# A student calculates the geometric mean of 8 values is 10.5. After rechecking it revealed that one value 15 is missing. Find the correct geometric mean?

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Feb 7, 2018

${15}^{\frac{1}{9}} \cdot {\left(10.5\right)}^{\frac{8}{9}}$.

#### Explanation:

Suppose that the $8$ observations are ${x}_{1} , {x}_{2} , \ldots , {x}_{8}$, where,

all ${x}_{i} > 0$.

Given that, the GM of $\left\{{x}_{i} : i = 1 , 2 , \ldots , 8\right\}$ is $10.5$

$\therefore \sqrt[8]{{x}_{1} \cdot {x}_{2} \cdot \ldots \cdot {x}_{8}} = {\left({x}_{1} \cdot {x}_{2} \cdot \ldots \cdot {x}_{8}\right)}^{\frac{1}{8}} = 10.5$

$\therefore \left({x}_{1} \cdot {x}_{2} \cdot \ldots \cdot {x}_{8}\right) = {\left(10.5\right)}^{8.} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left[\ast\right]$.

Now, taking into account of the missing entry $15$, we now have

$9$ observations, namely, $\left\{{x}_{i} : i = 1 , 2 , \ldots , 8\right\} \cup \left\{15\right\}$.

Hence, the corrected GM will be given by,

$\sqrt[9]{15 \cdot {x}_{1} \cdot {x}_{2} \cdot \ldots \cdot {x}_{8}}$,

$= {\left(15 \cdot {x}_{1} \cdot {x}_{2} \cdot \ldots \cdot {x}_{8}\right)}^{\frac{1}{9}}$,

$= {15}^{\frac{1}{9}} \cdot {\left({x}_{1} \cdot {x}_{2} \cdot \ldots \cdot {x}_{8}\right)}^{\frac{1}{9}}$,

=15^(1/9)*{(10.5)^8)}^(1/9),

$= {15}^{\frac{1}{9}} \cdot {\left(10.5\right)}^{\frac{8}{9}}$.

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