Maths

STATISTICS 265
Substituting the values of a, f i
d i
and f i
from Table 14.4, we get
x =
435
47.5 47.5 14.5 62
30
✁ ✂ ✁ ✂ .
Therefore, the mean of the marks obtained by the students is 62.
The method discussed above is called the Assumed Mean Method.
Activity 1 : From the Table 14.3 find the mean by taking each of x i
(i.e., 17.5, 32.5,
and so on) as ‘a’. What do you observe? You will find that the mean determined in
each case is the same, i.e., 62. (Why ?)
So, we can say that the value of the mean obtained does not depend on the
choice of ‘a’.
Observe that in Table 14.4, the values in Column 4 are all multiples of 15. So, if
we divide the values in the entire Column 4 by 15, we would get smaller numbers to
multiply with f i
. (Here, 15 is the class size of each class interval.)
So, let u i
=
xi
h
✄
a , where a is the assumed mean and h is the class size.
Now, we calculate u i
in this way and continue as before (i.e., find f i
u i
and
then f i
u i
). Taking h = 15, let us form Table 14.5.
Table 14.5
Class interval f i
x i
d i
= x i
– a u i
=
x i –a
h
f i
u i
10 - 25 2 17.5 –30 –2 –4
25 - 40 3 32.5 –15 –1 –3
40 - 55 7 47.5 0 0 0
55 - 70 6 62.5 15 1 6
70 - 85 6 77.5 30 2 12
85 - 100 6 92.5 45 3 18
Total f i
= 30 f i
u i
= 29
Let u =
f u
f
i ☎ i
☎
Here, again let us find the relation between u and x .
i