Let the assumed mean, A = 35, c = 10

Standard deviation σ = \(c\times \sqrt { \frac { \Sigma { f }_{ i }{ d }_{ i }^{ 2 } }{ N } -\left( \frac { \Sigma { f }_{ i }{ d }_{ i } }{ N } \right) ^{ 2 } } \)
σ = \(10\times \sqrt { \frac { 210 }{ 71 } -\left( \frac { 30 }{ 71 } \right) ^{ 2 } } =10\times \sqrt { \frac { 210 }{ 71 } -\frac { 900 }{ 5041 } } \)
= 10 x \(\sqrt { 2.779 } \); σ ≃ 16.67

Let the assumed mean, A = 8

Standard deviation
σ = \(\sqrt { \frac { \Sigma { f }_{ i }{ d }_{ i }^{ 2 } }{ N } -\left( \frac { \Sigma { f }_{ i }{ d }_{ i } }{ N } \right) ^{ 2 } } \)
= \(\sqrt { \frac { 240 }{ 29 } -\left( \frac { 4 }{ 29 } \right) ^{ 2 } } =\sqrt { \frac { 240\times 29-16 }{ 29\times 29 } } \)
σ = \(\sqrt { \frac { 6944 }{ 29\times 29 } } \); σ ≃ 2.87
Calculation of Standard deviation for continuous frequency distribution
(i) Mean method:
Standard deviation σ = \(\sqrt { \frac { \Sigma { f }_{ i }({ x }_{ i }-\bar { x } )^{ 2 } }{ N } } \)
Where, x_i = Middle value of the i th class
f_i = Frequency of the i th class
(ii) Shortcut method (or) Step deviation method:
To make the calculation simple, we provide the following formula. Let A be the assumed mean, x_i be the middle value of the i^th class and c is the width of the class interval.
Let d_i = \(\frac { { x }_{ i }-A }{ c } \)
σ = \(\sqrt { \frac { \Sigma { f }_{ i }{ d }_{ i }^{ 2 } }{ N } -\left( \frac { \Sigma { f }_{ i }{ d }_{ i } }{ N } \right) ^{ 2 } } \).

Mean
\(\bar { x } =\frac { \Sigma { x }_{ i }{ f }_{ i } }{ N } =\frac { 432 }{ 48 } \) = 9 (Since N = Σf_i)
Standard deviation
σ =\(\sqrt { \frac { \Sigma { f }_{ i }{ d }_{ i }^{ 2 } }{ N } } =\sqrt { \frac { 124 }{ 48 } } =\sqrt { 2.58 } \)
σ ≃ 1.6
Assumed Mean method:
Let x₁, x₂, x₃, ......x₄ be the given data with frequencies f₁, f₂, f₃, ... fn respectively.
Let \(\bar { x } \) be their mean and A be the assumed mean
di = x_i - A
Standard deviation σ = \(\sqrt { \frac { \Sigma { f }_{ i }{ d }_{ i }^{ 2 } }{ N } -\left( \frac { \Sigma { f }_{ i }{ d }_{ i } }{ N } \right) ^{ 2 } } \).

Mean \(\bar { x } \) = \(\frac { Sum\ of\ all\ observations }{ Number\ of\ observation } \)
= \(\frac { \Sigma x_{ i } }{ n } =\frac { 1+2+3+...+n }{ n } =\frac { n(n+1) }{ 2\times n } \)
Mean \(\bar { x } \) = \(\frac { n+1 }{ 2 } \)
Variance σ² = \(\frac { \Sigma x_{ i }^{ 2 } }{ n } -\left( \frac { \Sigma x_{ i } }{ n } \right) ^{ 2 }\left[ \begin{matrix} \Sigma x_{ i }^{ 2 }={ 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }+...+{ n }^{ 2 } \\ (\Sigma x_{ i })^{ 2 }=(1+2+3+...+n)2 \end{matrix} \right] \)
= \(\frac { n(n+1)(2n+1) }{ 6\times n } -\left[ \frac { n(n+1) }{ 2\times n } \right] ^{ 2 }\)
= \(\frac { 2n^{ 2 }+3n+1 }{ 6 } -\frac { { n }^{ 2 }+2n+1 }{ 4 } \)
Variance σ² = \(\frac { 4n^{ 2 }+6n+2-3n^{ 2 }-6n-3 }{ 12 } =\frac { { n }^{ 2 }-1 }{ 12 } \).

Given, n = 5

Standard deviation
σ = \(\sqrt { \frac { { \Sigma d }_{ i }^{ 2 } }{ n } -\left( \frac { { \Sigma d }_{ i } }{ n } \right) ^{ 2 } } \)
σ = \(\sqrt { \frac { 151 }{ 5 } -\left( \frac { 25 }{ 5 } \right) ^{ 2 } } =\sqrt { 30.2-25 } =\sqrt { 5.2 } \) ≃ 2.28
When we multiply each data by 4, we get the new values as 8, 12, 20, 28, 32.

 Standard deviation
σ = \(\sqrt { \frac { { \Sigma d }_{ i }^{ 2 } }{ n } -\left( \frac { { \Sigma d }_{ i } }{ n } \right) ^{ 2 } } \)
= \(\sqrt { \frac { 2416 }{ 5 } -\left( \frac { 100 }{ 5 } \right) ^{ 2 } } =\sqrt { 483.2-400 } =\sqrt { 83.2 } \)
σ = \(\sqrt { 16\times 5.2 } =4\sqrt { 5.2 } \) ≃ 9.12

Arranging the values in ascending order we get, 4, 7, 8, 10, 11 and n = 5

Standard deviation
σ = \(\sqrt { \frac { { \Sigma d }_{ i }^{ 2 } }{ n } -\left( \frac { { \Sigma d }_{ i } }{ n } \right) ^{ 2 } } \)
= \(\sqrt { \frac { 350 }{ 5 } -\left( \frac { 40 }{ 5 } \right) ^{ 2 } } \)
σ = \(\sqrt { 6 } \) ⋍ 2.45
When we add 3 to all the values, we get the new values as 7, 10, 11, 13, 14.

 Standard deviation
σ = \(\sqrt { \frac { { \Sigma d }_{ i }^{ 2 } }{ n } -\left( \frac { { \Sigma d }_{ i } }{ n } \right) ^{ 2 } } \)
= \(\sqrt { \frac { 635 }{ 5 } -\left( \frac { 55 }{ 5 } \right) ^{ 2 } } \)
σ = \(\sqrt { 6 } \) ⋍ 2.45

We note that all the observations are divisible by 5. Hence we can use the step deviation method. Let the Assumed mean A = 20, n = 8.

Standard deviation
σ = \(\sqrt { \frac { { \Sigma d }_{ i }^{ 2 } }{ n } -\left( \frac { { \Sigma d }_{ i } }{ n } \right) ^{ 2 } } \)xc
= \(\sqrt { \frac { 44 }{ 8 } -\left( \frac { 4 }{ 8 } \right) ^{ 2 } } \times 5=\sqrt { \frac { 11 }{ 2 } -\frac { 1 }{ 4 } \times 5 } \)
= \(\sqrt { 5.5-0.25 } \) = 2.29 x 5
σ ≃ 11.45

The mean of marks is 35.9 which is not an integer. Hence we take assumed mean, A=35,n=10.

Standard deviation
σ =\(\sqrt { \frac { { \Sigma d }_{ i }^{ 2 } }{ n } -\left( \frac { { \Sigma d }_{ i } }{ n } \right) ^{ 2 } } \)
= \(\sqrt { \frac { 453 }{ 10 } -\left( \frac { 9 }{ 10 } \right) ^{ 2 } } \)
= \(\sqrt { 45.3-0.81 } \)
= \(\sqrt { 44.49 } \)
σ ≃ 6.67
(ii) Step deviation method
Let x₁, x₂, x_n,... be the given data. Let A be the assumed mean.
Let c be the common divisor of x_i-A
Let d_i = \(\frac { x_{ i }-A }{ c } \)
Then x_i = d_ic + A             ....(1)
Σx_i = Σ(d_ic + A) = cΣd_i + A x n
\(\frac { \Sigma { x }_{ i } }{ n } =c\frac { \Sigma d_{ i } }{ n } +A\)
\(\bar { x } \) = c \(\bar { d } \)+A            ..(2)
\({ x }_{ i }-\bar { x } =c{ d }_{ i }+A-c\bar { d } -A=c(\bar { d_{ i } } -\bar { d } )\)(using (1) and (2))
σ = \(\sqrt { \frac { \Sigma ({ x }_{ i }-\bar { x } )^{ 2 } }{ n } } =\sqrt { \frac { \Sigma (c({ d }_{ i }-\bar { d } )^{ 2 } }{ n } } =\sqrt { \frac { { c }^{ 2 }\Sigma ({ d }_{ i }-\bar { d } )^{ 2 } }{ n } } \)
σ = \(c\times \sqrt { \frac { { \Sigma d }_{ i }^{ 2 } }{ n } -\left( \frac { { \Sigma d }_{ i } }{ n } \right) ^{ 2 } } \\ \)

Arranging the numbers in ascending order we get, 11.4, 12.5, 12.8, 16.3, 17.8, 19.2 Number of observations
n = 6
Mean = \(\frac { 11.4+12.5+12.8+16.3+17.8+19.2 }{ 6 } =\frac { 90 }{ 6 } \)=15

Standard deviation σ = \(\sqrt { \frac { { \Sigma d }_{ i }^{ 2 } }{ n } } \)
=\(\sqrt { \frac { 51.22 }{ 6 } } =\sqrt { 8.53 } \)
Hence, σ ≃2.9
Assumed Mean method:
When the mean value is not an integer (since calculations are very tedious in decimal form) then it is better to use the assumed mean method to find the standard deviation.
Ler x₁, x₂, x₃, .....xn be the given data values and let \(\bar { x } \) be their mean
Let di be the deviation of xi from the assumed mean A, which is usually the middle value or near the middle value of the given data.
d_i = x_i - A gives, x_i = d_i + A           ...(1)
Σd_i = Σ(x_i - A)
= Σx_i-(A + A + A + ..to n times)
Σd_i = Σxi - A x n
\(\frac { \Sigma { d }_{ i } }{ n } =\frac { \Sigma { x }_{ i } }{ n } \) - A
\(\bar { d } \) = \(\bar { x } \) - A (or) \(\bar { x } \) = \(\bar { d } \)+ A         ..(2)
Now, Standard deviation
σ = \(\\ \sqrt { \frac { \Sigma ({ x }_{ i }-\bar { x } )^{ 2 } }{ n } } =\sqrt { \frac { \Sigma ({ d }_{ i }+A-\bar { d } -A)^{ 2 } }{ n } } \) (using (1) and (2))
= \(\sqrt { \frac { \Sigma (d_{ i }-\bar { d } )^{ 2 } }{ n } } =\sqrt { \frac { \Sigma ({ d }_{ i }^{ 2 }+2d_{ i }-\bar { d } -\bar { d } ^{ 2 }) }{ n } } \)
= \(\sqrt { \frac { { \Sigma d }_{ i }^{ 2 } }{ n } -2\bar { d } \frac { \Sigma { d }_{ i } }{ n } +\frac { { \bar { d } }^{ 2 } }{ n } (1+1+1+...to\quad n\quad times) } \)
\(=\sqrt { \frac { { \Sigma d }_{ i }^{ 2 } }{ n } -2\bar { d } \times \bar { d } +\frac { { \bar { d } }^{ 2 } }{ n } \times n } \) (since \(\bar { d } \) is a constant)
= \(\sqrt { \frac { { \Sigma d }_{ i }^{ 2 } }{ n } -{ \bar { d } }^{ 2 } } \)
Standard deviationσ = \(\sqrt { \frac { { \Sigma d }_{ i }^{ 2 } }{ n } -\left( \frac { { \Sigma d }_{ i } }{ n } \right) ^{ 2 } } \).

Standard deviation
σ =\(\sqrt { \frac { \Sigma x_{ i }^{ 2 } }{ n } -\left( \frac { { \Sigma x }_{ i } }{ n } \right) ^{ 2 } } \)
=\(\sqrt { \frac { 623 }{ 7 } -\left( \frac { 63 }{ 7 } \right) ^{ 2 } } \)
=\(\\ \sqrt { 89-81 } =\sqrt { 8 } \)
Hence, σ ≃ 2.83
(ii) Mean method:
Another convenient way of finding standard deviation is to use the following formula.
Standard deviation (by mean method) σ = \(\sqrt { \frac { \Sigma ({ x }_{ i }-\bar { x } )^{ 2 } }{ n } } \)
If di = xi - \(\bar { x } \) are the deviations, then σ = \(\sqrt { \frac { { \Sigma d }_{ i }^{ 2 } }{ n } } \).