Advertisements
Advertisements
Question
Find the mean of the following frequency distribution using step-deviation method
Class | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 |
Frequency | 7 | 10 | 15 | 8 | 10 |
Solution
Let us choose a = 25, h = 10, then `d_i = x_i – 25 and u_i = (x_i−25)/10`
Using step-deviation method, the given data is shown as follows:
Class | Frequency `(f_i)` | Class mark `(x_i)` | `d_i = x_i `– 25 | `u_i =( x_i−25)/10` | `(f_i u_i)` |
0 –10 | 7 | 5 | -20 | -2 | -14 |
10 – 20 | 10 | 15 | -10 | -1 | -10 |
20 – 30 | 15 | 25 | 0 | 0 | 0 |
30 – 40 | 8 | 35 | 10 | 1 | 8 |
40 – 50 | 10 | 45 | 20 | 2 | 20 |
Total | `Ʃ f_i` = 50 | `Ʃ f_i u_i = 4` |
The mean of the data is given by,
x = a + `((sum _i f_i u_i)/(sum _i f_i)) xx h`
=`25+ 4/50 xx 10`
=`25+ 4/5`
=`(125+4)/5`
= `129/5`
= 25.8
Thus, the mean is 25.8.
APPEARS IN
RELATED QUESTIONS
Frequency distribution of daily commission received by 100 salesmen is given below :
Daily Commission (in Rs.) |
No. of Salesmen |
100-120 | 20 |
120-140 | 45 |
140-160 | 22 |
160-180 | 09 |
180-200 | 04 |
Find mean daily commission received by salesmen, by the assumed mean method.
The weekly observations on cost of living index in a certain city for the year 2004 - 2005 are given below. Compute the weekly cost of living index.
Cost of living Index | Number of Students |
1400 - 1500 | 5 |
1500 - 1600 | 10 |
1600 - 1700 | 20 |
1700 - 1800 | 9 |
1800 - 1900 | 6 |
1900 - 2000 | 2 |
Find the mean of the following frequency distribution is 57.6 and the total number of observation is 50.
Class | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 | 100-120 |
Frequency | 7 | `f_1` | 12 | `f_2` | 8 | 5 |
Consider the following distribution of daily wages of 50 workers of a factory:
Daily wages (in ₹) |
500-520 | 520-540 | 540-560 | 560-580 | 580-600 |
Number of workers | 12 | 14 | 8 | 6 | 10 |
Find the mean daily wages of the workers of the factory by using an appropriate method.
The mean of n observation is `overlineX` .f the first item is increased by 1, second by 2 and so on, then the new mean is
If for certain frequency distribution, Median = 156 and Mode = 180, Find the value of the Mean.
The measurements (in mm) of the diameters of the head of the screws are given below :
Diameter (in mm) | no. of screws |
33 - 35 | 9 |
36 - 38 | 21 |
39 - 41 | 30 |
42 - 44 | 22 |
45 - 47 | 18 |
Calculate the mean diameter of the head of a screw by the ' Assumed Mean Method'.
Find the mean of the following distribution:
x | 4 | 6 | 9 | 10 | 15 |
f | 5 | 10 | 10 | 7 | 8 |
Find the mean wage of a worker from the following data:
Wages (In ₹) | 1400 | 1450 | 1500 | 1550 | 1600 | 1650 | 1700 |
Number of workers | 15 | 20 | 18 | 27 | 15 | 3 | 2 |
Find the mean for the following distribution:
Class Interval | 20 - 40 | 40 - 60 | 60 - 80 | 80 - 100 |
Frequency | 4 | 7 | 6 | 3 |