# Step-deviation Method

When the values of the mid-values x_{i} in the different classes of a grouped data are large in magnitude, then the computation of the mean x becomes quite lengthy and tedious. In such a case, computation is simplified by 'short-cut method'. The procedure is,

Â

2. Subtract 'a' from the values of x

_{i}.

3. The reduced value (x

_{i }- a) is called the deviation of x

_{i}from 'a'.

4. Divide the deviations (x

_{i }- a) by a constant h (h is taken to be width of the class interval in the frequency-table (or) h = difference in two successive value of x

_{i}).

5. 'a' is taken some where in the middle of all values of x

_{i}.

Now, we can define

u_{i=}

h = difference in two successive values of x_{i}Â

Now we know that,

âˆ´Mean=

We can also calculate the mean by the formula.

x =

where 'a' is assumed mean and d_{i} represents the deviation from the assumed mean i.e.

d_{i }= x_{i }- a.

Â

For the following distribution, find out the mean-wage.

Wages in Rupees |
900 | 950 | 1000 | 1100 | 1260 | 1440 | 1500 |

No. of workers |
26 | 22 | 18 | 19 | 15 | 3 | 2 Â |

Let the assumed mean 'a' = 1100

Now we may arrange the given data as under:

Wages (in Rs) |
Frequency f_{i} |
Deviationd _{i}=x_{iÂ }- a =x_{iÂ }- 1100 |
f_{i}Â d_{i} |

900 | 26 | -200 | -5200 |

950 | 22 | -150 | -3300 |

1000 | 18 | -100 | -1800 |

1100 = a | 19 | 00 | 00 |

1260 | 15 | 160 | 2400 |

1440 | 3 | 340 | 1020 |

1500 | 2 | 400 | 800 |

Total | Î£f_{i}Â = 105 |
Â | Î£f_{i}x_{iÂ }= -6080 |

âˆ´Mean=Â = 1100 -Â Â = 1042.1.

Â

The frequency distribution of marks in mathematics are given in the table. Find the mean by short-cut method.

Marks | 5 | 10 | 15 | 20 | 25 | 30 | 35 | 40 | 45 | 50 |

No. ofÂ Students |
15 | 50 | 80 | 76 | 72 | 45 | 39 | 9 | 8 | 6 |

Let the assumed mean = a = 25

h = 10 - 5 = 15 - 10 = â€¦ = 50 - 45 = 5

Now we have the following table,

MarksÂ x_{i} |
No. of studentsÂ f_{i} |
x_{iÂ }- a |
u_{i}=Â Here h=5 |
f_{i}u_{i} |

5 | 15 | -20 | -4 | -60 |

10 | 50 | -15 | -3 | -150 |

15 | 80 | -10 | -2 | -160 |

20 | 76 | -5 | -1 | -76 |

25 = a | 72 | 0 | 0 | 0 |

30 | 45 | 5 | 1 | 45 |

35 | 39 | 10 | 2 | 78 |

40 | 9 | 15 | 3 | 27 |

45 | 8 | 20 | 4 | 32 |

50 | 6 | 25 | 5 | 30 |

Total | N = 400 | Â | Â | -234 |

NowÂ

Â Â Â Â Â Â Â Â =Â

Â Â Â Â Â Â Â Â =- 0.585Â Â Â

Â

âˆ´= 25 + (5Â Ã— âˆ’0.585)

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 25 - 2.925 = 22.075

âˆ´Mean marks = 22.075.

Â

Â

Â In a study on a certain disease, the following data was given. Find the average age of the first Â detection.

Age of first detection (in yrs) |
No. of patients |

4 - 8 | 2 |

8 - 12 | 12 |

12 - 16 | 15 |

16 - 20 | 25 |

20 - 24 | 18 |

24 - 28 | 12 |

28 - 32 | 3 |

32 - 36 | 1 |

Let the assumed mean a = 18 and h = 4.

Then we have the following table for the solution.

Â

Age inÂ Years |
Mid-pointx _{i} |
Frequencyf _{i} |
x_{iÂ }- a |
u_{i}=Â Here h = 4 |
f_{i}u_{i} |

4 - 8 | 6 | 2 | -12 | -3 | -6 |

8 - 12 | 10 | 12 | -8 | -2 | -24 |

12 - 16 | 14 | 15 | -4 | -1 | -15 |

16 - 20 | 18 = a | 25 | 0 | 0 | 0 |

20 - 24 | 22 | 18 | 4 | 1 | 18 |

24 - 28 | 26 | 12 | 8 | 2 | 24 |

28 - 32 | 30 | 3 | 12 | 3 | 9 |

32 - 36 | 34 | 1 | 16 | 4 | 4 |

Total | Â | Î£f_{i}Â = 88 |
Â | Â | Î£f_{i}u_{iÂ }= 10 |

Â Â Â Â Â (where N =Â Î£f_{i})

uÂ =Â __Â = 0.114__

Â = 18 + (4Â Ã—Â 0.114)

MeanÂ Â Â Â Â =18 + 0.46 = 18.46.

Â

Find the mean of the following data:

x | 4 | 7 | 10 | 13 | 16 | 19 | 22 |

f | 23 | 25 | 27 | 29 | 27 | 25 | 23 |

Let the assumed mean = a = 13, then we have the following table:

x_{i} |
f_{i} |
d_{iÂ }= x_{iÂ }- a |
d_{i}Â f_{i} |

4 | 23 | -9 | -207 |

7 | 25 | -6 | -150 |

10 | 27 | -3 | -81 |

13 | 29 | 00 | 00 |

16 | 27 | +3 | +81 |

19 | 25 | +6 | +150 |

22 | 23 | +9 | +207 |

Â | Î£f_{i}Â = 179 |
Â | Î£f_{i}d_{iÂ }= 0 |

Mean =Â Â = 13 +Â Â =13

Â

**Second Method of Solution**

Hint: a = 13, h = 3, u_{i }=Â

âˆ´

Here Î£f_{i} = 179 andÂ Î£f_{i}u_{i} = 0

âˆ´ = 13 + (3 Ã— ) = 13.