First slide

Introduction to statistics

Question

If the mean and standard deviation of $5$ observations $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ are $10$ and $3$ respectively, then the variance of $6$ observations $x_{1}, x_{2}, \dots, x_{5}$ and $- 50$ is equal to

Moderate

A

$582.5$
B

$507.5$
C

$586.5$
D

$509.5$

Solution

It is given that
   $\frac{x_{1} + x_{2} + x_{3} + x_{4} + x_{5}}{5} = 10$ and $\frac{1}{5} \sum_{i = 1}^{5} x_{i}^{2} - 10^{2} = 3^{2}$
$\Rightarrow \sum_{i = 1}^{5} x_{i} = 50$ and    $\frac{1}{5} \sum_{i = 1}^{5} x_{i}^{2} - 100 = 9$
$\Rightarrow \sum_{i = 1}^{5} x_{i} = 50$ and    $\sum_{i = 1}^{5} x_{i}^{2} = 545$
Required variance $= \frac{1}{6} \{\sum_{i = 1}^{5} x_{i}^{2} + (- 50)^{2}\} - {\{\frac{\sum_{i = 1}^{5} x_{i} + (- 50)}{6}\}}^{2}$
   $= \frac{1}{6} {545 + 2500} - {\{\frac{50 - 50}{6}\}}^{2} = \frac{3045}{6} = 507.5$

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