Question

Draw the necessary tables to find the standard deviation for the data
20, 14, 16, 30, 21 and 25.

Answer 1

Hint:
We will find the mean of data using the formula for the mean. Then, we will find the difference between the mean and each observation. We will square these differences. Then, we will find the mean of the squared differences. Finally, we will find the square root of the quantity that we have obtained and that will give us the standard deviation of the data.
Formula used: We will use the following formulas:
The mean \[\left( {\bar x} \right)\] is given by \[\bar x = \dfrac{{{x_1} + {x_2} + ... + {x_n}}}{n}\] where \[{x_1}\], \[{x_2}\] and \[{x_n}\] are the 1st, 2nd and \[{n^{th}}\] observation respectively and \[n\] is the total number of observation.
Standard deviation is given by \[S.D. = \sqrt {\dfrac{{\sum {{\left( {{x_i} - \bar x} \right)}^2}}}{n}} \] where \[{x_i}\] is the \[{i^{th}}\] observation of the data, \[\bar x\] is the mean and \[n\] is the total number of observations.

Complete step by step solution:
We will find the mean of the data. We will substitute 6 for \[n\], 20 for \[{x_1}\], 14 for \[{x_2}\], 16 for \[{x_3}\], 30 for \[{x_4}\], 21 for \[{x_5}\] and 25 for \[{x_6}\] in the formula \[\bar x = \dfrac{{{x_1} + {x_2} + ... + {x_n}}}{n}\].
\[\bar x = \dfrac{{20 + 14 + 16 + 30 + 21 + 25}}{6} = \dfrac{{126}}{6} = 21\]
We will make a table to represent \[{x_i} - \bar x,\left( {{x_i} - \bar x} \right)\] and \[{\left( {{x_i} - \bar x} \right)^2}\].

\[{x_i}\]	\[{x_i} - \bar x\]	\[{\left( {{x_i} - \bar x} \right)^2}\]
20	\[20 - 21 = - 1\]	\[{\left( { - 1} \right)^2} = 1\]
14	\[14 - 21 = - 7\]	\[{\left( { - 7} \right)^2} = 49\]
16	\[16 - 21 = - 5\]	\[{\left( { - 5} \right)^2} = 25\]
30	\[30 - 21 = 9\]	\[{9^2} = 81\]
21	\[21 - 21 = 0\]	\[{0^2} = 0\]
25	\[25 - 21 = 4\]	\[{4^2} = 16\]

Now, we will find \[\sum {\left( {{x_i} - \bar x} \right)^2}\]. The symbol \[\sum {} \]represents summation and the means that we have to add all the observations given in its parenthesis.
\[\sum {\left( {{x_i} - \bar x} \right)^2} = 1 + 49 + 25 + 81 + 0 + 16 = 172\]
We will substitute 172 for \[\sum {\left( {{x_i} - \bar x} \right)^2}\]and 6 for \[n\] in the formula \[S.D. = \sqrt {\dfrac{{\sum {{\left( {{x_i} - \bar x} \right)}^2}}}{n}} \].
\[S.D. = \sqrt {\dfrac{{172}}{6}} = \sqrt {28.667} = 5.35\]
The standard deviation of the data is 5.35.

Note:
The standard deviation is the square root of the variance of the data. Variance is given by \[V = \dfrac{{\sum {{\left( {{x_i} - \bar x} \right)}^2}}}{n}\] . We might get confused between the 2 formulas. A good way to remember when to use the square root is that Standard Deviation (S.D.) starts from S and square root also starts from S, so we have to use square root in the formula for standard deviation.