The mean and variance of 7 observations are 8 and 16 respectively. If one observation 14 is omitted and aaa and are respectively mean and variance of remaining 6 observations, then is equal to ______. [2023]

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Solution

Q.
The mean and variance of 7 observations are 8 and 16 respectively. If one observation 14 is omitted and $a$ and $b$ are respectively mean and variance of remaining 6 observations, then $a + 3 b - 5$ is equal to ______. [2023]

Ans.
(37)

$Mean = 7, Variance = 16$

Let the observations be $x_{1}, x_{2}, \dots, x_{7}$ .

Now, $\frac{x_{1} + x_{2} + \dots + x_{7}}{7} = 8,$ $x_{1} + x_{2} + \dots + x_{7} = 56$

When one observation $14$ is omitted, the mean is

$a = \frac{x_{1} + x_{2} + \dots + x_{7} - 14}{6} \Rightarrow a = \frac{56 - 14}{6} = 7$

$Variance = \frac{\sum_{i = 1}^{7} {(x_{i})}^{2}}{7} - 64 = 16 \Rightarrow \sum_{i = 1}^{7} {(x_{i})}^{2} = 560$

When one observation is omitted, then variance

$b = \frac{\sum_{i = 1}^{6} {(x_{i})}^{2}}{6} - 49 \Rightarrow b = \frac{560 - {(14)}^{2}}{6} - 49 = \frac{364}{6} - 49 = \frac{35}{3}$

So, $a + 3 b - 5 = 7 + 3 \times \frac{35}{3} - 5 = 37$

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