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13.Statistics
hard
The mean and variance of $7$ observations are $8$ and $16$ respectively. If one observation $14$ is omitted a and $b$ are respectively mean and variance of remaining $6$ observation, then $a+3 b-5$ is equal to $..........$.
A
$36$
B
$35$
C
$34$
D
$37$
(JEE MAIN-2023)
Solution
$\frac{x_1+x_2+\ldots .+x_7}{7}=8$
$\frac{x_1+x_2+x_3 \ldots .+x_6+14}{7}=8$
$\Rightarrow x_1+x_2+\ldots .+x_6=42$
$\therefore \frac{x_1+x_2 \ldots .+x_6}{6}=\frac{42}{6}=7=a$
$\frac{\sum x_i^2}{7}-8^2=16$
$\Rightarrow x^2=560$
$\Rightarrow x_1^2+x_2^2+\ldots+x_6^2=364$
$b=\frac{x_1^2+x_2^2+\ldots . .+x_6^2}{6}-7^2$
$=\frac{364}{6}-49$
$b=\frac{70}{6}$
$a+3 b-5=7+3 \times \frac{70}{6}-5$
$=37$
Standard 11
Mathematics
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