Question

The geometric mean of $6$ observations was calculated as $13.$ It was later observed that one of the observations was recorded as $28$ instead of $36.$ The correct geometric mean is

• A. $\left(\frac{9}{7}\right)^{\frac{1}{6}}$
• B. $3 \, \left(\frac{9}{7}\right)^{\frac{1}{6}}$
• C. $13 \, \left(\frac{9}{7}\right)^{\frac{1}{6}}$
• D. $13 \left(\frac{7}{9}\right)^{\frac{1}{6}}$

Answer 1

Answer: (C)

Let $x_{1}, \, x_{2}\ldots , \, x_{6}$ are the observations and $x_{1}=28$
$\Rightarrow \, \, \, 28. \, x_{2}\ldots x_{6}=13^{6}$
$\Rightarrow \, \, \, x_{2}\ldots x_{6}=\frac{13^{6}}{28}$
Now correct observation is $36$
$\Rightarrow \, \, \, 36.x_{2}\ldots x_{6}=\frac{13^{6}}{28}\times 36$
So, now correct geometric mean $=13 \, \left(\frac{9}{7}\right)^{\frac{1}{6}}$