The standard deviation for the scores \[1\] , \[2\], \[3\], \[4\], \[5\], \[6\] and \[7\] is \[2\]. Then, the standard deviation of \[12\], \[23\],\[34\], \[45\],\[56\], \[67\] and \[78\] is :
A. \[2\]
B. \[4\]
C. \[22\]
D. \[11\]

Hint: Here we will be using the formula of calculating the mean and standard deviation for a particular series of numbers. The formula as given below:
\[{\text{Mean}} = \dfrac{{{\text{sum of the terms}}}}{{{\text{number of terms}}}}\] and
\[{\text{S}}{\text{.D}} = \sqrt {\dfrac{{{{\left( {{x_1} - \mu } \right)}^2} + {{\left( {{x_2} - \mu } \right)}^2} + {{\left( {{x_3} - \mu } \right)}^2}..... + {{\left( {{x_n} - \mu } \right)}^2}}}{{{\text{number of terms}}}}} \], where
\[{x_1}\] is the first term of the series,
\[{x_2}\] known as the second term, and
\[{x_n}\] denotes the \[nth\].
\[\mu \] denotes the mean of that series.

Complete step-by-step solution:
Step 1: For calculating the standard deviation of the given series \[12\],\[23\], \[34\], \[45\], \[56\],
\[67\] and \[78\], at first we will be calculating the mean of that series as shown below:
\[{\text{Mean}} = \dfrac{{{\text{sum of the terms}}}}{{{\text{number of terms}}}}\]
By substituting the values of the sum of terms and number of terms which is \[7\], we get:
\[ \Rightarrow {\text{Mean}} = \dfrac{{12 + 23 + 34 + 45 + 56 + 67 + 78}}{7}\]
By doing the addition in the RHS side of the above expression we get:
\[ \Rightarrow {\text{Mean}} = \dfrac{{315}}{7}\]
By doing the final division in the above expression we get:
\[ \Rightarrow {\text{Mean}} = 45\]
Step 2: By using the formula of standard deviation we get:
\[ \Rightarrow {\text{S}}{\text{.D}} = \sqrt {\dfrac{{{{\left( {12 - 45} \right)}^2} + {{\left( {23 - 45} \right)}^2} + {{\left( {34 - 45} \right)}^2} + {{\left( {45 - 45} \right)}^2} + {{\left( {56 - 45} \right)}^2} + {{\left( {67 - 45} \right)}^2} + {{\left( {78 - 45} \right)}^2}}}{7}} \] , where
\[{x_1} = 12\],
\[{x_2} = 23\],
\[{x_3} = 34\],
\[{x_4} = 45\],
\[{x_5} = 56\],\[{x_6} = 67\],
\[{x_7} = 78\] and \[\mu ({\text{mean}}) = 45\].
Solving the brackets by doing addition and subtraction we get:
\[ \Rightarrow {\text{S}}{\text{.D}} = \sqrt {\dfrac{{{{\left( { - 33} \right)}^2} + {{\left( { - 22} \right)}^2} + {{\left( { - 11} \right)}^2} + {{\left( 0 \right)}^2} + {{\left( {11} \right)}^2} + {{\left( {22} \right)}^2} + {{\left( {33} \right)}^2}}}{7}} \]
By solving the powers of the particular terms in the above expression we get:
\[ \Rightarrow {\text{S}}{\text{.D}} = \sqrt {\dfrac{{1089 + 484 + 121 + 0 + 121 + 484 + 1089}}{7}} \]
By doing the addition in the numerator of the RHS side of the above expression, we get:
\[ \Rightarrow {\text{S}}{\text{.D}} = \sqrt {\dfrac{{3388}}{7}} \]
By dividing the RHS side of the above expression we get:
\[ \Rightarrow {\text{S}}{\text{.D}} = \sqrt {484} \]
Finally, by solving the root of the RHS side of the above expression we get:
\[ \Rightarrow {\text{S}}{\text{.D}} = 22\]

Option C is correct.

Note: Students should remember the formulas for calculating mean, median, mode, and standard deviation. The symbol of mean and standard deviation is \[\mu \] and \[\sigma \] respectively. Also, you should remember that for calculating the standard deviation for any series of numbers first we need to calculate the mean of that series.