13.Statistics
hard
The mean of the numbers $a, b, 8,5,10$ is $6$ and their variance is $6.8$. If $M$ is the mean deviation of the numbers about the mean, then $25\; M$ is equal to
A
$60$
B
$55$
C
$50$
D
$45$
(JEE MAIN-2022)
Solution
$\sigma^{2}=\frac{\sum\limits_{i=1}^{5}\left(x_{i}-\bar{x}\right)^{2}}{n}$
Mean $=6$
$\frac{a+b+8+5+10}{5}=6$
$a+b=7$
$b=7-a$
$6.8=\frac{(a-6)^{2}+(b-6)^{2}+(8-6)^{2}+(5-6)^{2}+(10-6)^{2}}{5}$
$34=(a-6)^{2}+(7-a-6)^{2}+4+1+18$
$a^{2}-7 a+12=0 \Rightarrow a=4$ or $a=3$
$a=4 \quad a=3$
$b=3 \quad b=4$
$M=\frac{\sum\limits_{i=1}^{5}\left|x_{i}-x\right|}{n}$
$M=\frac{|a-6|+|b-6|+|8-6|+|5-6|+|10-6|}{5}$
when $a =3, b =4 \quad$
$M =\frac{3+2+2+1+4}{5}$
$M =\frac{12}{5}$
when $a =4, b =3$
$ M =\frac{2+3+2+1+7}{5}$
$M =\frac{12}{5}$
$25\;M =25 \times \frac{12}{5}=60$
Standard 11
Mathematics
