Answer

Verified

348.6k+ views

**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.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Why Are Noble Gases NonReactive class 11 chemistry CBSE

Let X and Y be the sets of all positive divisors of class 11 maths CBSE

Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE

Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE

Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE

Trending doubts

Establish a relation between electric current and drift class 12 physics CBSE

Guru Purnima speech in English in 100 words class 7 english CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference Between Plant Cell and Animal Cell

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

Copper is not used as potentiometer wire because class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE