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Q. Consider the following distribution of daily wages of 50 workers
of a factory
Daily expenditure in(Rs) | 100 – 150 | 150 - 200 | 200 - 250 | 250 -300 | 300 – 350 |
No. of households | 4 | 5 | 12 | 2 | 2 |
Find the mean daily expenditure on food by a suitable method.
Sol:
Assumed mean, a = 225.
Class width(h) =
class interval
class mark
( xi )
frequency
( fi )
fiui
ui = xi –
h
4
5
12
2
2
125
175
225
275
325
– 2
– 1
0
1
2
– 8
– 5
0
2
4
Σ fi =
Σ fiui =
25
– 7
By step deviation method
Σ fiui
Σ fi
a +
h
∴ Mean = Rs. 211
Find the mean daily expenditure on food by a suitable method.
Daily expenditure
in(Rs)
No. of households
100 - 150
150 - 200
200 - 250
250 - 300
300 - 350
50
a
a
225
Total
=
225
+
– 7
25
50
1
2
= 225
–
14
By adding h to xi we get next xi
We assume any one value from xi column as ‘a’ (i.e.Assumed mean)
Preferably the middle value
4 x -2
5 x -1
12 x 0
2 x 1
2 x 2
Adding all fi
Adding all fiui
--
By subtracting two consecutive lower limits
Additional Example