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**Hint:**For solving this type of question we have to know the first standard deviation and mean deviation. The formula for finding it will be the square root of the variance. And for variance, it will be the square of the average of the squared differences from the mean. So for the standard deviation, it will be the square root of the variance.

**Formula used:**

The variance will be calculated by,

${\sigma ^2} = \dfrac{{{{\sum {\left( {x - \bar x} \right)} }^2}}}{{n - 1}}$

$\sigma $ , will be the variance

$x$ , will be the first term

$\bar x$ , will be the next term

$n$ , will be the number of terms

Standard deviation,

$Standard{\text{ deviation = }}\sqrt {Variance} $

Mean deviation,

$M.D = \dfrac{{\sum x }}{N}$

**Complete step-by-step answer:**So we have the series given as \[x = 160,160,161,162,163,163,164,164,170\]

So the, $\sum x = 160 + 160 + 161 + 162 + 163 + 163 + 164 + 164 + 170$

So on adding it, we will get the series as

\[\sum x = 1,467\]

Here, we have $N = 9$

Therefore, $M.D = \dfrac{{\sum x }}{N}$

On substituting the values, we get

$M.D = \dfrac{{\sum {1,467} }}{9}$

On solving the above division, we get

$M.D = 163$

Hence, the mean deviation will be equal to $163$ .

Now we will calculate the variance,

${\sigma ^2} = \dfrac{{{{\sum {\left( {x - \bar x} \right)} }^2}}}{{n - 1}}$

Now on substituting the values, we get

\[{\sigma ^2} = \dfrac{{{{\left( {160 - 160} \right)}^2} + {{\left( {161 - 160} \right)}^2} + {{\left( {162 - 161} \right)}^2} + {{\left( {163 - 162} \right)}^2} + {{\left( {163 - 163} \right)}^2} + {{\left( {164 - 163} \right)}^2} + {{\left( {164 - 164} \right)}^2} + {{\left( {170 - 164} \right)}^2}}}{{9 - 1}}\]

Now on solving it, we will get

$\Rightarrow$ ${\sigma ^2} = \dfrac{{{0^2} + {1^2} + {1^2} + {1^2} + {0^2} + {0^2} + {6^2}}}{8}$

Now on adding, we get

$\Rightarrow$ ${\sigma ^2} = \dfrac{{0 + 1 + 1 + 1 + 0 + 0 + 36}}{8}$

And on solving we get

$\Rightarrow$ ${\sigma ^2} = \dfrac{{39}}{8}$

So S.D will be calculated by the formula $Standard{\text{ deviation = }}\sqrt {Variance} $

Substituting the values, we get

$Standard{\text{ deviation = }}\sqrt {\sqrt {\dfrac{{39}}{8}} } $

Hence, the above will be the standard deviation.

**Note:**We should also know that the average deviation is known to be the mean deviation. Sometimes the question may come from the graph where we have to find the mean deviation and all. Here in this type of question variability will be different. The process will be the same as the above just we have to find out the numbers from the graph.

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