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**Hint:**Here the question is related to the statistics topic. To determine mean, variance and standard deviation of the given data we use three different methods namely, direct method, deviation method and step deviation method. Here we see a detailed explanation about step deviation method and we solve one problem related to this.

**Complete step by step answer:**

In the statistics we have two kinds of data namely, grouped data and ungrouped data. For the ungrouped data, the step deviation method is not implemented. The step deviation method is implemented for the grouped data. The grouped data should consist of the class interval. We have a formula for the mean, variance and standard deviation in the step deviation method and it is given as

**mean:**

\[mean = A + c \times \dfrac{{\sum {{f_i}{t_i}} }}{{\sum {{f_i}} }}\]

**Variance:**

\[Variance = \,{c^2}\left[ {\dfrac{{\sum {{f_i}{t_i}^2} }}{{\sum {{f_i}} }} - {{\left( {\dfrac{{\sum {{f_i}{t_i}} }}{{\sum {{f_i}} }}} \right)}^2}} \right]\]

**Standard deviation:**

\[S.D = \,c\sqrt {\left[ {\dfrac{{\sum {{f_i}{t_i}^2} }}{{\sum {{f_i}} }} - {{\left( {\dfrac{{\sum {{f_i}{t_i}} }}{{\sum {{f_i}} }}} \right)}^2}} \right]} \]

Here,

$A$ = mean of the \[{x_i}\]

$c$ = length of class interval

\[{f_{_i}}\] = frequency

\[{t_{_i}} = \dfrac{{{x_i} - A}}{{\sum {{f_i}} }}\]

These are the important formulas and we have to remember it.

Now we will consider one problem and we solve it by using the step deviation method.

Example: Calculate the mean and standard deviation for the following data:

Class-interval | 0-100 | 100-200 | 200-300 | 300-400 | 400-500 | 500-600 | 600-700 |

\[{f_i}\] | 9 | 17 | 32 | 23 | 40 | 18 | 1 |

The solution for the given example as follows:

Class-interval | Mid-points\[{x_i}\] | Frequency\[{f_i}\] | \[{t_{_i}} = \dfrac{{{x_i} - A}}{{\sum {{f_i}} }}\] | \[{t_{_i}}^2\] | \[{f_i}{t_{_i}}\] | \[{f_i}{t_{_i}}^2\] |

0 – 100 | 50 | 9 | -3 | 9 | -27 | 81 |

100-200 | 150 | 17 | -2 | 4 | -34 | 68 |

200-300 | 250 | 32 | -1 | 1 | -32 | 32 |

300-400 | 350 | 23 | 0 | 0 | 0 | 0 |

400-500 | 450 | 40 | 1 | 1 | 40 | 40 |

500-600 | 550 | 18 | 2 | 4 | 36 | 72 |

600-700 | 650 | 1 | 3 | 9 | 3 | 9 |

Total | 140 | -14 | 302 |

The mid-points are calculated by adding the end points of the class interval and then by dividing the sum by 2.

Here A is calculated by

\[A = \dfrac{{\sum {{x_i}} }}{n}\], n is the number of \[{x_{_i}}\].

So the value of A

\[ \Rightarrow A = \dfrac{{50 + 150 + 250 + 350 + 450 + 550 + 650}}{7}\]

On simplifying we have

\[ \Rightarrow A = \dfrac{{2450}}{7}\]

On dividing by 7 we have

\[ \Rightarrow A = 350\]

Now we calculate the mean

\[mean = A + c \times \dfrac{{\sum {{f_i}{t_i}} }}{{\sum {{f_i}} }}\]

On substituting the values we have

\[ \Rightarrow mean = 350 + 100 \times \dfrac{{ - 14}}{{140}}\]

On simplifying we have

\[ \Rightarrow mean = 340\]

Now we calculate the variance

\[Variance = \,{c^2}\left[ {\dfrac{{\sum {{f_i}{t_i}^2} }}{{\sum {{f_i}} }} - {{\left( {\dfrac{{\sum {{f_i}{t_i}} }}{{\sum {{f_i}} }}} \right)}^2}} \right]\]

On substituting the values we have

\[ \Rightarrow Variance = \,{100^2}\left[ {\dfrac{{302}}{{140}} - {{\left( {\dfrac{{ - 14}}{{140}}} \right)}^2}} \right]\]

On simplifying we have

\[ \Rightarrow Variance = \,{100^2}[2.1571 - 0.01]\]

\[ \Rightarrow Variance = \,10000 \times 2.1471\]

On multiplying we have

\[ \Rightarrow Variance = \,21471\]

Now we calculate the standard deviation

\[S.D = \,c\sqrt {\left[ {\dfrac{{\sum {{f_i}{t_i}^2} }}{{\sum {{f_i}} }} - {{\left( {\dfrac{{\sum {{f_i}{t_i}} }}{{\sum {{f_i}} }}} \right)}^2}} \right]} \]

On substituting the values

\[ \Rightarrow S.D = \,100\sqrt {\left[ {\dfrac{{302}}{{140}} - {{\left( {\dfrac{{ - 14}}{{140}}} \right)}^2}} \right]} \]

On simplifying we have

\[ \Rightarrow S.D = \,100\sqrt {\left[ {2.1571 - 0.01} \right]} \]

\[ \Rightarrow S.D = \,100 \times 1.4652\]

On multiplying we have

\[ \therefore S.D = 146.52\]

Hence we have determined the mean, variance and standard deviation by step deviation method.

**Note:**The students may not get confused by the deviation method and the step deviation method. Because the formulas are slightly similar. To solve these kinds of problems the students must know the tables of multiplication and simple arithmetic operations. To determine the standard deviation just apply the root for the variance.

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