Question

# Find S.D and M.D of the given value. $160,160,161,162,163,163,164,164,170$ .

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

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
$\Rightarrow$ ${\sigma ^2} = \dfrac{{{0^2} + {1^2} + {1^2} + {1^2} + {0^2} + {0^2} + {6^2}}}{8}$
$\Rightarrow$ ${\sigma ^2} = \dfrac{{0 + 1 + 1 + 1 + 0 + 0 + 36}}{8}$
$\Rightarrow$ ${\sigma ^2} = \dfrac{{39}}{8}$
So S.D will be calculated by the formula $Standard{\text{ deviation = }}\sqrt {Variance}$
$Standard{\text{ deviation = }}\sqrt {\sqrt {\dfrac{{39}}{8}} }$